OP Malhotra Class 11 Maths Solutions Chapter 12 Permutations and Combinations Exercise 12 (E)

Question 1. In how many ways can a committee of 8 be chosen from 10 individuals ?
Answer: To form a committee of 8 people from a group of 10 people, we use combinations. The order in which people are chosen does not matter for a committee. We calculate this as \( {}^{10}C_8 \). This means we are selecting 8 individuals from 10 available individuals. \[ {}^{10}C_8 = \frac{10!}{8!(10-8)!} = \frac{10!}{8!2!} \] \[ = \frac{10 \times 9 \times 8!}{8! \times 2 \times 1} \] \[ = \frac{10 \times 9}{2} = 45 \] There are 45 different ways to choose a committee of 8 from 10 people.
In simple words: To pick 8 people for a group from 10 available people, there are 45 different ways to do it. We use combination math because the order of picking doesn't matter.

🎯 Exam Tip: Remember to use combinations (\( {}^nC_r \)) when the order of selection doesn't matter, such as when forming committees or choosing groups, and permutations (\( {}^nP_r \)) when order does matter, like arranging items.

Question 2. In how many ways can a committee of five persons be formed out of 8 members when a particular member is taken every time ?
Answer: We need to form a committee of five persons from 8 members. Since one particular member must always be included, we can consider that person already selected. This reduces the problem to choosing the remaining 4 members from the remaining 7 members. \[ \text{Required number of ways} = {}^{7}C_4 \] \[ = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} \] \[ = \frac{7 \times 6 \times 5 \times 4!}{4! \times 3 \times 2 \times 1} \] \[ = \frac{7 \times 6 \times 5}{6} = 35 \] So, there are 35 ways to form the committee with that specific member always included.
In simple words: If you must pick a specific person for a 5-person team from 8 people, you only need to choose 4 more people from the remaining 7. There are 35 ways to do this.

🎯 Exam Tip: When specific members must always be included or excluded, adjust the total number of items and the number to be chosen *before* applying the combination or permutation formula.

Question 3. In how many ways can a committee of 4 be selected out of 12 persons so that a particular person may (i) always be taken, (ii) never be taken ?
Answer:
(i) If a particular person is always taken: We need to form a committee of 4 from 12 persons. If one specific person must always be included, then we only need to choose 3 more people from the remaining 11 persons. \[ \text{Required number of ways} = {}^{11}C_3 \] \[ = \frac{11!}{3!(11-3)!} = \frac{11!}{3!8!} \] \[ = \frac{11 \times 10 \times 9 \times 8!}{3 \times 2 \times 1 \times 8!} \] \[ = \frac{11 \times 10 \times 9}{6} = 11 \times 5 \times 3 = 165 \]
(ii) If a particular person is never taken: We need to form a committee of 4 from 12 persons. If one specific person must never be included, then we remove that person from the available group. We then choose all 4 committee members from the remaining 11 persons. \[ \text{Required number of ways} = {}^{11}C_4 \] \[ = \frac{11!}{4!(11-4)!} = \frac{11!}{4!7!} \] \[ = \frac{11 \times 10 \times 9 \times 8 \times 7!}{4 \times 3 \times 2 \times 1 \times 7!} \] \[ = \frac{11 \times 10 \times 9 \times 8}{24} = 11 \times 10 \times 3 = 330 \] It's interesting to note that removing someone permanently gives more options than fixing someone in.
In simple words: (i) If one person *must* be in the 4-person team, you pick the other 3 from the remaining 11 people, which is 165 ways. (ii) If one person *cannot* be in the team, you pick all 4 people from the remaining 11, which is 330 ways.

🎯 Exam Tip: Clearly identify whether a specific individual is included or excluded before determining the size of the pool and the number of remaining selections needed. This is a common pitfall.

Question 4. In how many ways can a team of 11 players be selected from 14 players when two of them can play as goalkeepers only ?
Answer: We need to select a team of 11 players from 14. There are 2 players who can *only* play as goalkeepers. Since a football team needs one goalkeeper, we must select 1 goalkeeper from these 2 specialized players. Number of ways to select 1 goalkeeper from 2 = \( {}^{2}C_1 = 2 \). After selecting 1 goalkeeper, we still need to select 10 more players for the team (11 - 1 = 10). The remaining players are \( 14 - 2 = 12 \) non-goalkeepers. Number of ways to select 10 players from these 12 = \( {}^{12}C_{10} \). \[ {}^{12}C_{10} = \frac{12!}{10!(12-10)!} = \frac{12!}{10!2!} \] \[ = \frac{12 \times 11 \times 10!}{10! \times 2 \times 1} \] \[ = \frac{12 \times 11}{2} = 66 \] To find the total number of ways, we multiply the ways to select the goalkeeper and the other players. Required number of ways = (Ways to select 1 goalkeeper) \( \times \) (Ways to select 10 other players) \[ = {}^{2}C_1 \times {}^{12}C_{10} = 2 \times 66 = 132 \] So, there are 132 ways to select the team.
In simple words: We need to pick 11 players from 14. Two players can only be goalkeepers, so we pick 1 of them. Then, we need 10 more players from the remaining 12 non-goalkeepers. This gives us 132 ways to choose the team.

🎯 Exam Tip: When specific roles or types of players are required, handle those selections first and then choose the remaining players from the reduced pools. Always consider if the roles restrict choices.

Question 5. A person has got 12 friends of whom 8 are relatives. In how many ways can he invite 7 guests such that 5 of them may be relatives ?
Answer: A person has 12 friends in total. Out of these, 8 are relatives and \( 12 - 8 = 4 \) are not relatives. The person wants to invite 7 guests, and exactly 5 of these guests must be relatives. First, we select 5 relatives from the 8 available relatives: Number of ways to select 5 relatives = \( {}^{8}C_5 \). \[ {}^{8}C_5 = \frac{8!}{5!(8-5)!} = \frac{8!}{5!3!} \] \[ = \frac{8 \times 7 \times 6 \times 5!}{5! \times 3 \times 2 \times 1} \] \[ = \frac{8 \times 7 \times 6}{6} = 56 \] Since a total of 7 guests are to be invited, and 5 relatives have already been chosen, we need to select \( 7 - 5 = 2 \) more guests. These 2 guests must be chosen from the friends who are *not* relatives. There are 4 such friends. Number of ways to select 2 non-relatives = \( {}^{4}C_2 \). \[ {}^{4}C_2 = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} \] \[ = \frac{4 \times 3 \times 2!}{2 \times 1 \times 2!} \] \[ = \frac{4 \times 3}{2} = 6 \] To find the total number of ways, we multiply the ways of selecting relatives and non-relatives: Required number of ways = \( {}^{8}C_5 \times {}^{4}C_2 = 56 \times 6 = 336 \). So, there are 336 ways to invite 7 guests with 5 of them being relatives.
In simple words: You have 12 friends (8 relatives, 4 others) and want to invite 7 guests, making sure 5 are relatives. You choose 5 relatives from 8 (56 ways), and then choose the remaining 2 guests from the 4 non-relatives (6 ways). Multiply these to get the total: 336 ways.

🎯 Exam Tip: When a selection has multiple categories with specific requirements, calculate the combinations for each category separately and then multiply them to get the total number of ways.

Question 6. How many diagonals are there in a polygon of (i) 8 sides, (ii) 10 sides ?
Answer: A diagonal connects two non-adjacent vertices of a polygon. The formula for the number of diagonals in a polygon with \( n \) sides is \( \frac{n(n-3)}{2} \) or \( {}^nC_2 - n \). This is because \( {}^nC_2 \) gives the total number of lines that can be formed by joining any two vertices, and \( n \) of these lines are the sides of the polygon, which are not diagonals.
(i) For a polygon with 8 sides: Here, \( n = 8 \). \[ \text{Required number of diagonals} = {}^{8}C_2 - 8 \] \[ = \frac{8!}{2!(8-2)!} - 8 = \frac{8!}{2!6!} - 8 \] \[ = \frac{8 \times 7 \times 6!}{2 \times 1 \times 6!} - 8 \] \[ = \frac{8 \times 7}{2} - 8 = 28 - 8 = 20 \] A polygon with 8 sides (an octagon) has 20 diagonals.
(ii) For a polygon with 10 sides: Here, \( n = 10 \). \[ \text{Required number of diagonals} = {}^{10}C_2 - 10 \] \[ = \frac{10!}{2!(10-2)!} - 10 = \frac{10!}{2!8!} - 10 \] \[ = \frac{10 \times 9 \times 8!}{2 \times 1 \times 8!} - 10 \] \[ = \frac{10 \times 9}{2} - 10 = 45 - 10 = 35 \] A polygon with 10 sides (a decagon) has 35 diagonals.
In simple words: To find the number of diagonals in a polygon, you first count all possible lines you can draw by connecting any two corners, and then you subtract the number of sides (because sides are not diagonals). (i) An 8-sided shape has 20 diagonals. (ii) A 10-sided shape has 35 diagonals.

🎯 Exam Tip: Remember the formula for diagonals, \( {}^nC_2 - n \), which calculates all possible pairs of vertices and then removes the polygon's sides. This is a fundamental concept in combinatorics related to geometry.

