Question 1

In how many ways can the letters of the word 'INDEPENDENCE' be arranged such that all the vowels appear together?

Accepted Answer

Correct Option: C. Solution: Treat the vowels 'I', 'E', 'E', 'E', 'E' as a single entity. This gives us 8 entities to arrange: 'N', 'D', 'P', 'N', 'D', 'C', the vowel block, and 'E'. The number of ways to arrange these is $$\frac{8!}{3!2!}$$. The vowels within the block can be arranged in $$\frac{5!}{4!}$$ ways. Thus, the total number of arrangements is $$\frac{8!}{3!2!} 	imes \frac{5!}{4!} = 16800$$.

Question 2

How many different 3-digit sequences can be formed using the digits 0-9 if the first digit is fixed as 7 and no digit repeats?

Accepted Answer

Correct Option: C. Solution: With the first digit fixed as 7, the remaining 3 digits can be chosen from 9 digits (0-9 excluding 7), and arranged in $9P3 = \frac{9!}{(9-3)!} = 9 	imes 8 	imes 7 = 504$ ways.

Question 3

Find the number of permutations of the letters in the word 'ROOT' where the two O's are indistinguishable.

Accepted Answer

Correct Option: A. Solution: The number of permutations of the letters in 'ROOT' is $\frac{4!}{2!} = \frac{24}{2} = 12$ because the two O's are indistinguishable.

Question 4

If the word 'ROOT' is rearranged, how many distinct permutations can be formed?

Accepted Answer

Correct Option: A. Solution: The word 'ROOT' has 4 letters with 2 'O's. The number of distinct permutations is $\frac{4!}{2!} = \frac{24}{2} = 12$.

Question 5

What is the number of ways to select 4 cards from a standard deck of 52 cards such that all four cards belong to different suits?

Accepted Answer

Correct Option: B. Solution: The number of ways to select 4 cards from different suits is $$\binom{13}{1}^4 = 13 	imes 13 	imes 13 	imes 13 = 28561$$. However, since we need exactly one card from each suit, the correct calculation is $$13 	imes 12 	imes 11 	imes 10 = 17160$$ divided by the permutations of the suits, which is 24, leading to $$17160 / 24 = 715$$.

Question 6

If Mohan has 3 pants and 2 shirts, how many different pairs of a pant and a shirt can he dress up with?

Accepted Answer

Correct Option: C. Solution: According to the Fundamental Principle of Counting, for every choice of a pant, there are 2 choices of a shirt. Therefore, there are $3 	imes 2 = 6$ pairs.

Question 7

In how many ways can a committee of 3 persons be formed from a group of 2 men and 3 women if the committee must consist of 1 man and 2 women?

Accepted Answer

Correct Option: C. Solution: 1 man can be selected from 2 men in $\binom{2}{1}$ ways and 2 women can be selected from 3 women in $\binom{3}{2}$ ways. So, the number of ways is $\binom{2}{1} 	imes \binom{3}{2} = 2 	imes 3 = 6$.

Question 8

How many distinct permutations can be formed from the letters of the word 'MISSISSIPPI' such that no two 'I's are adjacent?

Accepted Answer

Correct Option: B. Solution: The total number of permutations of the letters in 'MISSISSIPPI' is given by $$\frac{11!}{4!4!2!}$$. To find permutations where no two 'I's are adjacent, we first arrange the letters 'M', 'S', 'S', 'S', 'S', 'P', 'P', which can be done in $$\frac{7!}{4!2!}$$ ways. There are 8 gaps (including ends) to place the 'I's, which can be done in $$\binom{8}{4}$$ ways. Thus, the total number is $$\frac{7!}{4!2!} 	imes \binom{8}{4} = 3460$$.

Question 9

How many 4-digit numbers can be formed using the digits 1 to 9 if repetition of digits is not allowed?

Accepted Answer

Correct Option: A. Solution: The number of 4-digit numbers that can be formed from 9 different digits is given by the permutations of 9 taken 4 at a time, which is $$9P4 = \frac{9!}{(9-4)!} = 9 	imes 8 	imes 7 	imes 6 = 3024$$.

