GSEB Class 11 Maths Solutions Chapter 7 Permutations and Combinations Exercise 7.4

Question 1. If \( ^nC_8 = ^nC_2 \). Find \( ^nC_2 \).
Answer: We are given the relation \( ^nC_8 = ^nC_2 \).
Using the property \( ^nC_r = ^nC_{n-r} \), we can write \( ^nC_8 = ^nC_{n-8} \).
So, from the given equation, we have \( ^nC_2 = ^nC_{n-8} \).
This implies that \( 2 = n - 8 \) or \( n = 2 + 8 \), which gives \( n = 10 \).
Now we need to find \( ^nC_2 \), substituting \( n = 10 \).
\( ^nC_2 = ^{10}C_2 \)
\( ^{10}C_2 = \frac{10 \times 9}{1 \times 2} \)
\( ^{10}C_2 = 5 \times 9 \)
\( ^{10}C_2 = 45 \).
In simple words: When two combinations are equal like \( ^nC_a = ^nC_b \), it means either \( a = b \) or \( a + b = n \). Here, \( 8 \) and \( 2 \) are different, so we know \( 8 + 2 = n \), which tells us \( n \) is \( 10 \). Then, we just calculate \( ^{10}C_2 \) using the standard formula.

Exam Tip: Remember the key property \( ^nC_r = ^nC_{n-r} \). This property often simplifies combination problems by helping you find the value of 'n' or a missing 'r'.

Question 2. Determine n, if
(i) \( ^{2n}C_3 : ^nC_2 = 12:1 \)
(ii) \( ^{2n}C_3 : ^nC_3 = 11:1 \)
Answer:
(i) Given \( ^{2n}C_3 : ^nC_2 = 12:1 \).
\( \frac{^{2n}C_3}{^nC_2} = \frac{12}{1} \)
\( \frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{2!(n-2)!}} = 12 \)
\( \frac{2n(2n-1)(2n-2)}{3 \times 2 \times 1} \div \frac{n(n-1)}{2 \times 1} = 12 \)
\( \frac{2n(2n-1)2(n-1)}{6} \times \frac{2}{n(n-1)} = 12 \)
\( \frac{4n(2n-1)(n-1)}{6} \times \frac{2}{n(n-1)} = 12 \)
\( \frac{4(2n-1)}{3} = 12 \)
\( 4(2n-1) = 36 \)
\( 2n-1 = 9 \)
\( 2n = 10 \)
\( n = 5 \).
(ii) Given \( ^{2n}C_3 : ^nC_3 = 11:1 \).
\( \frac{^{2n}C_3}{^nC_3} = \frac{11}{1} \)
\( \frac{\frac{2n(2n-1)(2n-2)}{3 \times 2 \times 1}}{\frac{n(n-1)(n-2)}{3 \times 2 \times 1}} = 11 \)
\( \frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)} = 11 \)
\( \frac{2n(2n-1)2(n-1)}{n(n-1)(n-2)} = 11 \)
\( \frac{4(2n-1)}{n-2} = 11 \)
\( 4(2n-1) = 11(n-2) \)
\( 8n - 4 = 11n - 22 \)
\( 11n - 8n = 22 - 4 \)
\( 3n = 18 \)
\( n = 6 \).
In simple words: For both parts, we set up the ratio of the combinations as a fraction. Then, we expand the combination terms using the formula \( ^nC_r = \frac{n(n-1)...(n-r+1)}{r!} \). After simplifying the fractions, we solve the resulting algebraic equation for \( n \).

Exam Tip: When dealing with ratios of combinations, expand them into their factorial forms or product forms for easier simplification. Always make sure to cancel common terms carefully to avoid errors.

Question 3. How many chords can be drawn through 21 points on a circle?
Answer: A chord is formed by connecting any two distinct points on a circle. So, to find the number of chords, we simply need to choose 2 points from the available 21 points.
This is a combination problem, as the order of selecting the points does not matter.
The number of chords is given by \( ^{21}C_2 \).
\( ^{21}C_2 = \frac{21!}{2!(21 - 2)!} \)
\( = \frac{21!}{2!19!} \)
\( = \frac{21 \times 20 \times (19)!}{2 \times 1 \times (19)!} \)
\( = \frac{21 \times 20}{2} \)
\( = 21 \times 10 \)
\( = 210 \) chords.
In simple words: To make a chord, you need two points. Since we have 21 points, we choose any 2 of them. We use combinations for this because picking point A then B is the same as picking point B then A. The calculation gives us 210 possible chords.

