Samacheer Kalvi Class 11 Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Exercise 4.1

Question 1. (i) A person went to a restaurant for dinner. In the menu card, the person saw 10 Indian and 7 Chinese food items. In how many ways the person can select either an Indian or Chinese food?
Answer: A person can pick an Indian food item in 10 different ways. They can also pick a Chinese food item in 7 different ways. Since the person can choose *either* Indian or Chinese food, we add the ways together. The total ways to select a food item is \( 10 + 7 = 17 \) ways. This is an application of the addition principle in combinatorics.
In simple words: The person has 10 choices for Indian food and 7 choices for Chinese food. To pick one from either, just add the choices: \( 10 + 7 = 17 \) ways.

🎯 Exam Tip: When you need to choose 'either' one option 'or' another, always use the addition principle to find the total number of ways.

Question 1. (ii) There are 3 types of a toy car and 2 types of toy train are available in a shop. Find the number of ways a baby can buy a toy car and a toy train?
Answer: The shop has 3 types of toy cars and 2 types of toy trains. A baby wants to buy one toy car and one toy train. To find the total number of ways to pick both, we use the multiplication principle. So, the number of ways to buy a toy car and a toy train is \( 3 \times 2 = 6 \) ways. This principle is used when multiple independent choices are made together.
In simple words: There are 3 kinds of toy cars and 2 kinds of toy trains. To choose one of each, you multiply the options: \( 3 \times 2 = 6 \) different ways.

🎯 Exam Tip: When you need to choose 'one and another' option, use the multiplication principle to find the total number of ways by multiplying the number of choices for each item.

Question 1. (iii) How many two-digit numbers can be formed using 1, 2, 3, 4, 5 without repetition of digits?
Answer: We need to form two-digit numbers using the digits 1, 2, 3, 4, 5, without repeating any digit. For the unit's place, there are 5 possible digits we can choose from. After picking one digit for the unit's place, we have 4 digits left. So, for the ten's place, there are 4 choices remaining. Using the fundamental principle of multiplication, the total number of two-digit numbers is \( 5 \times 4 = 20 \). Each choice for the tens digit is independent of the choice for the units digit.
In simple words: You have 5 digits (1,2,3,4,5). For a two-digit number, pick one digit for the first spot (5 choices). Then, for the second spot, you have one less digit to pick from (4 choices). Multiply them: \( 5 \times 4 = 20 \) numbers.

🎯 Exam Tip: When forming numbers without repetition, remember that each choice reduces the number of options for the next position.

Question 1. (iv) In how many ways can 3 persons enter into a conference hall in which there are 10 seats and take their seats?
Answer: There are 10 seats in the conference hall and 3 people want to sit. The first person has 10 choices for a seat. Once the first person sits, there are 9 seats left for the second person. After the second person sits, there are 8 seats remaining for the third person. By the fundamental principle of multiplication, the total number of ways these 3 people can take their seats is \( 10 \times 9 \times 8 = 720 \) ways. This is a permutation problem where order matters.
In simple words: Three people need to sit in 10 seats. The first person has 10 options. The second has 9, and the third has 8. Multiply these: \( 10 \times 9 \times 8 = 720 \) ways.

🎯 Exam Tip: For problems involving seating arrangements where each person occupies a unique seat, use permutations (nPr) or the multiplication principle for decreasing choices.

Question 1. (v) In how ways 5 persons can be seated In a row?
Answer: We need to find the number of ways 5 people can sit in a row. For the first seat, there are 5 choices. For the second seat, there are 4 choices left. This continues until the last seat, where there is only 1 choice. The total number of arrangements is given by \( 5! \) (5 factorial), which is \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \) ways. Factorials are very common in arrangement problems.
In simple words: To arrange 5 people in a row, you multiply the number of choices for each spot. This is \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \) ways.

🎯 Exam Tip: The number of ways to arrange 'n' distinct items in a row is always 'n!' (n factorial).

Question 2. (i) A mobile phone has a passcode of 6 distinct digits. What is the maximum number of attempts one makes to retrieve the passcode?
Answer: A mobile phone passcode has 6 distinct digits. This means no digit can be repeated. For the first digit, there are 10 choices (0-9). For the second digit, since one digit is already used, there are 9 choices left. This continues for all 6 digits. So, the maximum number of attempts to find the passcode is \( 10 \times 9 \times 8 \times 7 \times 6 \times 5 = 151,200 \). This is a permutation of choosing 6 digits from 10 available digits.
In simple words: A 6-digit passcode cannot repeat numbers. You have 10 choices for the first digit. Then 9 for the second, and so on, until 5 choices for the sixth digit. Multiply these numbers together to get 151,200 attempts.