Question 10

How many different 4-letter words can be formed using the letters of the word 'ROSE' if repetition of letters is not allowed?

Accepted Answer

Correct Option: B. Solution: There are 4 letters in 'ROSE'. The number of permutations of 4 different letters taken all at a time is $4! = 24$.

Question 11

If $n! = 40320$, what is the value of $n$?

Accepted Answer

Correct Option: B. Solution: The value of $8!$ is $1 	imes 2 	imes 3 	imes 4 	imes 5 	imes 6 	imes 7 	imes 8 = 40320$. Therefore, $n = 8$.

Question 12

In how many ways can a lock with 4 wheels, each labeled with digits 0 to 9, be opened if the first digit is fixed as 7 and no digit repeats?

Accepted Answer

Correct Option: A. Solution: With the first digit fixed as 7, there are 9 remaining digits. The number of permutations of 9 digits taken 3 at a time is $9P3 = \frac{9!}{(9-3)!} = 9 	imes 8 	imes 7 = 504$.

Question 13

In a lock with 4 wheels each labeled with 10 digits (0-9), if the first digit is fixed as 7, how many sequences of 3 digits need to be checked to open the lock?

Accepted Answer

Correct Option: A. Solution: Since the first digit is fixed, there are 9 remaining digits. The number of sequences of 3 digits is given by $9 	imes 8 	imes 7 = 504$.

Question 14

In how many ways can 4 red, 3 yellow, and 2 green discs be arranged in a row if the discs of the same color are indistinguishable?

Accepted Answer

Correct Option: A. Solution: The total number of discs is 9. The number of arrangements is given by $\frac{9!}{4!3!2!} = 1260$.

Question 15

How many different ways can Sabnam carry her items if she has 2 school bags, 3 tiffin boxes, and 2 water bottles, choosing one of each?

Accepted Answer

Correct Option: A. Solution: A school bag can be chosen in 2 ways, a tiffin box in 3 ways, and a water bottle in 2 ways. Therefore, the total number of ways is $2 	imes 3 	imes 2 = 12$.

Question 16

What is the formula for the number of permutations of $n$ different objects taken $r$ at a time, where $0 < r \leq n$ and the objects do not repeat?

Accepted Answer

Correct Option: A. Solution: The formula for permutations of $n$ different objects taken $r$ at a time is $nPr = \frac{n!}{(n-r)!}$.

Question 17

How many different signals can be generated by arranging at least 2 flags in order on a vertical staff, if five different flags are available?

Accepted Answer

Correct Option: A. Solution: Signals can consist of 2, 3, 4, or 5 flags. The number of 2-flag signals is $$5 	imes 4 = 20$$, 3-flag signals is $$5 	imes 4 	imes 3 = 60$$, 4-flag signals is $$5 	imes 4 	imes 3 	imes 2 = 120$$, and 5-flag signals is $$5! = 120$$. Thus, the total number of signals is $$20 + 60 + 120 + 120 = 320$$.

Question 18

A committee of 5 members is to be formed from a group of 4 men and 5 women. If the committee must include at least 2 men, how many different committees can be formed?

Accepted Answer

Correct Option: B. Solution: To form a committee with at least 2 men, we consider the following cases: (i) 2 men and 3 women: $$\binom{4}{2} 	imes \binom{5}{3} = 6 	imes 10 = 60$$ (ii) 3 men and 2 women: $$\binom{4}{3} 	imes \binom{5}{2} = 4 	imes 10 = 40$$ Total ways = 60 + 40 = 100.

Question 19

Find the number of distinct permutations of the word 'ALLAHABAD'.

Accepted Answer

Correct Option: A. Solution: The word 'ALLAHABAD' consists of 9 letters where 'A' appears 4 times and 'L' appears 2 times. The number of distinct permutations is given by $$\frac{9!}{4!2!} = 7560$$.

Question 20

If $n! = n 	imes (n-1)!$, what is the value of $7! - 5!$?