Exam Tip: Remember that "selection" or "choosing" usually implies combinations (\( ^nC_r \)), while "arrangement" or "ordering" implies permutations (\( ^nP_r \)). For chords, only the selection of points matters, not their order.

Question 4. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Answer: We need to select 3 boys from a group of 5 boys and 3 girls from a group of 4 girls.
The number of ways to select 3 boys from 5 boys is \( ^5C_3 \).
\( ^5C_3 = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10 \) ways.
The number of ways to select 3 girls from 4 girls is \( ^4C_3 \).
\( ^4C_3 = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = \frac{4}{1} = 4 \) ways.
To find the total number of ways to select a team of 3 boys and 3 girls, we multiply the number of ways for each selection (as these are independent choices).
Total number of ways = (Ways to select boys) \( \times \) (Ways to select girls)
\( = ^5C_3 \times ^4C_3 \)
\( = 10 \times 4 \)
\( = 40 \) ways.
In simple words: To pick 3 boys from 5, there are 10 ways. To pick 3 girls from 4, there are 4 ways. Since we need both, we multiply these numbers together to get the total team combinations, which is 40.

Exam Tip: When selections are independent (like choosing boys and choosing girls from separate groups), you multiply the number of ways for each selection to get the total number of combined ways. Always break down complex selection problems into simpler parts.

Question 5. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls, if each selection consists of 3 balls of each colour?
Answer: We need to select a total of 9 balls, with 3 balls of each color.
There are 6 red balls, 5 white balls, and 5 blue balls available.
1. The number of ways to select 3 red balls from 6 red balls is \( ^6C_3 \).
\( ^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \) ways.
2. The number of ways to select 3 white balls from 5 white balls is \( ^5C_3 \).
\( ^5C_3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10 \) ways.
3. The number of ways to select 3 blue balls from 5 blue balls is \( ^5C_3 \).
\( ^5C_3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10 \) ways.
To find the total number of ways to select 3 balls of each color, we multiply the number of ways for each color selection.
Total ways \( = ^6C_3 \times ^5C_3 \times ^5C_3 \)
\( = 20 \times 10 \times 10 \)
\( = 2000 \) ways.
In simple words: We calculate how many ways to pick 3 red balls, how many ways to pick 3 white balls, and how many ways to pick 3 blue balls. Then, we multiply these numbers together to get the total number of ways to pick all 9 balls with 3 of each color.

Exam Tip: Break down problems involving selections from multiple categories into separate combination calculations for each category. Then, multiply the results to get the total number of combined selections.

Question 6. Determine the number of 5 cards combinations out of a deck of 52 cards, if there is exactly one ace in each combination?
Answer: A standard deck of 52 cards has 4 aces and 48 non-ace cards.
We need to form a 5-card combination with exactly one ace.
This means we must select 1 ace from the 4 available aces AND 4 other cards from the 48 non-ace cards.
1. Number of ways to select 1 ace from 4 aces is \( ^4C_1 \).
\( ^4C_1 = \frac{4!}{1!(4-1)!} = \frac{4}{1} = 4 \) ways.
2. Number of ways to select 4 non-ace cards from the remaining 48 cards is \( ^{48}C_4 \).
\( ^{48}C_4 = \frac{48!}{4!(48-4)!} = \frac{48!}{4!44!} \)
\( = \frac{48 \times 47 \times 46 \times 45 \times (44)!}{4 \times 3 \times 2 \times 1 \times (44)!} \)
\( = \frac{48}{4 \times 3 \times 2 \times 1} \times 47 \times 46 \times 45 \)
\( = 2 \times 47 \times 46 \times 45 \)
\( = 778320 \) ways.
Total number of 5-card combinations with exactly one ace \( = ^4C_1 \times ^{48}C_4 \)
\( = 4 \times 778320 \)
\( = 3113280 \) ways.
In simple words: We must choose one ace from the four aces available. Then, we choose four more cards from the remaining 48 cards (which are not aces). We multiply these two selection numbers together to find the total number of ways to get a 5-card hand with exactly one ace.

Exam Tip: When a problem specifies a fixed number of items from a certain category (e.g., "exactly one ace"), always calculate that selection first. Then, select the remaining items from the complementary group (e.g., non-aces) and multiply the results.