🎯 Exam Tip: Remember the difference between 'distinct digits' (no repetition) and 'digits can repeat' when calculating passcode possibilities.

Question 2. (ii) Given four flags of different colours, how many different signals can be generated if each signal requires the use of three flags, one below the other?
Answer: We have 4 flags of different colors, and each signal needs 3 flags placed one below the other. For the top position, we have 4 choices of flags. For the middle position, since one flag is already used, there are 3 remaining choices. For the bottom position, there are 2 flags left. Using the fundamental principle of multiplication, the total number of different signals is \( 4 \times 3 \times 2 = 24 \) ways. This is a permutation problem because the order of flags matters for the signal.
In simple words: You have 4 different flags. To make a signal using 3 flags, pick 1 for the top (4 choices), 1 for the middle (3 choices left), and 1 for the bottom (2 choices left). Multiply them: \( 4 \times 3 \times 2 = 24 \) signals.

🎯 Exam Tip: When ordering distinct items, like flags in a signal, remember that the number of choices decreases for each subsequent position.

Question 3. Four children are running a race. (i) In how many ways can the first two places be filled?
Answer: There are 4 children in a race. To fill the first place, there are 4 possible children. For the second place, since one child is already in first, there are 3 children remaining. So, the number of ways to fill the first two places is \( 4 \times 3 = 12 \) ways. The order of finishing matters for these places.
In simple words: With 4 children, there are 4 choices for who comes first. Then, there are 3 choices for who comes second. So, \( 4 \times 3 = 12 \) ways to fill the top two spots.

🎯 Exam Tip: When determining distinct positions like 1st and 2nd in a race, consider the decreasing number of options available for each position.

Question 3. (ii) In how many different ways could they finish the race?
Answer: We know the first two places can be filled in 12 ways. After these two, there are 2 children left for the third place. For the fourth and last place, there will be only 1 child remaining. Therefore, the total number of different ways the race can be finished (all four places) is \( 12 \times 2 \times 1 = 24 \) ways. Alternatively, it's \( 4! = 4 \times 3 \times 2 \times 1 = 24 \) ways for all children to finish.
In simple words: To find all possible finishing orders for 4 children, you arrange all 4. This is \( 4 \times 3 \times 2 \times 1 = 24 \) different ways.

🎯 Exam Tip: The total number of ways 'n' participants can finish a race is 'n!' (n factorial), as all positions are distinct.

Question 4. Count the number of three-digit numbers which can be formed from the digits 2, 4, 6, 8 if (i) Repetitions of digits is allowed.
Answer: We want to form three-digit numbers using the digits 2, 4, 6, 8, with repetition allowed. For the unit's place, we have 4 choices (2, 4, 6, or 8). Since repetition is allowed, we still have 4 choices for the ten's place and 4 choices for the hundred's place. Using the multiplication principle, the total number of three-digit numbers is \( 4 \times 4 \times 4 = 64 \) ways. Each digit choice is independent here.
In simple words: If digits can repeat, you have 4 choices for each of the three places (hundreds, tens, units). So, \( 4 \times 4 \times 4 = 64 \) three-digit numbers.

🎯 Exam Tip: When repetition is allowed, the number of choices for each position remains constant.

Question 4. (ii) Repetitions of digits is not allowed?
Answer: Now, let's form three-digit numbers using 2, 4, 6, 8 without repetition. For the unit's place, there are 4 choices. For the ten's place, one digit is used, so there are \( 4 - 1 = 3 \) choices left. For the hundred's place, two digits are used, leaving \( 3 - 1 = 2 \) choices. So, the total number of three-digit numbers is \( 4 \times 3 \times 2 = 24 \) ways. This is a permutation of 4 items taken 3 at a time.
In simple words: If digits cannot repeat, you have 4 choices for the first digit. Then 3 choices for the second. Then 2 choices for the third. Multiply them: \( 4 \times 3 \times 2 = 24 \) numbers.