Question 7. In how many ways can a committee, consisting of a chairman, secretary, treasurer and four ordinary members be chosen from eight persons ? (Committees with different chairmen, secretaries, treasurers count as different committees.)
Answer: We need to form a committee of \( 1+1+1+4 = 7 \) members from 8 persons, where specific roles are assigned. Since having different chairmen, secretaries, and treasurers counts as a different committee, the order of selection for these roles matters, indicating a permutation-like approach, or sequential combinations for distinct roles. 1. Select the chairman: There are 8 persons available, so we choose 1 for chairman. Number of ways = \( {}^{8}C_1 = 8 \). 2. Select the secretary: After choosing the chairman, 7 persons remain. We choose 1 for secretary. Number of ways = \( {}^{7}C_1 = 7 \). 3. Select the treasurer: After choosing the chairman and secretary, 6 persons remain. We choose 1 for treasurer. Number of ways = \( {}^{6}C_1 = 6 \). 4. Select the four ordinary members: After choosing the chairman, secretary, and treasurer, 5 persons remain (\( 8 - 3 = 5 \)). We need to choose 4 ordinary members from these 5. Number of ways = \( {}^{5}C_4 \). \[ {}^{5}C_4 = \frac{5!}{4!(5-4)!} = \frac{5!}{4!1!} = 5 \] To find the total number of ways to form this committee, we multiply the number of ways for each selection because these are sequential decisions. Required number of ways = (Ways to choose chairman) \( \times \) (Ways to choose secretary) \( \times \) (Ways to choose treasurer) \( \times \) (Ways to choose 4 ordinary members) \[ = {}^{8}C_1 \times {}^{7}C_1 \times {}^{6}C_1 \times {}^{5}C_4 \] \[ = 8 \times 7 \times 6 \times 5 \] \[ = 1680 \] Thus, there are 1680 ways to form the committee with these specific roles.
In simple words: To pick a committee of 7 from 8 people with special jobs (chairman, secretary, treasurer, and 4 regular members), we pick the chairman first (8 ways), then the secretary from the rest (7 ways), then the treasurer (6 ways). Finally, we pick 4 regular members from the last 5 people (5 ways). Multiplying these gives 1680 total ways.

🎯 Exam Tip: When positions within a committee are distinct (like chairman, secretary), treat them as separate selections or use permutations. If all members are equal (ordinary members), use combinations. Multiply the results for each step.

Question 8. (a) In how many ways can a student choose 5 courses out of 9 courses if 2 courses are compulsory for every student ? (b) In how many ways ca we select a cricket eleven from 17 players in which 5 players can bowl ? Each cricket team must include 2 bowlers.
Answer:
(a) A student needs to choose 5 courses out of 9, with 2 courses being compulsory. Since 2 courses are compulsory, they are already chosen. The student needs to select \( 5 - 2 = 3 \) more courses. The remaining courses available for selection are \( 9 - 2 = 7 \). So, the student must choose 3 courses from these 7 remaining courses. \[ \text{Required number of ways} = {}^{7}C_3 \] \[ = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} \] \[ = \frac{7 \times 6 \times 5 \times 4!}{3 \times 2 \times 1 \times 4!} \] \[ = \frac{7 \times 6 \times 5}{6} = 35 \] There are 35 ways a student can choose the courses.
(b) We need to select a cricket team of 11 players from 17, where 5 players can bowl, and the team must include exactly 2 bowlers. First, select the 2 bowlers from the 5 available bowlers: Number of ways to select 2 bowlers = \( {}^{5}C_2 \). \[ {}^{5}C_2 = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} \] \[ = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{5 \times 4}{2} = 10 \] Next, we need to select the remaining \( 11 - 2 = 9 \) players for the team. These players must be chosen from the non-bowlers. The total number of players is 17. Since 5 are bowlers, \( 17 - 5 = 12 \) are non-bowlers. Number of ways to select 9 non-bowlers from 12 = \( {}^{12}C_9 \). \[ {}^{12}C_9 = \frac{12!}{9!(12-9)!} = \frac{12!}{9!3!} \] \[ = \frac{12 \times 11 \times 10 \times 9!}{3 \times 2 \times 1 \times 9!} \] \[ = \frac{12 \times 11 \times 10}{6} = 2 \times 11 \times 10 = 220 \] To find the total number of ways to form the team, we multiply the ways to select bowlers and non-bowlers. Required number of ways = (Ways to select bowlers) \( \times \) (Ways to select non-bowlers) \[ = {}^{5}C_2 \times {}^{12}C_9 = 10 \times 220 = 2200 \] So, there are 2200 ways to select the cricket team.
In simple words: (a) For courses, if 2 are a must, you only need to pick 3 more from the 7 remaining choices. That's 35 ways. (b) For cricket, you must pick 2 bowlers from 5 (10 ways), then 9 regular players from the remaining 12 (220 ways). Multiply these for a total of 2200 ways to make the team.

🎯 Exam Tip: Break down complex selection problems into smaller, independent choices. For "compulsory" items, reduce the pool and the number to be chosen. For "at least/at most" conditions, consider mutually exclusive cases and sum their combinations.

Question 9. How many committees of 5 members each can be formed with 8 officials and 4 non-official members in the following cases : (i) each consists of 3 officials and 2 non-official members ; (ii) each contains at least two non-official members ; (iii) a particular official member is never included ; (iv) a particular non-official member is always included ?
Answer: We have 8 official members and 4 non-official members, and we need to form a committee of 5 members.
(i) Each committee consists of 3 officials and 2 non-official members: Number of ways to select 3 officials from 8 = \( {}^{8}C_3 \). \[ {}^{8}C_3 = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \] Number of ways to select 2 non-official members from 4 = \( {}^{4}C_2 \). \[ {}^{4}C_2 = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6 \] Total number of such committees = \( {}^{8}C_3 \times {}^{4}C_2 = 56 \times 6 = 336 \).
(ii) Each committee contains at least two non-official members: "At least two non-official members" means the committee can have: * 2 non-official members and 3 official members * 3 non-official members and 2 official members * 4 non-official members and 1 official member (We cannot have 5 non-official members as there are only 4 available.) Number of ways for (2 non-officials, 3 officials) = \( {}^{4}C_2 \times {}^{8}C_3 = 6 \times 56 = 336 \). Number of ways for (3 non-officials, 2 officials) = \( {}^{4}C_3 \times {}^{8}C_2 \). \[ {}^{4}C_3 = \frac{4!}{3!1!} = 4 \] \[ {}^{8}C_2 = \frac{8!}{2!6!} = \frac{8 \times 7}{2 \times 1} = 28 \] So, \( 4 \times 28 = 112 \). Number of ways for (4 non-officials, 1 official) = \( {}^{4}C_4 \times {}^{8}C_1 \). \[ {}^{4}C_4 = 1 \] \[ {}^{8}C_1 = 8 \] So, \( 1 \times 8 = 8 \). Total number of committees with at least two non-official members = \( 336 + 112 + 8 = 456 \).
(iii) A particular official member is never included: If one particular official member is never included, we exclude him from the pool of members. New pool: 7 official members and 4 non-official members. We still need to form a committee of 5. We need to consider all possible compositions of the 5-member committee from this reduced pool: * 1 non-official and 4 official members: \( {}^{4}C_1 \times {}^{7}C_4 = 4 \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 4 \times 35 = 140 \) * 2 non-official and 3 official members: \( {}^{4}C_2 \times {}^{7}C_3 = 6 \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 6 \times 35 = 210 \) * 3 non-official and 2 official members: \( {}^{4}C_3 \times {}^{7}C_2 = 4 \times \frac{7 \times 6}{2 \times 1} = 4 \times 21 = 84 \) * 4 non-official and 1 official member: \( {}^{4}C_4 \times {}^{7}C_1 = 1 \times 7 = 7 \) * 0 non-official and 5 official members: \( {}^{4}C_0 \times {}^{7}C_5 = 1 \times \frac{7 \times 6}{2 \times 1} = 1 \times 21 = 21 \) Total number of ways = \( 140 + 210 + 84 + 7 + 21 = 462 \).
(iv) A particular non-official member is always included: If one particular non-official member is always included, then that person is already in the committee. We need to select \( 5 - 1 = 4 \) more members. The remaining members available for selection are: - 8 official members - \( 4 - 1 = 3 \) non-official members Total remaining members = \( 8 + 3 = 11 \). We need to choose 4 members from these 11. \[ \text{Required number of ways} = {}^{11}C_4 \] \[ = \frac{11!}{4!7!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} \] \[ = 11 \times 10 \times 3 = 330 \] It's simpler to consider that by fixing one non-official member, we effectively choose 4 members from the remaining 11 people.
In simple words: You're forming a 5-person committee from 8 officials and 4 non-officials. (i) To have 3 officials and 2 non-officials, there are 336 ways. (ii) To have at least two non-officials, sum the ways for 2, 3, or 4 non-officials, totaling 456 ways. (iii) If one specific official is never picked, you choose 5 people from the remaining 11 (7 officials + 4 non-officials), totaling 462 ways. (iv) If one specific non-official is always picked, you choose the remaining 4 people from the other 11 (8 officials + 3 non-officials), totaling 330 ways.

🎯 Exam Tip: For "at least" or "at most" scenarios, list all possible valid combinations of members. For "always included" or "never included" cases, adjust the total pool size and the number of selections needed *before* applying the combinations formula.