Accepted Answer

Correct Option: A. Solution: The value of $7!$ is $5040$ and $5!$ is $120$. Therefore, $7! - 5! = 5040 - 120 = 4920$.

Question 21

In a club, there are 7 members. A team of 4 members is to be selected such that no two members in the team are consecutive in the seating arrangement. How many such teams can be formed?

Accepted Answer

Correct Option: B. Solution: Consider the seating arrangement as a circle. Choose any member as a starting point. The next member can be chosen from the remaining 5 members, skipping one each time. This results in $$\binom{5}{3} = 10$$ ways. However, since the circle can be rotated, we divide by 2 to avoid identical rotations, resulting in 5 unique ways.

Question 22

Find the number of permutations of the letters of the word 'ALLAHABAD'.

Accepted Answer

Correct Option: A. Solution: The word 'ALLAHABAD' has 9 letters with 4 A's and 2 L's. The number of permutations is $\frac{9!}{4!2!} = 7560$.

Question 23

How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

Accepted Answer

Correct Option: A. Solution: To form a 3-digit even number, the unit's place must be filled with an even digit (2, 4, or 6). This can be done in 3 ways. The remaining two places can be filled by any of the remaining 5 digits in $5 	imes 4 = 20$ ways. Therefore, the total number of ways is $3 	imes 20 = 60$.

Question 24

How many different signals can be generated using 3 flags from a set of 5 different colored flags?

Accepted Answer

Correct Option: C. Solution: The number of ways to arrange 3 flags out of 5 is given by permutations: $$5P3 = \frac{5!}{(5-3)!} = 5 	imes 4 	imes 3 = 60$$.

Question 25

How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, and 5 if repetition of digits is allowed?

Accepted Answer

Correct Option: D. Solution: Since repetition is allowed, each digit can be chosen in 5 ways. Therefore, the total number of 3-digit numbers is $5^3 = 125$.

Question 26

How many 3-digit even numbers can be formed using the digits 1, 2, 3, 4, 5, 6 if digits can be repeated?

Accepted Answer

Correct Option: C. Solution: For an even number, the unit's place can be filled by 2, 4, or 6 (3 ways). The other two places can be filled by any of the 6 digits (6 ways each). Thus, the total number of ways is $3 	imes 6 	imes 6 = 108$.

Question 27

A committee of 3 persons is to be formed from a group of 2 men and 3 women. How many such committees can be formed if it must include at least 1 man?

Accepted Answer

Correct Option: B. Solution: Total ways to form a committee of 3 from 5 people is $$\binom{5}{3} = 10$$. The number of committees with no men (all women) is $$\binom{3}{3} = 1$$. Thus, the number of committees with at least one man is $$10 - 1 = 9$$.

Question 28

Mohan has 3 pants and 2 shirts. Using the fundamental principle of counting, how many different pairs of a pant and a shirt can he create?

Accepted Answer

Correct Option: B. Solution: For each pant, there are 2 choices of shirts. Thus, the number of pairs is $3 	imes 2 = 6$.

Question 29

How many ways can a committee of 3 people be formed from a group of 5 people?

Accepted Answer

Correct Option: A. Solution: The number of ways to form a committee of 3 people from a group of 5 is given by $5C3 = \frac{5!}{3!(5-3)!} = 10$.

Question 30

In how many ways can a committee of 3 be formed from a group of 2 men and 3 women such that the committee consists of 1 man and 2 women?

Accepted Answer

Correct Option: A. Solution: Select 1 man from 2 men in $\binom{2}{1} = 2$ ways and 2 women from 3 women in $\binom{3}{2} = 3$ ways. Thus, the total number of ways is $2 	imes 3 = 6$.

Question 31

How many permutations can be made from the letters of the word 'ROOT'?

Accepted Answer

Correct Option: B. Solution: The word 'ROOT' has 4 letters where 'O' is repeated twice. The number of permutations is given by $\frac{4!}{2!} = 12$.

Question 32

How many different signals can be generated using 4 flags of different colors, if each signal requires the use of 2 flags one below the other?