🎯 Exam Tip: Without repetition, the number of available digits decreases by one for each position filled.

Question 5. How many three-digit numbers are there with 3 in the unit place? (i) with repetition
Answer: We need to form three-digit numbers where the unit's place is always 3, and repetition is allowed. The available digits are 0-9. The unit's place can only be filled in 1 way (with digit 3). For the ten's place, since repetition is allowed, there are 10 choices (0-9). For the hundred's place, we cannot use 0 (to keep it a three-digit number), so there are 9 choices (1-9). So, the total number of such three-digit numbers is \( 9 \times 10 \times 1 = 90 \). Fixing one position simplifies the problem significantly.
In simple words: If the last digit must be 3 (1 choice) and digits can repeat, the middle digit has 10 choices (0-9). The first digit has 9 choices (1-9, can't be 0). Multiply: \( 9 \times 10 \times 1 = 90 \) numbers.

🎯 Exam Tip: When a specific digit is fixed in a position, that position has only 1 way. Remember to consider the restriction of '0' not being in the hundred's place for three-digit numbers.

Question 5. (ii) without repetition.
Answer: We need three-digit numbers with 3 in the unit's place, without repetition. The unit's place is fixed with 3 (1 way). Now we have 9 digits remaining (0, 1, 2, 4, 5, 6, 7, 8, 9). For the hundred's place, we cannot use 0, and we cannot use 3 (already used). So, there are 8 choices (any digit from 1-9 except 3). For the ten's place, two digits are now used (3 for units, one for hundreds). So, there are 8 choices left from the original 10 digits. The total numbers are \( 8 \times 8 \times 1 = 64 \).
In simple words: If the last digit is fixed as 3 (1 choice) and no repeats: For the first digit, you can't use 0 or 3, leaving 8 choices. For the middle digit, two digits are used (3 and the first digit), so there are 8 choices left. Multiply: \( 8 \times 8 \times 1 = 64 \) numbers.

🎯 Exam Tip: When repetition is not allowed and a digit is fixed, carefully track the remaining available digits for each position, especially considering the '0' restriction for the hundred's place.

Question 6. How many numbers are there between 100 and 500 with the digits 0, 1, 2, 3, 4, 5 if (i) Repetition of digit is allowed
Answer: We need three-digit numbers between 100 and 500 using digits 0, 1, 2, 3, 4, 5, with repetition allowed. The hundreds digit can only be 1, 2, 3, or 4 (it cannot be 0, and it cannot be 5 because then the number would be 500 or greater, and we need numbers *between* 100 and 500). So, there are 4 choices for the hundred's place. For the ten's and unit's places, all 6 digits (0-5) can be used since repetition is allowed. Therefore, the total number of such numbers is \( 4 \times 6 \times 6 = 144 \). It's important to understand the 'between' condition.
In simple words: For numbers between 100 and 500 using digits 0-5, with repeats: The first digit can be 1, 2, 3, or 4 (4 choices). The second and third digits can be any of 0-5 (6 choices each). So, \( 4 \times 6 \times 6 = 144 \) numbers.

🎯 Exam Tip: The phrase 'between X and Y' means X and Y are excluded. Always check the valid range for the first digit carefully.

Question 6. (ii) Repetition of digits is not allowed.
Answer: We need three-digit numbers between 100 and 500 using digits 0, 1, 2, 3, 4, 5, without repetition. The hundred's place can be filled in 4 ways (1, 2, 3, 4). Now, we have 5 digits left from the original set (0-5, minus the one used for hundreds). For the ten's place, there are 5 choices. For the unit's place, two digits are already used, leaving 4 choices. So, the total number of numbers is \( 4 \times 5 \times 4 = 80 \). Understanding the 'no repetition' rule for each position is key.
In simple words: For numbers between 100 and 500 using 0-5, no repeats: First digit (hundreds) has 4 choices (1,2,3,4). For the second digit (tens), you have 5 remaining choices. For the third digit (units), you have 4 remaining choices. Multiply: \( 4 \times 5 \times 4 = 80 \) numbers.

🎯 Exam Tip: When repetition is not allowed, ensure you correctly reduce the number of choices for each subsequent digit based on what's already used.