Question 10. In a college team there are 15 players of whom 3 are teachers. In how many ways can a team of 11 players be selected so as to include (i) only one teacher, (ii) at least one teacher ?
Answer: We have a total of 15 players, consisting of 3 teachers and \( 15 - 3 = 12 \) non-teacher players. We need to form a team of 11 players.
(i) The team must include only one teacher: First, select 1 teacher from the 3 available teachers: Number of ways = \( {}^{3}C_1 = 3 \). Since we need a team of 11 players and 1 teacher is selected, we need to choose \( 11 - 1 = 10 \) more players. These 10 players must be chosen from the 12 non-teacher players. Number of ways to select 10 non-teachers = \( {}^{12}C_{10} \). \[ {}^{12}C_{10} = \frac{12!}{10!(12-10)!} = \frac{12!}{10!2!} \] \[ = \frac{12 \times 11 \times 10!}{10! \times 2 \times 1} \] \[ = \frac{12 \times 11}{2} = 66 \] Total number of ways to form the team with only one teacher = \( {}^{3}C_1 \times {}^{12}C_{10} = 3 \times 66 = 198 \).
(ii) The team must include at least one teacher: "At least one teacher" means the team can have 1 teacher, 2 teachers, or 3 teachers. * Case 1: 1 teacher and 10 non-teachers Number of ways = \( {}^{3}C_1 \times {}^{12}C_{10} = 3 \times 66 = 198 \) (calculated above). * Case 2: 2 teachers and 9 non-teachers Number of ways = \( {}^{3}C_2 \times {}^{12}C_9 \). \[ {}^{3}C_2 = \frac{3!}{2!1!} = 3 \] \[ {}^{12}C_9 = \frac{12!}{9!3!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 2 \times 11 \times 10 = 220 \] So, \( 3 \times 220 = 660 \). * Case 3: 3 teachers and 8 non-teachers Number of ways = \( {}^{3}C_3 \times {}^{12}C_8 \). \[ {}^{3}C_3 = 1 \] \[ {}^{12}C_8 = \frac{12!}{8!4!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495 \] So, \( 1 \times 495 = 495 \). Total number of ways to form the team with at least one teacher = Sum of ways from all cases \[ = 198 + 660 + 495 = 1353 \] An alternative method for "at least one teacher" is to calculate the total number of ways to form a team of 11 from 15 players (no restrictions), and subtract the number of ways to form a team with no teachers. Total ways to choose 11 players from 15 = \( {}^{15}C_{11} \). Ways to choose 11 players with no teachers (i.e., all from 12 non-teachers) = \( {}^{12}C_{11} \). \( {}^{15}C_{11} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 15 \times 7 \times 13 = 1365 \) \( {}^{12}C_{11} = \frac{12!}{11!1!} = 12 \) So, \( 1365 - 12 = 1353 \), which matches the previous calculation.
In simple words: Out of 15 players (3 teachers, 12 others), we pick 11. (i) If only one teacher is picked, you choose 1 teacher from 3 (3 ways) and 10 others from 12 (66 ways), total 198 ways. (ii) If at least one teacher is picked, you add up ways for 1, 2, or 3 teachers. This gives 198 + 660 + 495 = 1353 ways.

🎯 Exam Tip: For "at least one X" problems, consider using the complementary counting principle: (Total ways) - (Ways with no X). This can often simplify calculations, especially when there are many "at least" cases.

Question 11. How many different groups can be selected for playing tennis out of 4 ladies and 3 gentlemen, there being one lady and one gentleman on each side ?
Answer: We have 4 ladies (L) and 3 gentlemen (G). We need to form two tennis teams (groups), with each team having one lady and one gentleman. Let's consider forming the first group (Team 1). Number of ways to select 1 lady for Team 1 from 4 ladies = \( {}^{4}C_1 = 4 \). Number of ways to select 1 gentleman for Team 1 from 3 gentlemen = \( {}^{3}C_1 = 3 \). So, the total number of ways to form Team 1 = \( {}^{4}C_1 \times {}^{3}C_1 = 4 \times 3 = 12 \). After forming Team 1, we are left with \( 4 - 1 = 3 \) ladies and \( 3 - 1 = 2 \) gentlemen. Now, let's form the second group (Team 2) from the remaining players. Number of ways to select 1 lady for Team 2 from 3 remaining ladies = \( {}^{3}C_1 = 3 \). Number of ways to select 1 gentleman for Team 2 from 2 remaining gentlemen = \( {}^{2}C_1 = 2 \). So, the total number of ways to form Team 2 = \( {}^{3}C_1 \times {}^{2}C_1 = 3 \times 2 = 6 \). If we consider the teams to be distinct (Team 1 vs. Team 2), the total number of ways to form both teams would be \( 12 \times 6 = 72 \). However, the question asks for "different groups" (teams) and usually, when forming groups for a game, the order in which the groups are formed does not matter (i.e., Group A vs. Group B is the same as Group B vs. Group A). Since the two groups are of the same size and structure (one lady and one gentleman), forming {L1, G1} and {L2, G2} is considered the same as forming {L2, G2} and {L1, G1}. We have overcounted by a factor of \( 2! \). Therefore, the required number of ways to form two distinct groups = \( \frac{72}{2!} = \frac{72}{2} = 36 \). This accounts for the fact that the two "sides" are not inherently labeled "Side 1" and "Side 2" but are just two sets of players.
In simple words: To make two tennis teams, each with one lady and one gentleman from 4 ladies and 3 gentlemen: First, pick one lady and one gentleman for the first team (12 ways). Then, pick one lady and one gentleman from the remaining players for the second team (6 ways). Since the two teams are not named differently, we divide the total by 2, which gives 36 different ways to form the groups.

🎯 Exam Tip: When forming multiple identical groups from a larger set, calculate the number of ways to form ordered groups, then divide by the factorial of the number of groups to correct for overcounting if the groups themselves are indistinguishable.

Question 12. If \( {}^nC_{10} = {}^nC_{14} \), find the value of \( {}^nC_{20} \) and \( {}^{25}C_n \).
Answer: We are given the combination equation: \( {}^nC_{10} = {}^nC_{14} \). A key property of combinations states that if \( {}^nC_r = {}^nC_s \), then either \( r = s \) or \( r + s = n \). In this case, \( 10 \neq 14 \), so we must use the second condition: \( r + s = n \). So, \( 10 + 14 = n \) \[ n = 24 \] Now we need to find the values of \( {}^nC_{20} \) and \( {}^{25}C_n \) using \( n = 24 \). First, calculate \( {}^nC_{20} \), which is \( {}^{24}C_{20} \): We use the property \( {}^nC_r = {}^nC_{n-r} \). So, \( {}^{24}C_{20} = {}^{24}C_{24-20} = {}^{24}C_4 \). \[ {}^{24}C_4 = \frac{24!}{4!(24-4)!} = \frac{24!}{4!20!} \] \[ = \frac{24 \times 23 \times 22 \times 21 \times 20!}{4 \times 3 \times 2 \times 1 \times 20!} \] \[ = \frac{24 \times 23 \times 22 \times 21}{24} \] \[ = 23 \times 22 \times 21 = 10626 \] Next, calculate \( {}^{25}C_n \), which is \( {}^{25}C_{24} \): Using the property \( {}^nC_r = {}^nC_{n-r} \). So, \( {}^{25}C_{24} = {}^{25}C_{25-24} = {}^{25}C_1 \). The value of \( {}^nC_1 \) is simply \( n \). Therefore, \( {}^{25}C_1 = 25 \). So, \( n = 24 \), \( {}^{24}C_{20} = 10626 \), and \( {}^{25}C_{24} = 25 \).
In simple words: When two combinations like \( {}^nC_{10} \) and \( {}^nC_{14} \) are equal, it means the sum of the bottom numbers (10 and 14) is equal to the top number \( n \). So \( n \) is 24. Then, we use this value of \( n \) to find \( {}^{24}C_{20} \), which is 10626, and \( {}^{25}C_{24} \), which is 25.

🎯 Exam Tip: Master the fundamental properties of combinations, especially \( {}^nC_r = {}^nC_s \implies r=s \text{ or } r+s=n \) and \( {}^nC_r = {}^nC_{n-r} \). These properties frequently simplify complex calculations and are crucial for solving such problems quickly.

Question 13. In how many ways can I invite one or more of six friends to a dinner ?
Answer: You have 6 friends, and for each friend, you have two choices: either invite them or not invite them. If there are \( n \) items, and for each item there are 2 choices (include or exclude), the total number of ways to make a selection is \( 2^n \). In this case, \( n = 6 \) friends. So, the total number of ways to invite friends (including the case where no one is invited) is \( 2^6 \). \[ 2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64 \] The question specifies "one or more" friends. This means we must exclude the case where none of the friends are invited. There is only 1 way to invite no friends. So, the required number of ways = (Total ways to invite friends) - (Ways to invite no friends) \[ = 2^6 - 1 \] \[ = 64 - 1 = 63 \] Therefore, you can invite one or more of your six friends in 63 ways.
In simple words: For each of your 6 friends, you can either invite them or not. This means there are 2 choices for each friend. So, for 6 friends, there are \( 2^6 = 64 \) ways to invite. But since you must invite "one or more," you subtract the one way where you invite nobody. This leaves 63 ways.

🎯 Exam Tip: For selection problems where each item can either be chosen or not chosen, the total number of options is \( 2^n \). If the condition specifies "at least one" selection, subtract 1 (for the case of no selections) from \( 2^n \).

Question 14. In how many ways can 10 marbles be divided between two boys so that one of them may get 2 and the other 8 ?
Answer: We have 10 marbles to divide between two distinct boys (Boy 1 and Boy 2) such that one gets 2 marbles and the other gets 8 marbles. This problem can be interpreted in two ways: **Interpretation 1: Assign specific counts to specific boys.** If Boy 1 gets 2 marbles and Boy 2 gets 8 marbles: Ways to choose 2 marbles for Boy 1 from 10 = \( {}^{10}C_2 \). The remaining \( 10 - 2 = 8 \) marbles must go to Boy 2. Ways to choose 8 marbles for Boy 2 from 8 = \( {}^{8}C_8 \). Total ways = \( {}^{10}C_2 \times {}^{8}C_8 = 45 \times 1 = 45 \). If Boy 1 gets 8 marbles and Boy 2 gets 2 marbles: Ways to choose 8 marbles for Boy 1 from 10 = \( {}^{10}C_8 \). The remaining \( 10 - 8 = 2 \) marbles must go to Boy 2. Ways to choose 2 marbles for Boy 2 from 2 = \( {}^{2}C_2 \). Total ways = \( {}^{10}C_8 \times {}^{2}C_2 = 45 \times 1 = 45 \). Since the problem states "one of them may get 2 and the other 8", it implies we need to consider both scenarios where either Boy 1 or Boy 2 gets 2 marbles. So, total ways = \( 45 + 45 = 90 \). **Interpretation 2: Choose the set of 8 marbles, and then assign it to one of the boys.** First, we select a set of 8 marbles from the 10 available marbles. This set will go to one boy, and the remaining 2 marbles will go to the other boy. Number of ways to select 8 marbles from 10 = \( {}^{10}C_8 \). \[ {}^{10}C_8 = \frac{10!}{8!(10-8)!} = \frac{10!}{8!2!} = \frac{10 \times 9}{2} = 45 \] Once these 8 marbles are selected, there are 2 boys to whom this group of 8 marbles can be given (the other boy will automatically receive the remaining 2 marbles). Number of ways to choose which boy gets the 8 marbles = \( {}^{2}C_1 = 2 \). Total number of ways = \( {}^{10}C_8 \times {}^{2}C_1 = 45 \times 2 = 90 \). Both interpretations lead to the same result, 90 ways.
In simple words: To divide 10 marbles between two boys so that one gets 2 and the other gets 8, we first choose which boy gets the 8 marbles (2 ways). Then, we select 8 marbles out of 10 for that boy (45 ways). The remaining 2 marbles automatically go to the other boy. So, there are \( 2 \times 45 = 90 \) ways.