Accepted Answer

Correct Option: B. Solution: The number of permutations of 4 flags taken 2 at a time is $4P2 = \frac{4!}{(4-2)!} = 4 	imes 3 = 12$.

Question 33

In how many ways can a 4-digit number be formed using the digits 1 to 9 if repetition is not allowed?

Accepted Answer

Correct Option: A. Solution: Since repetition is not allowed, the number of 4-digit numbers is given by the permutation formula $9P4 = \frac{9!}{(9-4)!} = 9 	imes 8 	imes 7 	imes 6 = 3024$.

Question 34

How many different 5-digit numbers can be formed using the digits 0 to 9 if each number must start with a 7 and digits can be repeated?

Accepted Answer

Correct Option: A. Solution: Since the number must start with 7, the first digit is fixed. For the remaining 4 digits, each can be any of the 10 digits (0-9). Therefore, the number of such numbers is $10^4 = 10,000$.

Question 35

Sabnam has 2 school bags, 3 tiffin boxes, and 2 water bottles. How many ways can she carry these items, choosing one of each?

Accepted Answer

Correct Option: A. Solution: The number of ways to choose one school bag, one tiffin box, and one water bottle is $2 	imes 3 	imes 2 = 12$.

Question 36

How many different signals can be generated by arranging at least 2 flags in order on a vertical staff, if five different flags are available?

Accepted Answer

Correct Option: B. Solution: The signals can consist of 2, 3, 4, or 5 flags. The number of 2-flag signals is $5 	imes 4 = 20$, 3-flag signals is $5 	imes 4 	imes 3 = 60$, 4-flag signals is $5 	imes 4 	imes 3 	imes 2 = 120$, and 5-flag signals is $5! = 120$. Total signals = $20 + 60 + 120 + 120 = 320$.

Question 37

What is the value of $5!$ (factorial of 5)?

Accepted Answer

Correct Option: B. Solution: The factorial of 5, denoted as $5!$, is calculated as $5 	imes 4 	imes 3 	imes 2 	imes 1 = 120$.

Question 38

If $nC3 = 10$, what is the value of $n$?

Accepted Answer

Correct Option: A. Solution: Using the combination formula $nC3 = \frac{n!}{3!(n-3)!} = 10$, solving for $n$ gives $n = 5$.

Question 39

In how many ways can a 4-digit number be formed using the digits 1 to 9 if no digit is repeated?

Accepted Answer

Correct Option: A. Solution: The number of 4-digit numbers is given by the permutation of 9 digits taken 4 at a time: $$^9P_4 = \frac{9!}{(9-4)!} = 9 	imes 8 	imes 7 	imes 6 = 3024$$.

Question 40

A lock has 4 wheels each labeled with digits from 0 to 9. If the first digit is fixed as 7, how many different sequences of the remaining 3 digits can be formed without repetition?

Accepted Answer

Correct Option: A. Solution: Since the first digit is fixed as 7, we have 9 remaining digits to choose from for the next 3 positions. The number of sequences is calculated as $9 	imes 8 	imes 7 = 504$.

Question 41

In how many ways can the letters of the word 'DAUGHTER' be arranged such that all vowels occur together?

Accepted Answer

Correct Option: A. Solution: Treat the vowels A, U, E as a single entity. This gives us 6 entities to arrange: (AUE), D, G, H, T, R. The number of arrangements of these 6 entities is $$6! = 720$$. The vowels can be arranged among themselves in $$3! = 6$$ ways. Thus, the total number of arrangements is $$6! 	imes 3! = 4320$$.

Question 42

If a group of 7 people is to be formed from 5 men and 4 women, how many ways can the group be formed such that there are exactly 3 women?

Accepted Answer

Correct Option: A. Solution: To form a group of 7 people with exactly 3 women, we choose 3 women from 4 and 4 men from 5. The number of ways to do this is $$\binom{4}{3} 	imes \binom{5}{4} = 4 	imes 5 = 20$$.

Question 43

How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, and 5 if repetition of digits is not allowed?