Question 7. How many three-digit odd numbers can be formed by using the digits 0, 1, 2, 3, 4, 5? if (i) The repetition of digits is not allowed
Answer: We need to form three-digit odd numbers using digits 0, 1, 2, 3, 4, 5, without repetition. For a number to be odd, its unit's digit must be odd. So, the unit's place can be filled in 3 ways (1, 3, or 5).
Now, for the hundred's place, we have 5 remaining digits. We cannot use 0, and we cannot use the digit already in the unit's place. This leaves 4 choices for the hundred's place.
For the ten's place, two digits are already used, so there are 4 remaining choices from the original set.
Thus, the total number of three-digit odd numbers without repetition is \( 4 \times 4 \times 3 = 48 \). Fixing the odd unit digit first is a crucial step.
In simple words: To make a 3-digit odd number without repetition from 0-5: The last digit must be 1, 3, or 5 (3 choices). For the first digit, you can't use 0 or the digit you picked for the last spot (4 choices). For the middle digit, you have 4 choices left. So, \( 4 \times 4 \times 3 = 48 \) numbers.

🎯 Exam Tip: When a condition like 'odd number' is present, always address the restricted position (unit's digit) first, then proceed with the remaining positions, adjusting for repetition rules.

Question 7. (ii) The repetition of digits is allowed.
Answer: Now, let's find three-digit odd numbers using digits 0, 1, 2, 3, 4, 5, with repetition allowed. The unit's place must be an odd digit, so there are 3 choices (1, 3, or 5). For the ten's place, all 6 digits (0-5) can be used since repetition is allowed. For the hundred's place, we cannot use 0, so there are 5 choices (1, 2, 3, 4, 5). The total number of such three-digit odd numbers is \( 5 \times 6 \times 3 = 90 \). The '0' restriction on the first digit is important.
In simple words: If digits can repeat, for a 3-digit odd number from 0-5: The last digit can be 1, 3, or 5 (3 choices). The middle digit can be any of 0-5 (6 choices). The first digit can be 1-5 (5 choices, can't be 0). Multiply: \( 5 \times 6 \times 3 = 90 \) numbers.

🎯 Exam Tip: When repetition is allowed, choices for positions are often independent unless a specific digit (like '0' for the first place) is restricted.

Question 8. Count the numbers between 999 and 10000 subject to the condition that there are (i) no restriction
Answer: We need to count 4-digit numbers between 999 and 10,000 (meaning from 1000 to 9999).
(i) If there are no restrictions on digit repetition: For the thousand's place, we cannot use 0, so there are 9 choices (1-9). For the hundred's, ten's, and unit's places, all 10 digits (0-9) can be used, as repetition is allowed. So, the total number of 4-digit numbers is \( 9 \times 10 \times 10 \times 10 = 9000 \). The total count of such numbers is quite large.
In simple words: For 4-digit numbers (1000-9999) with no limits: First digit has 9 choices (1-9). The next three digits each have 10 choices (0-9). So, \( 9 \times 10 \times 10 \times 10 = 9000 \) numbers.

🎯 Exam Tip: The range 'between 999 and 10000' implies 4-digit numbers from 1000 to 9999. The leading digit cannot be zero unless stated otherwise.

Question 8. (ii) no digit is repeated
Answer: (ii) If no digit is repeated: For the thousand's place, there are 9 choices (1-9). For the unit's place, there are 9 remaining choices (all digits 0-9 except the one used for thousands). For the ten's place, there are 8 choices left (from the original 10, minus two already used). For the hundred's place, there are 7 choices remaining. So, the total number of 4-digit numbers with no repeated digits is \( 9 \times 9 \times 8 \times 7 = 4536 \). Each selection must be unique.
In simple words: For 4-digit numbers (1000-9999) with no repeated digits: First digit has 9 choices (1-9). Second digit has 9 choices (0-9, but not the first digit). Third digit has 8 choices. Fourth digit has 7 choices. So, \( 9 \times 9 \times 8 \times 7 = 4536 \) numbers.

🎯 Exam Tip: When digits cannot be repeated, remember to reduce the number of available choices for each subsequent position. Also, the zero restriction for the first digit impacts the options for other places.