🎯 Exam Tip: When distributing distinct items into distinct groups or individuals, consider whether the groups/individuals are distinguishable. If they are, choices for distribution multiply. When the specific *count* is given for *each* group without naming the recipient, then multiply the combinations by the number of ways to assign those sets to the recipients.

Question 15. In how many ways can a selection be made out of 5 oranges, 8 mangoes and 7 plantains ?
Answer: We need to find the number of ways to make a selection from items that are alike of one kind. Let's assume the oranges are identical, the mangoes are identical, and the plantains are identical within their respective types. For 5 oranges, the number of ways to select oranges is 0, 1, 2, 3, 4, or 5. This gives \( 5 + 1 = 6 \) choices. For 8 mangoes, the number of ways to select mangoes is 0, 1, ..., 8. This gives \( 8 + 1 = 9 \) choices. For 7 plantains, the number of ways to select plantains is 0, 1, ..., 7. This gives \( 7 + 1 = 8 \) choices. To find the total number of selections, including the case where no items are selected at all, we multiply the number of choices for each type of fruit. Total number of selections (including selecting zero of each fruit) = \( (5+1) \times (8+1) \times (7+1) = 6 \times 9 \times 8 = 432 \). However, this count includes the case where zero oranges, zero mangoes, AND zero plantains are selected. This usually means "making no selection at all", which is often excluded when asking for "a selection". If the question implies making at least one selection (i.e., not selecting zero of all types), we subtract 1 from the total. Required number of selections = \( (6 \times 9 \times 8) - 1 = 432 - 1 = 431 \). The " - 1" removes the case where nothing is selected, meaning at least one fruit is picked. Selecting nothing is also a form of selection.
In simple words: To pick fruits from 5 oranges, 8 mangoes, and 7 plantains, you have 6 options for oranges (0 to 5), 9 for mangoes (0 to 8), and 8 for plantains (0 to 7). Multiply these options together (\( 6 \times 9 \times 8 = 432 \)). Then subtract 1 to remove the case where you pick no fruits at all, leaving 431 ways to make a selection.

🎯 Exam Tip: When selecting from distinct groups of identical items, the number of ways to select items from a group of \( p \) identical items is \( p+1 \) (including the choice of selecting none). Multiply these \( p+1 \) values for all groups. If "no selection" is to be excluded, remember to subtract 1 from the final product.

Question 16. In how many ways can 20 articles be packed in the three parcels so that the first contains 8 articles, the second 7 and the third 5 ?
Answer: We have 20 distinct articles to be packed into three distinct parcels (Parcel I, Parcel II, Parcel III) with specified quantities in each. 1. Pack Parcel I: We need to choose 8 articles for the first parcel from the 20 available articles. Number of ways = \( {}^{20}C_8 \). 2. Pack Parcel II: After packing 8 articles into Parcel I, \( 20 - 8 = 12 \) articles remain. We need to choose 7 articles for the second parcel from these 12 remaining articles. Number of ways = \( {}^{12}C_7 \). 3. Pack Parcel III: After packing 8 articles and 7 articles into the first two parcels, \( 12 - 7 = 5 \) articles remain. We need to choose 5 articles for the third parcel from these 5 remaining articles. Number of ways = \( {}^{5}C_5 \). To find the total number of ways to pack the articles, we multiply the number of ways for each step: Required number of ways = \( {}^{20}C_8 \times {}^{12}C_7 \times {}^{5}C_5 \). \[ = \frac{20!}{8!(20-8)!} \times \frac{12!}{7!(12-7)!} \times \frac{5!}{5!(5-5)!} \] \[ = \frac{20!}{8!12!} \times \frac{12!}{7!5!} \times \frac{5!}{5!0!} \] Notice that the intermediate factorials cancel out, simplifying the expression: \[ = \frac{20!}{8!7!5!} \] This formula is also known as the multinomial coefficient for distributing distinct items into distinct groups. The question doesn't require calculation to a number, so leaving it as the factorial expression is acceptable. This calculates the number of ways to partition 20 distinct articles into three distinct groups of sizes 8, 7, and 5.
In simple words: To pack 20 different items into three parcels (8 in the first, 7 in the second, 5 in the third), you first choose 8 items for the first parcel from 20. Then, you choose 7 items for the second parcel from the remaining 12. Finally, you choose 5 items for the third parcel from the last 5. The total number of ways is found by multiplying these choices.

🎯 Exam Tip: For distributing distinct items into distinct groups with specific sizes, the formula is \( \frac{n!}{n_1! n_2! \dots n_k!} \), where \( n \) is the total number of items and \( n_i \) is the size of each group. Remember that \( 0! = 1 \).

Question 17. Find the number of four letter arrangements of the letters of the word 'SHOOT'. How many of them begin with O?
Answer: The word 'SHOOT' has 5 letters: S, H, O, O, T. There are 2 'O's, and S, H, T are distinct. We need to form 4-letter arrangements (permutations). First, let's find the total number of 4-letter arrangements from 'SHOOT'. We consider cases based on repeated letters: **Case 1: All 4 letters are distinct.** The distinct letters are S, H, O, T (4 distinct letters). Number of permutations = \( {}^4P_4 = 4! = 24 \). **Case 2: 2 letters are alike (the two O's) and 2 others are distinct.** We must choose the two O's. We need to choose 2 more distinct letters from (S, H, T). This can be done in \( {}^3C_2 = 3 \) ways (e.g., OO S H, OO S T, OO H T). For each choice of 4 letters (e.g., OOSH), the number of permutations is \( \frac{4!}{2!} \) (because 2 letters are alike). \[ \frac{4!}{2!} = \frac{24}{2} = 12 \] Since there are 3 such combinations of letters, total permutations for this case = \( 3 \times 12 = 36 \). Total number of 4-letter arrangements = \( 24 + 36 = 60 \). This represents all possible 4-letter arrangements from the word SHOOT. Next, we find how many of these arrangements begin with 'O'. If the arrangement begins with 'O', we fix one 'O' at the first position. We then need to arrange the remaining 3 letters in the other 3 positions. The available letters are S, H, O, T. If one 'O' is fixed at the start, the remaining letters are S, H, T, and one 'O'. So, we have the remaining letters: O, S, H, T (where one O is already used). Now we need to arrange 3 letters from {S, H, T, O} (the remaining O is treated as a distinct letter from the first fixed O if we consider them distinct initially, but here the 'O's are identical in nature). Let's consider the composition of the remaining 3 letters from the set {S, H, T, O} (where the 'O' is the second 'O'). **Case 1: The other 'O' is used among the remaining 3 positions.** This means the letters are O, S, H, T (where the first O is fixed, and we choose 3 from S, H, T, O). If the second 'O' is used, then the remaining 3 letters are {O, S, H}, {O, S, T}, {O, H, T}. For {O, S, H}, arrangements are \( \frac{3!}{2!} \) if we treat the remaining O as distinct from the initial O. Wait, the original logic for "beginning with O" is simpler: Fix 'O' at the first position. We need to fill 3 more positions using the remaining letters: S, H, T, and the second 'O'. So, we have 4 letters (S, H, T, O) and 3 positions to fill. All these 4 letters are distinct (S, H, T, and the remaining 'O' which we can conceptually think of as a "different" O). Number of permutations of 3 letters chosen from these 4 distinct letters = \( {}^4P_3 \). \[ {}^4P_3 = \frac{4!}{(4-3)!} = \frac{4!}{1!} = 4 \times 3 \times 2 = 24 \] So, 24 arrangements begin with O. This calculation implicitly treats the 'O's as distinct initially, then correctly accounts for the fixed 'O' and the remaining distinct 'O' along with S, H, T. The key is that after fixing one 'O', the *remaining* 'O' is effectively unique among the selection pool of {S, H, T, O (the second)}. Let's confirm with the source's logic: "To fix beginning with O, the no. of 4-letter arrangement in which 4 places to be filled by 3 letters in 4C3 -3 ways i.e. 24 ways." This is a bit ambiguous in the source. My calculation of \( {}^4P_3 \) for 3 distinct remaining letters from {S, H, T, O} seems to match the '24 ways' figure provided without requiring the `4C3 - 3` syntax. The wording `4C3 - 3` in the source looks like an OCR error or misinterpretation. \( {}^4P_3 = 24 \) is the correct way to get 24 arrangements when 1 O is fixed and 3 places are filled from the 4 remaining distinct letters (S,H,T, the other O).
In simple words: The word 'SHOOT' has two 'O's. There are a total of 60 different 4-letter words that can be made from its letters. If we want to count only the words that start with 'O', we put one 'O' first. Then, we arrange any 3 of the remaining 4 letters (S, H, T, and the second 'O') in the next three spots. This gives us 24 words that begin with 'O'.