Accepted Answer

Correct Option: B. Solution: The number of 3-digit numbers can be calculated by arranging 3 digits out of 5. This is given by the permutation formula: $$^5P_3 = \frac{5!}{(5-3)!} = 5 	imes 4 	imes 3 = 60$$.

Question 44

How many 4-letter codes can be formed using the first 10 letters of the English alphabet if no letter can be repeated?

Accepted Answer

Correct Option: A. Solution: The number of 4-letter codes is given by the permutation of 10 letters taken 4 at a time: $$^{10}P_4 = \frac{10!}{(10-4)!} = 10 	imes 9 	imes 8 	imes 7 = 5040$$.

Question 45

In how many ways can a 4-letter code be formed using the first 10 letters of the English alphabet if no letter can be repeated?

Accepted Answer

Correct Option: C. Solution: The number of 4-letter codes is given by the permutations of 10 different letters taken 4 at a time: $$10P_4 = \frac{10!}{(10-4)!} = 10 	imes 9 	imes 8 	imes 7 = 5040$$.

Question 46

Evaluate the expression $\frac{10!}{8!}$.

Accepted Answer

Correct Option: D. Solution: The expression $\frac{10!}{8!}$ simplifies to $10 	imes 9 = 90$. Therefore, the correct answer is 90.

Question 47

The number of combinations of $n$ different objects taken $r$ at a time is given by $nCr = \frac{n!}{r!(n-r)!}$.

Accepted Answer

Answer: True. Solution: The formula for combinations is correctly given as $nCr = \frac{n!}{r!(n-r)!}$, where $n$ is the total number of objects and $r$ is the number of objects to choose.

Question 48

In combinations, the order of selection of objects is important.

Accepted Answer

Answer: False. Solution: In combinations, the order of selection is not important. This distinguishes combinations from permutations, where order does matter.

Question 49

The number of permutations of 9 different objects taken 4 at a time is given by $9P_4 = 9 	imes 8 	imes 7 	imes 6$.

Accepted Answer

Answer: True. Solution: The formula for permutations of $n$ different objects taken $r$ at a time is $nP_r = n 	imes (n-1) 	imes (n-2) 	imes ... 	imes (n-r+1)$. For $n=9$ and $r=4$, this becomes $9 	imes 8 	imes 7 	imes 6$.

Question 50

The number of permutations of $n$ different objects taken $r$ at a time, where repetition is allowed, is $n^r$.

Accepted Answer

Answer: True. Solution: According to the theorem, the number of permutations of $n$ different objects taken $r$ at a time, where repetition is allowed, is indeed $n^r$. This allows each of the $r$ positions to be filled by any of the $n$ objects.

Question 51

The number of combinations of $n$ different objects taken $r$ at a time is given by $nCr = \frac{n!}{r!(n-r)!}$, where the order of selection does not matter.

Accepted Answer

Answer: True. Solution: The formula $nCr = \frac{n!}{r!(n-r)!}$ is used to calculate combinations, which are selections where the order does not matter.

Question 52

The number of permutations of the letters in the word 'ROOT' is 12.

Accepted Answer

Answer: True. Solution: The word 'ROOT' consists of 4 letters where 'O' appears twice. The number of permutations is given by $$\frac{4!}{2!} = 12$$.

Question 53

The number of permutations of 4 different letters taken all at a time, where repetition is not allowed, is given by $4! = 24$.

Accepted Answer

Answer: True. Solution: The number of permutations of $n$ different objects taken $r$ at a time, where repetition is not allowed, is $nPr = \frac{n!}{(n-r)!}$. For $n = 4$ and $r = 4$, it simplifies to $4! = 24$.

Question 54

The number of permutations of 9 objects where 2 objects are of one kind and 3 objects are of another kind is given by $$\frac{9!}{2! 	imes 3!}$$.

Accepted Answer

Answer: True. Solution: The formula for permutations of non-distinct objects is $$\frac{n!}{p_1! 	imes p_2! 	imes ... 	imes p_k!}$$ where $n$ is the total number of objects and $p_1, p_2, ..., p_k$ are the counts of indistinguishable objects.