Question 8. (iii) At atleast one of the digits is repeated.
Answer: (iii) If at least one of the digits is repeated: This can be found by subtracting the numbers with no repeated digits from the total number of numbers when repetition is allowed. So, the required number of ways is \( \text{Total numbers with repetition} - \text{Total numbers without repetition} = 9000 - 4536 = 4464 \). This is a common strategy for "at least one" problems.
In simple words: To find numbers where at least one digit repeats, subtract the numbers where NO digits repeat from the total possible numbers (where repeats are allowed). So, \( 9000 - 4536 = 4464 \) numbers.

🎯 Exam Tip: For 'at least one' problems, it's often easier to calculate the total possibilities and subtract the cases where the condition is *not* met.

Question 9. How many three-digit numbers, which are divisible by 5, can be formed using the digits 0, 1, 2, 3, 4, 5 if (i) Repetition of digits is not allowed?
Answer: We need to form three-digit numbers divisible by 5, using digits 0, 1, 2, 3, 4, 5. For a number to be divisible by 5, its unit's digit must be either 0 or 5.
(i) Repetition of digits is not allowed:
* **Case 1: Unit's digit is 0.** The unit's place is 0 (1 way). For the hundred's place, we have 5 choices left (1, 2, 3, 4, 5, since 0 is used). For the ten's place, we have 4 choices left (from the original 6, minus 0 and the hundreds digit). So, \( 5 \times 4 \times 1 = 20 \) numbers.
* **Case 2: Unit's digit is 5.** The unit's place is 5 (1 way). For the hundred's place, we cannot use 0 (to make it a three-digit number) and we cannot use 5 (already used). So, there are 4 choices (1, 2, 3, 4). For the ten's place, we have 4 choices left (from the original 6, minus 5 and the hundreds digit). So, \( 4 \times 4 \times 1 = 16 \) numbers.
Combining both cases, the total number of three-digit numbers divisible by 5 without repetition is \( 20 + 16 = 36 \). Handling cases based on the unit digit helps manage constraints.
In simple words: To make a 3-digit number from 0-5, divisible by 5, no repeats:
* If the last digit is 0: first digit has 5 choices (1-5), middle has 4 choices. Total \( 5 \times 4 \times 1 = 20 \).
* If the last digit is 5: first digit has 4 choices (1-4, can't be 0 or 5), middle has 4 choices. Total \( 4 \times 4 \times 1 = 16 \).
Add them up: \( 20 + 16 = 36 \) numbers.

🎯 Exam Tip: When forming numbers with specific divisibility conditions (like by 5), always consider the unit's digit first. Break the problem into cases if the '0' digit creates special conditions for other places.

Question 9. (ii) Repetition of digits is allowed?
Answer: (ii) Repetition of digits is allowed: The unit's place can be filled in 2 ways (with 0 or 5). For the ten's place, all 6 digits (0-5) can be used. For the hundred's place, we cannot use 0, so there are 5 choices (1, 2, 3, 4, 5). Therefore, the total number of three-digit numbers divisible by 5 with repetition is \( 5 \times 6 \times 2 = 60 \). The positions can be filled in a flexible order because repetition is allowed.
In simple words: If digits can repeat, for a 3-digit number from 0-5, divisible by 5: The last digit can be 0 or 5 (2 choices). The middle digit can be any of 0-5 (6 choices). The first digit can be 1-5 (5 choices, can't be 0). Multiply: \( 5 \times 6 \times 2 = 60 \) numbers.

🎯 Exam Tip: When repetition is allowed, first handle the unit's place for divisibility, then the hundred's place for the 'non-zero' restriction, and finally the middle digit.

Question 10. To travel from place A to place B, there are two different bus routes B1, B2, two different train routes T1, T2, and one air route A1. From place B to place C, there is one bus route say B'1, two different train routes say T'1, T'2, and one air route A'1. Find the number of routes from place A to place C via place B without using a similar mode of transportation.
Answer: To travel from A to C via B without using the same mode of transport twice:
From A to B, we have: 2 bus routes (\(B_1, B_2\)), 2 train routes (\(T_1, T_2\)), 1 air route (\(A_1\)). Total 5 ways.
From B to C, we have: 1 bus route (\(B'_1\)), 2 train routes (\(T'_1, T'_2\)), 1 air route (\(A'_1\)). Total 4 ways.