🎯 Exam Tip: When dealing with permutations of words with repeated letters, first determine the distinct letters and their frequencies. For arrangements with fixed positions, fill those positions first, then calculate permutations for the remaining letters from the modified pool. Be careful when repeated letters are involved, applying \( \frac{n!}{p!q!\dots} \) appropriately.

Question 18. (i) Find the number of permutations of the letters of the word 'MATHEMATICS', taken 4 at a time. (ii) How many four-letter words can be formed using the letters of the word 'INEFFECTIVE' ? (iii) Find the number of ways in which (a) a selection, (6) an arrangement of four letters can be made from the letters of the word 'PROPORTION' ?
Answer: This question asks for permutations and selections of 4 letters from words with repeated letters.
(i) Word: 'MATHEMATICS' (11 letters) Letters and their frequencies: M(2), A(2), T(2), H(1), E(1), I(1), C(1), S(1). We need to find the number of permutations of 4 letters taken at a time. We consider different cases for the composition of the 4 letters: * **Case 1: All 4 letters are distinct.** The distinct letters are M, A, T, H, E, I, C, S (8 distinct letters). Number of ways to choose 4 distinct letters = \( {}^{8}C_4 \). \[ {}^{8}C_4 = \frac{8!}{4!4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \] For each set of 4 distinct letters, the number of permutations is \( 4! = 24 \). Total permutations for this case = \( 70 \times 24 = 1680 \). * **Case 2: 2 letters are alike and the other 2 are distinct.** The pairs of alike letters are (M, M), (A, A), (T, T) - there are 3 such pairs. Number of ways to choose one pair of alike letters = \( {}^{3}C_1 = 3 \). (e.g., MM) The remaining 2 distinct letters need to be chosen from the other 7 distinct letters (after picking one pair, say MM, the distinct letters left are A, T, H, E, I, C, S). Number of ways to choose 2 distinct letters from 7 = \( {}^{7}C_2 \). \[ {}^{7}C_2 = \frac{7!}{2!5!} = \frac{7 \times 6}{2} = 21 \] So, number of combinations of 4 letters with one pair alike and two distinct = \( 3 \times 21 = 63 \). For each such combination (e.g., MMAH), the number of permutations is \( \frac{4!}{2!} = \frac{24}{2} = 12 \). Total permutations for this case = \( 63 \times 12 = 756 \). * **Case 3: Two pairs of alike letters.** Number of ways to choose 2 pairs from the 3 available pairs (M, M), (A, A), (T, T) = \( {}^{3}C_2 = 3 \). (e.g., MMAA) For each such combination (e.g., MMAA), the number of permutations is \( \frac{4!}{2!2!} = \frac{24}{4} = 6 \). Total permutations for this case = \( 3 \times 6 = 18 \). Total number of permutations of 4 letters from 'MATHEMATICS' = \( 1680 + 756 + 18 = 2454 \).
(ii) Word: 'INEFFECTIVE' (11 letters) Letters and their frequencies: I(2), N(1), E(3), F(2), C(1), T(1), V(1). We need to form 4-letter words (permutations). * **Case 1: All 4 letters are distinct.** Distinct letters are I, N, E, F, C, T, V (7 distinct letters). Number of ways to choose 4 distinct letters = \( {}^{7}C_4 = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \). Number of permutations = \( 35 \times 4! = 35 \times 24 = 840 \). * **Case 2: 2 letters are alike (one pair) and 2 others are distinct.** Alike pairs: (I, I), (F, F) - 2 pairs. (E has 3, so EEE, not EE). We must use E for the "3 alike" case. Number of ways to choose one pair of alike letters = \( {}^{2}C_1 \) (either II or FF) = 2. Number of ways to choose 2 distinct letters from the remaining 6 distinct letters (e.g., if II is chosen, remaining distinct are N, E, F, C, T, V) = \( {}^{6}C_2 = \frac{6 \times 5}{2} = 15 \). So, combinations = \( 2 \times 15 = 30 \). Permutations for each (e.g., IINC) = \( \frac{4!}{2!} = 12 \). Total permutations for this case = \( 30 \times 12 = 360 \). * **Case 3: Two pairs of alike letters.** Pairs are (I, I) and (F, F). So only 1 way to choose two pairs: IIFF. Combinations = \( {}^{2}C_2 = 1 \). Permutations for IIFF = \( \frac{4!}{2!2!} = 6 \). Total permutations for this case = \( 1 \times 6 = 6 \). * **Case 4: 3 letters are alike and 1 is distinct.** Only 'E' occurs 3 times (EEE). Number of ways to choose 3 alike letters = \( {}^{1}C_1 \) (EEE) = 1. Number of ways to choose 1 distinct letter from the remaining 6 distinct letters (I, N, F, C, T, V) = \( {}^{6}C_1 = 6 \). So, combinations = \( 1 \times 6 = 6 \). (e.g., EEEN) Permutations for each (e.g., EEEN) = \( \frac{4!}{3!} = 4 \). Total permutations for this case = \( 6 \times 4 = 24 \). Total number of 4-letter words from 'INEFFECTIVE' = \( 840 + 360 + 6 + 24 = 1230 \). *Self-correction: The provided solution has intermediate calculations that sum to 1422. Let's re-evaluate the source's 'ABCD', 'AACD', 'AABB', 'AAAB' categories and values.* Source categories: (a) ABCD (4 distinct): \( {}^{7}C_4 = 35 \). Permutations: \( 35 \times 4! = 35 \times 24 = 840 \). (Matches) (b) AACD (2 alike, 2 distinct): Alike pairs: II, EE, FF. E appears 3 times, so EE is available. Total 3 alike pairs (II, EE, FF). Choose 1 alike pair: \( {}^{3}C_1 = 3 \). Remaining distinct letters: 6 (N, C, T, V, plus the non-chosen alike pairs, or single E). Let's list available letters: I, I, E, E, E, F, F, N, C, T, V. Distinct types: I, E, F, N, C, T, V (7 types). Choose 1 pair (say II): Remaining distinct types are E, F, N, C, T, V (6 types). Choose 2 distinct from these 6: \( {}^{6}C_2 = 15 \). So, combinations = \( {}^{3}C_1 \times {}^{6}C_2 = 3 \times 15 = 45 \). (Matches) Permutations: \( 45 \times \frac{4!}{2!} = 45 \times 12 = 540 \). (Matches) (c) AABB (Two pairs alike): Pairs are II, FF, EE. Choose 2 pairs from these 3: \( {}^{3}C_2 = 3 \). (Matches) Permutations: \( 3 \times \frac{4!}{2!2!} = 3 \times 6 = 18 \). (Matches) (d) AAAB (3 alike, 1 distinct): Only E occurs 3 times (EEE). Choose EEE: \( {}^{1}C_1 = 1 \). Remaining distinct letters: I, N, F, C, T, V (6 types). Choose 1 from these: \( {}^{6}C_1 = 6 \). (Matches) Combinations = \( 1 \times 6 = 6 \). Permutations: \( 6 \times \frac{4!}{3!} = 6 \times 4 = 24 \). (Matches) Total permutations = \( 840 + 540 + 18 + 24 = 1422 \). (Matches the source's sum). My initial calculation for Case 2 was \( 30 \times 12 = 360 \) which was due to counting 2 alike pairs (II, FF) instead of 3 (II, EE, FF). The E's can also form a pair. The source's split into II, EE, FF for "alike pairs" is correct.
(iii) Word: 'PROPORTION' (10 letters) Letters and their frequencies: P(2), R(2), O(3), T(1), I(1), N(1). We need to find (a) selections of 4 letters, and (b) arrangements of 4 letters. (a) Number of ways to make a selection of 4 letters (combinations): * **Case 1: 4 distinct letters.** Distinct letters are P, R, O, T, I, N (6 distinct letters). Number of ways to choose 4 distinct letters = \( {}^{6}C_4 = \frac{6!}{4!2!} = \frac{6 \times 5}{2} = 15 \). * **Case 2: 2 letters are alike and 2 others are distinct.** Alike pairs: P(2), R(2), O(3). So pairs are PP, RR, OO. (3 pairs). Number of ways to choose one pair of alike letters = \( {}^{3}C_1 = 3 \). Remaining distinct types: 5 (e.g., if PP is chosen, distinct types from R, O, T, I, N are 5). Number of ways to choose 2 distinct letters from 5 = \( {}^{5}C_2 = \frac{5 \times 4}{2} = 10 \). Total combinations = \( {}^{3}C_1 \times {}^{5}C_2 = 3 \times 10 = 30 \). * **Case 3: Two pairs of alike letters.** Pairs are PP, RR, OO. Choose 2 pairs from these 3: \( {}^{3}C_2 = 3 \). (e.g., PPOO, PPRR, RROO). * **Case 4: 3 letters are alike and 1 is distinct.** Only 'O' occurs 3 times (OOO). Choose OOO: \( {}^{1}C_1 = 1 \). Remaining distinct types: 5 (P, R, T, I, N). Choose 1 from these: \( {}^{5}C_1 = 5 \). Total combinations = \( {}^{1}C_1 \times {}^{5}C_1 = 1 \times 5 = 5 \). Total number of selections (combinations) = \( 15 + 30 + 3 + 5 = 53 \). (b) Number of ways to make an arrangement of 4 letters (permutations): We use the combinations from part (a) and apply permutation formulas for each case. * **Case 1: 4 distinct letters.** (15 combinations) Permutations = \( 15 \times 4! = 15 \times 24 = 360 \). * **Case 2: 2 letters are alike and 2 others are distinct.** (30 combinations) Permutations = \( 30 \times \frac{4!}{2!} = 30 \times 12 = 360 \). * **Case 3: Two pairs of alike letters.** (3 combinations) Permutations = \( 3 \times \frac{4!}{2!2!} = 3 \times 6 = 18 \). * **Case 4: 3 letters are alike and 1 is distinct.** (5 combinations) Permutations = \( 5 \times \frac{4!}{3!} = 5 \times 4 = 20 \). Total number of arrangements (permutations) = \( 360 + 360 + 18 + 20 = 758 \).
In simple words: (i) For 'MATHEMATICS', forming 4-letter arrangements involves checking different kinds of letter groups (all different, one pair same, two pairs same). Summing these up gives 2454 total arrangements. (ii) For 'INEFFECTIVE', applying the same logic for 4-letter arrangements (all different, one pair same, two pairs same, three letters same) gives 1422 arrangements. (iii) For 'PROPORTION', to pick 4 letters (selection), you get 53 ways. To arrange these 4 letters, you get 758 ways.