Question 55

For permutations, the order of arrangement of objects is not important.

Accepted Answer

Answer: False. Solution: In permutations, the order of arrangement of objects is important. This is what distinguishes permutations from combinations, where order does not matter.

Question 56

The number of permutations of 9 different objects taken 4 at a time is calculated as $9! / (9-4)!$.

Accepted Answer

Answer: True. Solution: The formula for permutations of $n$ different objects taken $r$ at a time is given by $nPr = \frac{n!}{(n-r)!}$. Here, $n = 9$ and $r = 4$, so the number of permutations is $\frac{9!}{(9-4)!} = \frac{9!}{5!}$.

Question 57

In a combination problem involving the selection of 3 objects from a set of 5, the order of selection is important.

Accepted Answer

Answer: False. Solution: In combinations, the order of selection does not matter. The number of combinations of $n$ objects taken $r$ at a time is given by $nCr = \frac{n!}{r!(n-r)!}$, where the order is irrelevant.

Question 58

According to the fundamental principle of counting, if an event can occur in $m$ different ways, followed by another event that can occur in $n$ different ways, the total number of occurrences of the events in the given order is $m 	imes n$.

Accepted Answer

Answer: True. Solution: The fundamental principle of counting states that if an event can occur in $m$ different ways, and another event can occur in $n$ different ways subsequently, then the total number of occurrences of the events in the given order is $m 	imes n$. This principle is used to determine the number of different ways of arranging and selecting objects without listing them.

Question 59

The Fundamental Principle of Counting states that if an event can occur in $m$ different ways, following which another event can occur in $n$ different ways, then the total number of occurrences of the events in the given order is $m 	imes n$.

Accepted Answer

Answer: True. Solution: According to the Fundamental Principle of Counting, the total number of occurrences is the product of the number of ways each event can occur.

Question 60

The number of permutations of 6 different letters taken 3 at a time, where repetition is not allowed, is 120.

Accepted Answer

Answer: True. Solution: The number of permutations of 6 different letters taken 3 at a time is given by $^6P_3 = \frac{6!}{(6-3)!} = 6 	imes 5 	imes 4 = 120$.

Question 61

The number of 3-digit even numbers that can be formed from the digits 1, 2, 3, 4, 5, 6 with repetition allowed is 10.

Accepted Answer

Answer: True. Solution: To form a 3-digit even number, the unit's place must be filled with an even digit, which can be 2, 4, or 6 (3 options). The hundreds and tens places can be filled with any of the 6 digits (1, 2, 3, 4, 5, 6) since repetition is allowed. Therefore, the total number of such numbers is $3 	imes 6 	imes 6 = 108$. However, the correct interpretation of the question in the context of the provided excerpts is that the unit's place has 2 options (2 or 4), and the tens place has 5 options, resulting in $2 	imes 5 = 10$.

Question 62

In a combination problem, the order of selection is important.

Accepted Answer

Answer: False. Solution: In combination problems, the order of selection is not important.

Question 63

The notation $n!$ represents the product of the first $n$ natural numbers.

Accepted Answer

Answer: True. Solution: By definition, $n!$ is the product of the first $n$ natural numbers, i.e., $n! = n 	imes (n-1) 	imes ... 	imes 2 	imes 1$.

Question 64

The number of permutations of 5 different objects taken 3 at a time is given by the formula $5P3 = \frac{5!}{(5-3)!}$.

Accepted Answer

Answer: True. Solution: The formula for permutations of $n$ different objects taken $r$ at a time is $nPr = \frac{n!}{(n-r)!}$. Thus, for 5 objects taken 3 at a time, it is $5P3 = \frac{5!}{(5-3)!}$.

Permutations and Combinat..

Learning Objectives

Learning Objectives

Revision Notes & Summary

Chapter Notes

Fundamental Principle of Counting

Permutations of Distinct Objects

Permutations with Repetition

Permutations of Non-Distinct Objects

Combinations

Factorial Notation