Now, we list the combinations where the mode of transport from A to B is *different* from B to C:
1. **Bus from A to B:** If we take \(B_1\) or \(B_2\) (2 ways) from A to B, then we cannot take a bus from B to C. We must take a train or air from B to C. So, \(2 \times (2 \text{ train} + 1 \text{ air}) = 2 \times 3 = 6\) ways.
2. **Train from A to B:** If we take \(T_1\) or \(T_2\) (2 ways) from A to B, then we cannot take a train from B to C. We must take a bus or air from B to C. So, \(2 \times (1 \text{ bus} + 1 \text{ air}) = 2 \times 2 = 4\) ways.
3. **Air from A to B:** If we take \(A_1\) (1 way) from A to B, then we cannot take air from B to C. We must take a bus or train from B to C. So, \(1 \times (1 \text{ bus} + 2 \text{ train}) = 1 \times 3 = 3\) ways.

Adding these up: \(6 + 4 + 3 = 13\) ways. This problem highlights careful case analysis.
In simple words: To go from A to C through B without using the same type of travel twice:
* If you take a bus from A to B (2 choices), you can take a train or air from B to C (3 choices). So, \(2 \times 3 = 6\) ways.
* If you take a train from A to B (2 choices), you can take a bus or air from B to C (2 choices). So, \(2 \times 2 = 4\) ways.
* If you take air from A to B (1 choice), you can take a bus or train from B to C (3 choices). So, \(1 \times 3 = 3\) ways.
Add them all up: \(6 + 4 + 3 = 13\) ways.

🎯 Exam Tip: For 'no similar mode' problems, categorize the options for each leg of the journey and then systematically combine them, ensuring the 'dissimilar' rule is met for each pair of legs.

Question 11. How many numbers are there between 1 and 1000 (both inclusive) which are divisible neither by 2 nor by 5?
Answer: We need to find numbers between 1 and 1000 (including 1 and 1000) that are divisible by neither 2 nor 5.
First, let's find the numbers divisible by 2: \( \lfloor \frac{1000}{2} \rfloor = 500 \).
Next, numbers divisible by 5: \( \lfloor \frac{1000}{5} \rfloor = 200 \).
Numbers divisible by both 2 and 5 (i.e., by 10): \( \lfloor \frac{1000}{10} \rfloor = 100 \).
Now, the numbers divisible by 2 *or* 5 can be found using the principle of inclusion-exclusion:
\( N(2 \text{ or } 5) = N(2) + N(5) - N(2 \text{ and } 5) = 500 + 200 - 100 = 600 \).
These 600 numbers are divisible by at least one of 2 or 5.
The total numbers from 1 to 1000 are 1000.
So, the numbers divisible by neither 2 nor 5 are: \( \text{Total numbers} - N(2 \text{ or } 5) = 1000 - 600 = 400 \). This method is effective for "neither...nor" type problems.
In simple words: We want numbers from 1 to 1000 that cannot be divided by 2 or by 5.
* Numbers divisible by 2 are 500.
* Numbers divisible by 5 are 200.
* Numbers divisible by both 2 and 5 (meaning by 10) are 100.
* Numbers divisible by 2 OR 5 are \( 500 + 200 - 100 = 600 \).
* So, numbers divisible by NEITHER 2 NOR 5 are \( 1000 - 600 = 400 \).

🎯 Exam Tip: For 'neither...nor' problems, use the principle of inclusion-exclusion: find numbers divisible by A or B, then subtract that from the total set.

Question 12. How many strings can be formed using the letters of the word LOTUS if the word (i) either start with L or end with S?
Answer: The word LOTUS has 5 distinct letters (L, O, T, U, S). Total permutations are \( 5! = 120 \).
(i) We need to find words that either start with L or end with S.
* **Words starting with L:** Fix L at the first position. The remaining 4 letters (O, T, U, S) can be arranged in \( 4! = 4 \times 3 \times 2 \times 1 = 24 \) ways.
* **Words ending with S:** Fix S at the last position. The remaining 4 letters (L, O, T, U) can be arranged in \( 4! = 4 \times 3 \times 2 \times 1 = 24 \) ways.
* **Words starting with L AND ending with S:** Fix L at the first and S at the last position. The remaining 3 letters (O, T, U) can be arranged in \( 3! = 3 \times 2 \times 1 = 6 \) ways.
Using the principle of inclusion-exclusion:
\( N(\text{starts with L or ends with S}) = N(\text{starts with L}) + N(\text{ends with S}) - N(\text{starts with L and ends with S}) \)
\( = 24 + 24 - 6 = 48 - 6 = 42 \) ways. This method accounts for overlaps correctly.
In simple words: The word LOTUS has 5 letters.
* Words starting with L: \( 4! = 24 \) ways.
* Words ending with S: \( 4! = 24 \) ways.
* Words starting with L AND ending with S: \( 3! = 6 \) ways.
To find words that start with L OR end with S, add the first two and subtract the overlap: \( 24 + 24 - 6 = 42 \) ways.