🎯 Exam Tip: When letters repeat in a word and you need to form shorter permutations or combinations, break the problem into cases based on the number and type of repeating letters chosen (e.g., all distinct, one pair alike, two pairs alike, three alike). Calculate combinations for letter selection, then permutations for arrangement within each case.

Question 19. A table has 7 seats, 4 being on one side facing the window and 3 being on the opposite side. In how many ways can 7 people be seated at the table. (i) if 2 people, X and Y, must sit on the same side ; (ii) X and Y must sit on opposite sides ; (iii) if 3 people, X, Y and Z, must sit on the side facing the window?
Answer: We have a table with 7 seats: 4 on the window side and 3 on the opposite side. There are 7 people to be seated.
(i) 2 people, X and Y, must sit on the same side: This can happen in two scenarios: X and Y sit on the window side OR X and Y sit on the opposite side. **Scenario 1: X and Y sit on the window side (4 seats).** - Number of ways to choose 2 seats for X and Y from 4 window seats = \( {}^{4}C_2 = 6 \). - Number of ways to arrange X and Y in these 2 chosen seats = \( 2! = 2 \). - So, X and Y can be seated in \( 6 \times 2 = 12 \) ways on the window side. - The remaining \( 7 - 2 = 5 \) people need to be seated in the remaining \( 7 - 2 = 5 \) seats (2 on the window side, 3 on the opposite side). - Number of ways to arrange the remaining 5 people = \( 5! = 120 \). - Total ways for Scenario 1 = \( 12 \times 120 = 1440 \). **Scenario 2: X and Y sit on the opposite side (3 seats).** - Number of ways to choose 2 seats for X and Y from 3 opposite seats = \( {}^{3}C_2 = 3 \). - Number of ways to arrange X and Y in these 2 chosen seats = \( 2! = 2 \). - So, X and Y can be seated in \( 3 \times 2 = 6 \) ways on the opposite side. - The remaining 5 people need to be seated in the remaining 5 seats (4 on the window side, 1 on the opposite side). - Number of ways to arrange the remaining 5 people = \( 5! = 120 \). - Total ways for Scenario 2 = \( 6 \times 120 = 720 \). Total number of ways if X and Y sit on the same side = (Ways for Scenario 1) + (Ways for Scenario 2) \[ = 1440 + 720 = 2160 \]
(ii) X and Y must sit on opposite sides: One person (say X) sits on the window side, and the other (Y) sits on the opposite side. - Number of ways to choose a seat for X from 4 window seats = \( {}^{4}C_1 = 4 \). - Number of ways to choose a seat for Y from 3 opposite seats = \( {}^{3}C_1 = 3 \). - Number of ways to arrange X and Y in their chosen seats (since X could be in Y's chosen seat if Y was picked first, etc. so \( 2! \) is for assigning X to chosen window seat, Y to chosen opposite seat, or vice versa) = \( 2! = 2 \). - The remaining \( 7 - 2 = 5 \) people need to be seated in the remaining \( 7 - 2 = 5 \) seats. - Number of ways to arrange the remaining 5 people = \( 5! = 120 \). Total number of ways = (Ways to choose seat for X) \( \times \) (Ways to choose seat for Y) \( \times \) (Ways to assign X, Y to seats) \( \times \) (Ways to arrange remaining 5 people) \[ = {}^{4}C_1 \times {}^{3}C_1 \times 2! \times 5! \] \[ = 4 \times 3 \times 2 \times 120 \] \[ = 12 \times 2 \times 120 = 24 \times 120 = 2880 \]
(iii) 3 people, X, Y and Z, must sit on the side facing the window: There are 4 seats on the window side. - Number of ways to choose 3 seats for X, Y, Z from 4 window seats = \( {}^{4}C_3 = 4 \). - Number of ways to arrange X, Y, Z in these 3 chosen seats = \( 3! = 6 \). - So, X, Y, Z can be seated in \( 4 \times 6 = 24 \) ways on the window side. - The remaining \( 7 - 3 = 4 \) people need to be seated in the remaining \( 7 - 3 = 4 \) seats (1 on the window side, 3 on the opposite side). - Number of ways to arrange the remaining 4 people = \( 4! = 24 \). Total number of ways = (Ways to seat X, Y, Z) \( \times \) (Ways to arrange remaining 4 people) \[ = ({}^{4}C_3 \times 3!) \times 4! \] \[ = 24 \times 24 = 576 \] The combination of choosing seats and arranging people is key here. For the 3 people X, Y, Z on the window side, you choose 3 seats out of 4, and then arrange those 3 people in the chosen seats. For the remaining 4 people, they can be arranged in the remaining 4 seats in 4! ways.
In simple words: We have 7 people and a 7-seat table (4 window, 3 opposite). (i) If X and Y must sit on the same side, there are 2160 ways. (ii) If X and Y must sit on opposite sides, there are 2880 ways. (iii) If X, Y, and Z must sit on the window side, there are 576 ways. Each calculation involves picking seats for specific people, arranging them, and then arranging everyone else.

🎯 Exam Tip: Break seating arrangement problems into distinct steps: first, choose seats for specified individuals/groups based on conditions; second, arrange those individuals/groups; and third, arrange the remaining individuals in the remaining seats. Multiply the results from each step.

Question 20. Seven cards, each bearing a letter, can be arranged to spell the word “DOUBLES”. How many three-letter code-words can be formed from these cards ? How many of these words (i) contain the letter S ; (ii) do not contain the letter O ; (iii) consist of a vowel between two consonants ?
Answer: The word "DOUBLES" has 7 distinct letters: D, O, U, B, L, E, S. Total number of three-letter code-words that can be formed from these 7 distinct letters is the number of permutations of 7 items taken 3 at a time. \[ {}^7P_3 = \frac{7!}{(7-3)!} = \frac{7!}{4!} = 7 \times 6 \times 5 = 210 \]
(i) Number of three-letter code-words that contain the letter S: There are two ways to calculate this: **Method 1: Direct calculation.** If 'S' is included, we need to choose 2 more letters from the remaining 6 letters (D, O, U, B, L, E). This can be done in \( {}^{6}C_2 = \frac{6 \times 5}{2} = 15 \) ways. Now we have 3 letters (S and the two chosen ones). These 3 letters can be arranged in \( 3! = 6 \) ways. So, total words containing S = \( 15 \times 6 = 90 \). **Method 2: Total permutations minus permutations without S.** Total 3-letter words = 210. If the word does not contain 'S', we choose 3 letters from the remaining 6 letters (D, O, U, B, L, E) and arrange them. Number of 3-letter words without S = \( {}^6P_3 = \frac{6!}{(6-3)!} = \frac{6!}{3!} = 6 \times 5 \times 4 = 120 \). Number of 3-letter words containing S = Total words - Words without S \[ = 210 - 120 = 90 \]
(ii) Number of three-letter code-words that do not contain the letter O: If the word does not contain 'O', we choose 3 letters from the remaining 6 letters (D, U, B, L, E, S) and arrange them. Number of 3-letter words without O = \( {}^6P_3 \). \[ {}^6P_3 = \frac{6!}{(6-3)!} = \frac{6!}{3!} = 6 \times 5 \times 4 = 120 \]
(iii) Number of three-letter code-words that consist of a vowel between two consonants: First, identify vowels and consonants in "DOUBLES": Vowels: O, U, E (3 vowels) Consonants: D, B, L, S (4 consonants) We need to form words like C V C (Consonant-Vowel-Consonant). 1. Choose 2 consonants from 4 and arrange them in the two consonant positions. Number of ways to choose and arrange 2 consonants = \( {}^4P_2 = \frac{4!}{(4-2)!} = \frac{4!}{2!} = 4 \times 3 = 12 \). 2. Choose 1 vowel from 3 and place it in the middle position. Number of ways to choose and place 1 vowel = \( {}^3P_1 = 3 \). To find the total number of such code-words, multiply the ways for each step. Total number of words (CVC) = (Ways to choose and arrange 2 consonants) \( \times \) (Ways to choose and place 1 vowel) \[ = {}^{4}P_2 \times {}^{3}P_1 = 12 \times 3 = 36 \] So, 36 words consist of a vowel between two consonants.
In simple words: The word "DOUBLES" has 7 different letters. You can make 210 different 3-letter words from these. (i) 90 of these words will contain the letter 'S'. (ii) 120 of these words will not contain the letter 'O'. (iii) 36 of these words will have a vowel in the middle and consonants on both sides.

🎯 Exam Tip: For "contain X" problems, subtract "do not contain X" from the total. For "do not contain X" problems, simply remove X from the pool and proceed. For structural problems like "CVC," fill positions sequentially (choose for C1, choose for V, choose for C2). Always distinguish between permutations (order matters) and combinations (order doesn't matter).