Question 7. In how many ways, can we select a cricket eleven from 17 players in which only 5 players can bowl, if each cricket eleven must include exactly 4 bowlers?
Answer: We have a total of 17 players. Among these, 5 are bowlers, and the remaining \( 17 - 5 = 12 \) are non-bowlers (or specialist batsmen/wicket-keepers).
We need to select a cricket eleven (11 players) that must include exactly 4 bowlers.
1. **Select the bowlers:** We need to choose 4 bowlers from the 5 available bowlers.
Number of ways \( = ^5C_4 \)
\( ^5C_4 = ^5C_{5-4} = ^5C_1 = 5 \) ways.
2. **Select the remaining players:** Since the team needs 11 players and we have already chosen 4 bowlers, we need to select \( 11 - 4 = 7 \) more players.
These 7 players must be selected from the non-bowlers pool, which consists of 12 players.
Number of ways \( = ^{12}C_7 \)
\( ^{12}C_7 = ^{12}C_{12-7} = ^{12}C_5 \)
\( ^{12}C_5 = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} \)
\( = \frac{12}{4 \times 3} \times \frac{10}{5 \times 2 \times 1} \times 11 \times 9 \times 8 \)
\( = 1 \times 1 \times 11 \times 9 \times 8 \)
\( = 792 \) ways.
3. **Total number of ways:** To form the complete team, we multiply the ways to select bowlers and non-bowlers.
Total ways \( = ^5C_4 \times ^{12}C_7 \)
\( = 5 \times 792 \)
\( = 3960 \) ways.
In simple words: First, we pick exactly 4 bowlers from the 5 available bowlers. Then, we pick the remaining 7 players for the team from the 12 non-bowler players. We multiply these two numbers to find all the different ways to create the team.

Exam Tip: Clearly separate the selection process for different roles (e.g., bowlers vs. non-bowlers). Use the complement rule \( ^nC_r = ^nC_{n-r} \) to simplify calculations when 'r' is large, as it often reduces the number of terms to multiply.

Question 8. A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected?
Answer: We have a bag with 5 black balls and 6 red balls.
We need to select 2 black balls and 3 red balls.
1. **Select black balls:** The number of ways to choose 2 black balls from 5 black balls is \( ^5C_2 \).
\( ^5C_2 = \frac{5 \times 4}{2 \times 1} = 10 \) ways.
2. **Select red balls:** The number of ways to choose 3 red balls from 6 red balls is \( ^6C_3 \).
\( ^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \) ways.
To find the total number of ways to select 2 black and 3 red balls, we multiply the number of ways for each color selection, as these are independent events.
Total ways \( = ^5C_2 \times ^6C_3 \)
\( = 10 \times 20 \)
\( = 200 \) ways.
In simple words: We calculate how many ways to pick 2 black balls from 5, and how many ways to pick 3 red balls from 6. Then, we multiply these two results to find the total number of ways to choose both sets of balls together.

Exam Tip: For selection problems involving different types of items (like different colored balls), calculate the combinations for each type separately and then multiply them to get the total number of ways to make the combined selection.

Question 9. In how many ways, can a student choose a programme of 5 courses, if 9 courses are available and 2 courses are compulsory for every student?
Answer: A student needs to choose a program of 5 courses.
There are 9 courses available in total.
Out of these 9 courses, 2 courses are compulsory, meaning the student has no choice but to take these two.
Since 2 courses are compulsory, these are already selected. The student must choose the remaining courses from the non-compulsory ones.
Number of courses to be chosen by the student \( = 5 \text{ (total courses)} - 2 \text{ (compulsory courses)} = 3 \) courses.
Number of available non-compulsory courses \( = 9 \text{ (total available)} - 2 \text{ (compulsory courses)} = 7 \) courses.
So, the student needs to choose 3 courses from these 7 non-compulsory courses.
The number of ways to do this is given by \( ^7C_3 \).
\( ^7C_3 = \frac{7!}{3!(7 - 3)!} \)
\( = \frac{7!}{3!4!} \)
\( = \frac{7 \times 6 \times 5 \times (4)!}{3 \times 2 \times 1 \times (4)!} \)
\( = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \)
\( = 7 \times 5 \)
\( = 35 \) ways.
In simple words: Two courses are mandatory, so they are automatically chosen. The student needs to pick 3 more courses to reach a total of 5. These 3 courses must come from the 7 courses that are not mandatory. So, we find the number of ways to pick 3 courses out of 7.

Exam Tip: When some choices are "compulsory" or "fixed," subtract them from the total number of items available and from the total number of items to be selected. Then, apply the combination formula to the remaining choices.