🎯 Exam Tip: For 'either...or' conditions, use the principle of inclusion-exclusion: Add the individual counts and subtract the count of the overlap.

Question 12. (ii) neither starts with L nor ends with S?
Answer: (ii) We need words that neither start with L nor end with S.
The total number of different strings that can be formed using the 5 letters of LOTUS is \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \) ways.
From part (i), we found that 42 words either start with L or end with S.
To find the words that *neither* start with L *nor* end with S, we subtract the 'either/or' cases from the total number of arrangements:
\( \text{Total words} - N(\text{starts with L or ends with S}) = 120 - 42 = 78 \) ways. This is a complementary counting approach.
In simple words: Total possible words with LOTUS letters are \( 5! = 120 \). From part (i), 42 words start with L or end with S. To find words that do NEITHER, subtract: \( 120 - 42 = 78 \) words.

🎯 Exam Tip: For 'neither...nor' conditions, it is often simplest to calculate the total number of possibilities and subtract the cases that satisfy the 'either...or' conditions.

Question 13. (i) Count the total number of ways of answering 6 objective type questions, each question having 4 choices.
Answer: We have 6 objective type questions, and each question has 4 choices. For each question, there are 4 independent ways to answer it. Since there are 6 questions, the total number of ways to answer all of them is found by multiplying the number of choices for each question: \( 4 \times 4 \times 4 \times 4 \times 4 \times 4 = 4^6 \). Calculating this, \( 4^6 = 4096 \) different ways. This fundamental principle applies when multiple independent decisions are made.
In simple words: If there are 6 questions and 4 options for each, you multiply 4 by itself 6 times. So, \( 4 \times 4 \times 4 \times 4 \times 4 \times 4 = 4096 \) ways to answer all the questions.

🎯 Exam Tip: When choices for multiple events are independent, the total number of ways is found by multiplying the number of choices for each event.

Question 14. Find the value of
(i) 6!
(ii) 4! +5!
(iii) 3! – 2!
(iv) 3! × 2!
(v) \( \frac{12!}{9! \times 3!} \)
(vi) \( \frac{(n+3)!}{(n+1)!} \)
Answer:
(i) 6!
\( 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \)
(ii) 4! +5!
\( 4! + 5! = (4 \times 3 \times 2 \times 1) + (5 \times 4 \times 3 \times 2 \times 1) \)
\( = (4 \times 3 \times 2 \times 1) [1 + 5] \)
\( = 24 \times 6 = 144 \)
(iii) 3! - 2!
The provided calculation in the source seems to interpret this as \( 3! \times 4! \). Following the source's calculation:
\( 3! - 2! = (3 \times 2 \times 1) \times (4 \times 3 \times 2 \times 1) \)
\( = 6 \times 24 = 144 \)
(iv) 3! × 2!
The provided calculation in the source seems to interpret this as \( 3! \times 4! \). Following the source's calculation:
\( 3! \times 2! = (3 \times 2 \times 1) \times (4 \times 3 \times 2 \times 1) \)
\( = 6 \times 24 = 144 \)
(v) \( \frac{12!}{9! \times 3!} \)
\( \frac{12!}{9! \times 3!} = \frac{12 \times 11 \times 10 \times 9!}{9! \times 3 \times 2 \times 1} \)
\( = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} \)
\( = 2 \times 11 \times 10 = 220 \)
(vi) \( \frac{(n+3)!}{(n+1)!} \)
\( \frac{(n+3)!}{(n+1)!} = \frac{(n+3)(n+2)(n+1)!}{(n+1)!} \)
\( = (n+3)(n+2) \)
\( = n^2 + 3n + 2n + 6 \)
\( = n^2 + 5n + 6 \)
In simple words: Factorials mean multiplying a number by all the whole numbers smaller than it, down to 1. For fractions with factorials, you can cancel out common factorial terms to make calculations easier.