Question 21. How many triangles may be formed by joining any three of the nine points when (i) no three of them are collinear; (ii) five of them are collinear ?
Answer: We have 9 points. A triangle is formed by joining any three non-collinear points.
(i) No three of them are collinear: If no three points are collinear, then any choice of 3 points will form a triangle. Number of ways to choose 3 points from 9 = \( {}^{9}C_3 \). \[ {}^{9}C_3 = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} \] \[ = \frac{9 \times 8 \times 7 \times 6!}{3 \times 2 \times 1 \times 6!} \] \[ = \frac{9 \times 8 \times 7}{6} = 3 \times 4 \times 7 = 84 \] So, 84 triangles can be formed if no three points are collinear.
(ii) Five of them are collinear: When some points are collinear, choosing any 3 points from these collinear points will not form a triangle. Total number of ways to choose 3 points from 9 (without considering collinearity) = \( {}^{9}C_3 = 84 \) (as calculated above). Now, identify the combinations that do *not* form triangles. These are the combinations of 3 points chosen from the 5 collinear points. Number of ways to choose 3 points from the 5 collinear points = \( {}^{5}C_3 \). \[ {}^{5}C_3 = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} \] \[ = \frac{5 \times 4 \times 3!}{3! \times 2 \times 1} \] \[ = \frac{5 \times 4}{2} = 10 \] These 10 combinations of points will not form triangles. To find the number of triangles, subtract these invalid combinations from the total combinations of 3 points. Required number of triangles = (Total combinations of 3 points) - (Combinations of 3 collinear points) \[ = {}^{9}C_3 - {}^{5}C_3 = 84 - 10 = 74 \] So, 74 triangles can be formed when five of the nine points are collinear.
In simple words: (i) If no three of the 9 points are in a straight line, you can make 84 triangles by picking any 3 points. (ii) If 5 of the 9 points are in a straight line, you first find all possible 3-point groups (84). Then, you remove the groups of 3 points that are from the straight line (10 groups), because they don't form triangles. This leaves 74 actual triangles.

🎯 Exam Tip: When dealing with collinear points in geometry combination problems, always calculate the total possible combinations (e.g., of points for triangles or lines) and then subtract the invalid combinations formed by the collinear points. For lines, remember to add 1 back for the single line formed by the collinear points.

Question 22. A committee of 5 is to be formed from a group of 12 students consisting of 8 boys and 4 girls. In how many ways can be committee be formed if it (i) consists of exactly 3 boys and 2 girls ; (ii) contains at least 3 girls ?
Answer: We have 12 students: 8 boys and 4 girls. We need to form a committee of 5 members.
(i) The committee consists of exactly 3 boys and 2 girls: First, select 3 boys from the 8 available boys: Number of ways to select boys = \( {}^{8}C_3 \). \[ {}^{8}C_3 = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \] Next, select 2 girls from the 4 available girls: Number of ways to select girls = \( {}^{4}C_2 \). \[ {}^{4}C_2 = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6 \] To find the total number of ways to form the committee with exactly 3 boys and 2 girls, multiply these combinations: Required number of committees = \( {}^{8}C_3 \times {}^{4}C_2 = 56 \times 6 = 336 \).
(ii) The committee contains at least 3 girls: "At least 3 girls" means the committee can have 3 girls or 4 girls (since there are only 4 girls available). The committee size is 5. * **Case 1: 3 girls and 2 boys.** Number of ways to select 3 girls from 4 = \( {}^{4}C_3 = 4 \). Number of ways to select 2 boys from 8 = \( {}^{8}C_2 = \frac{8 \times 7}{2 \times 1} = 28 \). Total for Case 1 = \( {}^{4}C_3 \times {}^{8}C_2 = 4 \times 28 = 112 \). * **Case 2: 4 girls and 1 boy.** Number of ways to select 4 girls from 4 = \( {}^{4}C_4 = 1 \). Number of ways to select 1 boy from 8 = \( {}^{8}C_1 = 8 \). Total for Case 2 = \( {}^{4}C_4 \times {}^{8}C_1 = 1 \times 8 = 8 \). Total number of committees with at least 3 girls = (Total for Case 1) + (Total for Case 2) \[ = 112 + 8 = 120 \] So, 120 committees can be formed with at least 3 girls.
In simple words: From 8 boys and 4 girls, we need to make a 5-person team. (i) To have exactly 3 boys and 2 girls, there are 336 ways. (ii) To have at least 3 girls, you can have either 3 girls and 2 boys (112 ways) or 4 girls and 1 boy (8 ways). Adding these up gives a total of 120 ways.

🎯 Exam Tip: For "exactly" conditions, calculate the combinations for each group and multiply. For "at least" conditions, list all possible valid compositions that satisfy the condition, calculate combinations for each composition, and then sum them up. Ensure you don't exceed the available number of members in any category.

Question 23. There are 5 gentlemen and 4 ladies to dine at a round table. In how many ways can they seat themselves so that no two ladies are together ?
Answer: We have 5 gentlemen (G) and 4 ladies (L) to be seated at a round table such that no two ladies sit together. To ensure no two ladies sit together, we must first seat the gentlemen. Since it's a round table, the number of ways to arrange \( n \) distinct items in a circle is \( (n-1)! \). 1. **Arrange the 5 gentlemen around the round table:** Number of ways to seat 5 gentlemen = \( (5-1)! = 4! = 24 \). These 5 gentlemen will create 5 spaces between them where the ladies can sit. G _ G _ G _ G _ G _ 2. **Arrange the 4 ladies in these spaces:** There are 5 available spaces. We need to seat 4 ladies in 4 of these 5 spaces. The ladies are distinct. Number of ways to choose 4 spaces from 5 and arrange the 4 ladies in them = \( {}^5P_4 \). \[ {}^5P_4 = \frac{5!}{(5-4)!} = \frac{5!}{1!} = 5 \times 4 \times 3 \times 2 = 120 \] To find the total number of ways, multiply the ways to arrange the gentlemen and the ways to arrange the ladies. Total number of ways = (Ways to arrange gentlemen) \( \times \) (Ways to arrange ladies) \[ = 4! \times {}^5P_4 = 24 \times 120 = 2880 \] Therefore, there are 2880 ways to seat 5 gentlemen and 4 ladies at a round table so that no two ladies are together.
In simple words: To seat 5 gentlemen and 4 ladies around a round table so no two ladies sit next to each other, first seat the 5 gentlemen (24 ways). This creates 5 spots between them. Then, pick 4 of these 5 spots for the 4 ladies and arrange them (120 ways). Multiply these two numbers to get the total of 2880 ways.

🎯 Exam Tip: For "no two X together" problems, first arrange the "non-X" items. This creates spaces. Then, arrange the "X" items in a selection of these spaces. Remember the \( (n-1)! \) formula for circular permutations.

Question 24. There are 12 points in a plane, of which 5 are collinear. Find (i) the number of triangles that can be formed with vertices at these points ; (ii) the number of straight lines obtained by joining these points in pairs.
Answer: We have 12 points in a plane, and 5 of these points are collinear (lie on the same straight line).
(i) The number of triangles that can be formed with vertices at these points: A triangle is formed by choosing any 3 non-collinear points. First, calculate the total number of ways to choose 3 points from 12, assuming no three are collinear: Total combinations of 3 points = \( {}^{12}C_3 \). \[ {}^{12}C_3 = \frac{12!}{3!9!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 2 \times 11 \times 10 = 220 \] Next, identify combinations of 3 points that do NOT form a triangle. These are the combinations where all 3 chosen points are from the 5 collinear points. Number of ways to choose 3 points from the 5 collinear points = \( {}^{5}C_3 \). \[ {}^{5}C_3 = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10 \] These 10 combinations do not form triangles. Number of triangles = (Total combinations of 3 points) - (Combinations of 3 collinear points) \[ = {}^{12}C_3 - {}^{5}C_3 = 220 - 10 = 210 \]
(ii) The number of straight lines obtained by joining these points in pairs: A straight line is formed by joining any 2 points. First, calculate the total number of ways to choose 2 points from 12, assuming no three are collinear: Total combinations of 2 points = \( {}^{12}C_2 \). \[ {}^{12}C_2 = \frac{12!}{2!10!} = \frac{12 \times 11}{2 \times 1} = 66 \] Next, identify combinations of 2 points that lie on the collinear line. When we choose 2 points from the 5 collinear points, they form a single line, not multiple distinct lines. So we subtract the extra lines counted for the collinear points and add 1 for the unique line they all form. Number of ways to choose 2 points from the 5 collinear points = \( {}^{5}C_2 \). \[ {}^{5}C_2 = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10 \] When we initially calculated \( {}^{12}C_2 \), these 10 pairs of points (from the 5 collinear points) were counted as 10 distinct lines. However, all these 10 pairs actually lie on the *same single line*. So, we subtract these 10 "lines" and add back 1 for the actual single line formed by the 5 collinear points. Number of straight lines = (Total combinations of 2 points) - (Combinations of 2 collinear points) + 1 (for the single line formed by the collinear points) \[ = {}^{12}C_2 - {}^{5}C_2 + 1 \] \[ = 66 - 10 + 1 = 57 \] So, 57 straight lines can be formed.
In simple words: You have 12 points, and 5 of them are on a straight line. (i) To find how many triangles can be made, first find all ways to pick 3 points from 12 (220 ways). Then, subtract the groups of 3 points that are all on the same straight line (10 ways), because they don't form triangles. This leaves 210 triangles. (ii) To find how many straight lines can be made, first find all ways to pick 2 points from 12 (66 ways). Then, subtract the pairs of points that are on the same straight line (10 ways), because they only make one line, not 10. Add 1 back for that single line. This gives 57 lines.

🎯 Exam Tip: For problems involving collinear points: for triangles, subtract combinations of 3 collinear points from the total; for straight lines, subtract combinations of 2 collinear points from the total, then add 1 (for the single line formed by all collinear points) back to the result.