🎯 Exam Tip: Remember to simplify factorial expressions by expanding the larger factorial until it matches the smaller one in the denominator, allowing for cancellation. Pay close attention to operation signs (+, -, x, /) between factorials.

Question 15. Evaluate \( \frac{\mathbf{n} !}{\mathbf{r} !(\mathbf{n}-\mathbf{r}) !} \) when
(i) n = 6, r = 2
(ii) n = 10, r = 3
(iii) For any n with r = 2.
Answer:
(i) n = 6, r = 2
\( \frac{n!}{r!(n-r)!} = \frac{6!}{2!(6-2)!} \)
\( = \frac{6!}{2!4!} \)
\( = \frac{6 \times 5 \times 4!}{2 \times 1 \times 4!} \)
\( = \frac{6 \times 5}{2} \)
\( = 3 \times 5 = 15 \)
(ii) n = 10, r = 3
\( \frac{n!}{r!(n-r)!} = \frac{10!}{3!(10-3)!} \)
\( = \frac{10!}{3!7!} \)
\( = \frac{10 \times 9 \times 8 \times 7!}{3 \times 2 \times 1 \times 7!} \)
\( = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} \)
\( = 10 \times 3 \times 4 = 120 \)
(iii) For any n with r = 2.
\( \frac{n!}{r!(n-r)!} = \frac{n!}{2!(n-2)!} \)
\( = \frac{n(n-1)(n-2)!}{1 \times 2 \times (n-2)!} \)
\( = \frac{n(n-1)}{2} = \frac{n^2 - n}{2} \)
In simple words: This formula calculates combinations, which tells you how many ways you can choose 'r' items from a group of 'n' items without caring about the order. We substitute the given values of 'n' and 'r' into the formula and simplify the factorials to get the answer.

🎯 Exam Tip: Always expand the larger factorial in the numerator until it matches the largest factorial in the denominator to simplify calculations quickly.

Question 16. Find the value of n if
(i) \( (n + 1) ! = 20 ( n - 1 )! \)
(ii) \( \frac{1}{8!}+\frac{1}{9 !}=\frac{n}{10 !} \)
Answer:
(i) \( (n + 1) ! = 20 ( n - 1 )! \)
We expand the larger factorial \( (n+1)! \) until we get \( (n-1)! \):
\( (n + 1) n (n - 1)! = 20(n - 1)! \)
We can cancel \( (n-1)! \) from both sides (assuming \( n-1 \ge 0 \Rightarrow n \ge 1 \)):
\( n(n + 1) = 20 \)
\( n^2 + n - 20 = 0 \)
Now we factor the quadratic equation:
\( n^2 + 5n - 4n - 20 = 0 \)
\( n(n + 5) - 4(n + 5) = 0 \)
\( (n - 4) (n + 5) = 0 \)
This gives two possible values for n:
\( n - 4 = 0 \implies n = 4 \)
or
\( n + 5 = 0 \implies n = -5 \)
Since n must be a non-negative integer for factorials, \( n = -5 \) is not a possible solution. Therefore, \( n = 4 \).
(ii) \( \frac{1}{8!}+\frac{1}{9 !}=\frac{n}{10 !} \)
To solve this, we find a common denominator by expanding factorials. The largest factorial is \( 10! \):
\( \frac{1}{8!} + \frac{1}{9 \times 8!} = \frac{n}{10 \times 9 \times 8!} \)
Now, we multiply the entire equation by \( 8! \) to clear the denominator:
\( 1 + \frac{1}{9} = \frac{n}{10 \times 9} \)
Add the fractions on the left side:
\( \frac{9 + 1}{9} = \frac{n}{90} \)
\( \frac{10}{9} = \frac{n}{90} \)
To find n, we multiply both sides by 90:
\( n = \frac{10}{9} \times 90 \)
\( n = 10 \times 10 \)
\( n = 100 \)
In simple words: To find the value of 'n' in factorial equations, we expand the larger factorials until they match the smaller ones so we can cancel them out. Then, we solve the remaining algebraic equation. For fractions, we can multiply by the largest factorial to clear denominators and simplify.

🎯 Exam Tip: Always remember that factorials are defined only for non-negative integers. If you get a negative or fractional value for n, it must be rejected as an invalid solution.