Question 25. A committee of 5 is to be formed from a group of 10 people, consisting of 4 single men, 4 single women and a married couple, the committee is to consist of a chairman, who must be a single man, 2 other men and 2 women, (i) Find the total number of committees possible. (ii) HOW many of these would include the married couple ?
Answer: We have a group of 10 people: * 4 single men (SM) * 4 single women (SW) * 1 married couple (M+W) This means there are \( 4+1=5 \) men in total and \( 4+1=5 \) women in total. The committee size is 5, consisting of: 1 chairman (must be a single man), 2 other men, and 2 women.
(i) Find the total number of committees possible: 1. **Select the chairman:** The chairman must be a single man. There are 4 single men available. Number of ways = \( {}^{4}C_1 = 4 \). 2. **Select 2 other men:** After the chairman (a single man) is selected, there are \( 4-1=3 \) single men left, plus the 1 married man. So, there are \( 3+1=4 \) men remaining from whom to choose the 2 other men. Number of ways = \( {}^{4}C_2 = 6 \). 3. **Select 2 women:** There are 4 single women and 1 married woman. So, there are \( 4+1=5 \) women available. Number of ways = \( {}^{5}C_2 = 10 \). To find the total number of committees, multiply the number of ways for each selection: Total committees = (Ways to choose chairman) \( \times \) (Ways to choose 2 other men) \( \times \) (Ways to choose 2 women) \[ = {}^{4}C_1 \times {}^{4}C_2 \times {}^{5}C_2 = 4 \times 6 \times 10 = 240 \]
(ii) How many of these would include the married couple? If the married couple (M and W) must be included, then the husband is one of the "2 other men" and the wife is one of the "2 women". 1. **Select the chairman:** The chairman must be a single man. This choice is independent of the married couple. Number of ways = \( {}^{4}C_1 = 4 \). 2. **Select 2 other men (including the married man):** One of the "2 other men" is already the married man. So we need to select \( 2-1=1 \) more man. This man must be chosen from the remaining single men (\( 4-1=3 \) single men, since one single man is chairman). Number of ways to choose 1 more man = \( {}^{3}C_1 = 3 \). 3. **Select 2 women (including the married woman):** One of the "2 women" is already the married woman. So we need to select \( 2-1=1 \) more woman. This woman must be chosen from the remaining 4 single women. Number of ways to choose 1 more woman = \( {}^{4}C_1 = 4 \). Total committees including the married couple = (Ways to choose chairman) \( \times \) (Ways to choose 1 other man) \( \times \) (Ways to choose 1 other woman) \[ = {}^{4}C_1 \times {}^{3}C_1 \times {}^{4}C_1 = 4 \times 3 \times 4 = 48 \] So, 48 of these committees would include the married couple.
In simple words: You're forming a 5-person committee from 10 people (4 single men, 4 single women, 1 married couple). The committee needs a single man as chairman, 2 more men, and 2 women. (i) There are 240 total ways to form such a committee. (ii) If the married couple must be in the committee, you still pick the chairman from single men (4 ways). Then, you pick 1 more man from the remaining single men (3 ways) and 1 more woman from the single women (4 ways). This totals 48 committees that include the married couple.

🎯 Exam Tip: When specific individuals (like a married couple) are mandatory, adjust the number of positions to be filled and the pool of available candidates *before* making the selections. Distinguish clearly between general roles (e.g., "men") and specific types (e.g., "single men", "married men").

Question 26. A committee of 5 persons is to be formed from a group of 6 gentlemen and 4 ladies. In how many ways can this be done if the committee is to include atleast one lady ?
Answer: We have 6 gentlemen (G) and 4 ladies (L). We need to form a committee of 5 persons that must include at least one lady. "At least one lady" means the committee can have 1 lady, 2 ladies, 3 ladies, or 4 ladies (since there are only 4 ladies available). We'll consider each case: * **Case 1: 1 lady and 4 gentlemen.** Number of ways = \( {}^{4}C_1 \times {}^{6}C_4 \). \[ {}^{4}C_1 = 4 \] \[ {}^{6}C_4 = \frac{6!}{4!2!} = \frac{6 \times 5}{2 \times 1} = 15 \] So, \( 4 \times 15 = 60 \) ways. * **Case 2: 2 ladies and 3 gentlemen.** Number of ways = \( {}^{4}C_2 \times {}^{6}C_3 \). \[ {}^{4}C_2 = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6 \] \[ {}^{6}C_3 = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \] So, \( 6 \times 20 = 120 \) ways. * **Case 3: 3 ladies and 2 gentlemen.** Number of ways = \( {}^{4}C_3 \times {}^{6}C_2 \). \[ {}^{4}C_3 = 4 \] \[ {}^{6}C_2 = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15 \] So, \( 4 \times 15 = 60 \) ways. * **Case 4: 4 ladies and 1 gentleman.** Number of ways = \( {}^{4}C_4 \times {}^{6}C_1 \). \[ {}^{4}C_4 = 1 \] \[ {}^{6}C_1 = 6 \] So, \( 1 \times 6 = 6 \) ways. To find the total number of ways to form the committee with at least one lady, sum the ways from all cases: Total ways = \( 60 + 120 + 60 + 6 = 246 \). Alternatively, using the complementary principle: Total ways to form a 5-person committee from 10 people (6 gentlemen + 4 ladies) = \( {}^{10}C_5 \). \[ {}^{10}C_5 = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 2 \times 3 \times 2 \times 7 \times 3 = 252 \] Number of ways to form a committee with no ladies (i.e., all 5 are gentlemen) = \( {}^{6}C_5 \). \[ {}^{6}C_5 = 6 \] Number of ways to form a committee with at least one lady = (Total ways) - (Ways with no ladies) \[ = 252 - 6 = 246 \] Both methods yield the same result.
In simple words: To form a 5-person committee from 6 gentlemen and 4 ladies, making sure there's at least one lady, you can sum the ways for having 1, 2, 3, or 4 ladies. This adds up to 246 ways. A quicker way is to find all possible committees (252 ways) and subtract the committees that have no ladies (6 ways), which also gives 246.

🎯 Exam Tip: When solving "at least" problems, always consider if direct summation of cases is easier or if the complementary principle (total ways minus "none" cases) simplifies the calculation. Choose the most efficient method to save time.

Question 27. Out of 3 books on Economics, 4 books on Political Science and 5 books on Geography, how many collections can be made, if each collection consists of
(i) exactly one book on each subject;
(ii) at least one book on each subject?
Answer:
(i) To select exactly one book from each subject:
For Economics, the number of ways to choose one book from 3 is \( 3C_1 \).
For Political Science, the number of ways to choose one book from 4 is \( 4C_1 \).
For Geography, the number of ways to choose one book from 5 is \( 5C_1 \).
So, the total number of collections is \( 3C_1 \times 4C_1 \times 5C_1 = 3 \times 4 \times 5 = 60 \). Each collection will have one book from each subject.
(ii) To select at least one book from each subject:
For Economics, the number of ways to choose at least one book from 3 is \( 3C_1 + 3C_2 + 3C_3 = 2^3 - 1 = 7 \) ways. This means we can choose 1, 2, or all 3 books.
For Political Science, the number of ways to choose at least one book from 4 is \( 4C_1 + 4C_2 + 4C_3 + 4C_4 = 2^4 - 1 = 16 - 1 = 15 \) ways.
For Geography, the number of ways to choose at least one book from 5 is \( 5C_1 + 5C_2 + 5C_3 + 5C_4 + 5C_5 = 2^5 - 1 = 31 \) ways.
Using the fundamental principle of counting, the required number of ways to select at least one book from each subject is \( 7 \times 15 \times 31 = 3255 \).
In simple words: When choosing exactly one from each subject, multiply the number of options for each subject. When choosing at least one, calculate all possible choices for each subject (excluding choosing none) and then multiply those totals.

🎯 Exam Tip: Remember that "at least one" means all possible selections except selecting zero items. This is often calculated as \( 2^n - 1 \), where n is the total number of items.

Question 28. Find the number of words which can be formed by taking two alike and two different letters from the word "COMBINATION".
Answer:
The word 'COMBINATION' has 11 letters. When we look closely, we find repeating letters: two I's, two O's, and two N's. The other letters (C, M, B, A, T) appear only once. So, there are 3 pairs of alike letters and 5 distinct single letters.
We need to form 4-letter words with two alike and two different letters. This involves two steps:
1. Select two alike letters: We can choose one pair out of the three available pairs (II, OO, NN). This can be done in \( 3C_1 \) ways.
2. Select two different letters: After picking one pair of alike letters, there are now 7 unique letters left (the remaining 2 pairs are treated as unique single letters, plus the original 5 distinct letters). We need to choose 2 different letters from these 7. This can be done in \( 7C_2 \) ways.
So, the total number of combinations is \( 3C_1 \times 7C_2 = 3 \times \frac{7!}{5!2!} = 3 \times \frac{7 \times 6}{2} = 3 \times 21 = 63 \).
Now, for each of these 63 combinations, we need to arrange the letters to form words. Each combination has 4 letters, with 2 of them being alike. The number of permutations for such a set of letters is given by \( \frac{4!}{2!1!1!} \), which equals \( \frac{24}{2} = 12 \) words.
Therefore, the total number of words formed is \( 63 \times 12 = 756 \).
In simple words: First, figure out how many ways you can pick 4 letters with 2 being the same and 2 being different. Then, for each of those groups of 4 letters, find all the different words you can make by arranging them. Finally, multiply these two numbers to get the total words.

🎯 Exam Tip: When forming words with repeated letters, remember to first calculate the combinations of letters and then the permutations for each combination. The formula for permutations with repeated items is \( \frac{n!}{p!q!...} \).