Question 1

In how many ways can 5 different books be arranged on a shelf such that a specific book (say, Book A) is always at one of the end positions?

Accepted Answer

Step 1: Book A must be at one of the two end positions. There are 2 choices for placing Book A (left end or right end). Step 2: The remaining 4 books can be arranged in the remaining 4 positions in 4! = 24 ways. Step 3: Total arrangements = 2 × 24 = 48.

Question 2

In how many ways can 5 different books be arranged on a shelf such that a specific book always remains in the middle position?

Accepted Answer

Step 1: Since one specific book must always be in the middle position (3rd position out of 5), its position is fixed and cannot be changed.
Step 2: The remaining 4 books can be arranged in the other 4 positions (1st, 2nd, 4th, and 5th) in any order.
Step 3: Number of ways to arrange 4 books in 4 positions = 4! = 4 × 3 × 2 × 1 = 24
Step 4: Since the middle position book is fixed, total arrangements = 1 × 24 = 24
Therefore, the answer is 24.

Question 3

In how many ways can 5 different books be arranged on a shelf such that a specific book (say, Book A) is always at one of the end positions?

Accepted Answer

Step 1: Book A must occupy either the leftmost or rightmost position. There are 2 choices for placing Book A. Step 2: After placing Book A, the remaining 4 books can be arranged in the remaining 4 positions in 4! = 24 ways. Step 3: Total arrangements = 2 × 24 = 48.

Question 4

In how many ways can the letters of the word 'GARDEN' be arranged such that the vowels always occupy the odd positions?

Accepted Answer

Step 1: Identify the letters in 'GARDEN' — G, A, R, D, E, N. Total 6 letters. Vowels: A, E (2 vowels). Consonants: G, R, D, N (4 consonants). Step 2: Odd positions in a 6-letter arrangement are positions 1, 3, 5 — that's 3 odd positions. We need to place 2 vowels in these 3 odd positions. Number of ways = P(3,2) = 3×2 = 6. Step 3: Remaining 4 consonants occupy the remaining 4 positions. Number of ways = 4! = 24. Step 4: Total arrangements = 6 × 24 = 144. Therefore, answer is 144.

Question 5

A committee of 5 members is to be formed from a group of 6 men and 4 women. In how many ways can the committee be formed such that it contains at least 2 women?

Accepted Answer

Step 1: Identify valid compositions: at least 2 women means 2, 3, or 4 women. Step 2: Case 1 (2 women, 3 men): C(4,2) × C(6,3) = 6 × 20 = 120. Step 3: Case 2 (3 women, 2 men): C(4,3) × C(6,2) = 4 × 15 = 60. Step 4: Case 3 (4 women, 1 man): C(4,4) × C(6,1) = 1 × 6 = 6. Step 5: Total = 120 + 60 + 6 = 186.

Question 6

A committee of 5 members is to be formed from a group of 6 men and 4 women. In how many ways can the committee be formed such that it contains at least 2 women?

Accepted Answer

Step 1: Identify valid compositions: at least 2 women means 2, 3, or 4 women. Step 2: Case 1 (2 women, 3 men): C(4,2) × C(6,3) = 6 × 20 = 120. Step 3: Case 2 (3 women, 2 men): C(4,3) × C(6,2) = 4 × 15 = 60. Step 4: Case 3 (4 women, 1 man): C(4,4) × C(6,1) = 1 × 6 = 6. Step 5: Total = 120 + 60 + 6 = 186.

Question 7

A committee of 5 members is to be formed from a group of 6 men and 5 women. In how many ways can the committee be formed such that it contains at least 2 women and at least 1 man, but the number of women must not exceed the number of men?

Accepted Answer

Step 1: Identify valid compositions. Committee size = 5. Constraint: at least 2 women, at least 1 man, women ≤ men. This means: (women, men) can be (2,3) or (1,4) — but (1,4) violates 'at least 2 women'. So only (2,3) is valid. Step 2: Calculate C(5,2) × C(6,3) = 10 × 20 = 200. Therefore, the answer is 200.

Question 8

A committee of 5 members is to be formed from a group of 6 men and 5 women. In how many ways can the committee be formed such that it contains at least 2 women and at least 1 man, but the number of women must not exceed the number of men?

Accepted Answer

Step 1: Identify valid compositions where women ≤ men, women ≥ 2, and men ≥ 1. Since total = 5, valid cases are: (2W, 3M) and (1W, 4M) — but 1W violates women ≥ 2, so only (2W, 3M) is valid. Step 2: Calculate C(5,2) × C(6,3) = 10 × 20 = 200. Therefore, the answer is 200.

Question 9

In how many ways can the letters of the word 'BOOK' be arranged?

Accepted Answer

Step 1: The word 'BOOK' has 4 letters with 'O' repeated twice. Step 2: Use the formula for permutations with repetition: n! / (n₁! × n₂! × ... × nₖ!), where n₁, n₂, etc. are frequencies of repeated elements. Step 3: Number of arrangements = 4! / 2! = 24 / 2 = 12. Therefore, the answer is 12.

Question 10

How many ways can a committee of 3 people be selected from a group of 8 people?

Accepted Answer

Step 1: We need to select 3 people from 8 where order does not matter (committee selection). Step 2: Use combination formula: C(n,r) = n! / (r!(n-r)!). Step 3: C(8,3) = 8! / (3! × 5!) = (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56. Therefore, the answer is 56.

Question 11

In how many ways can 5 different books be arranged on a shelf?

Accepted Answer

Step 1: We need to arrange 5 different books in a line. Step 2: Use the permutation formula for n distinct objects: n! = 5! = 5 × 4 × 3 × 2 × 1 = 120. Therefore, the answer is 120.

Question 12

In how many ways can 4 red balls and 3 blue balls be arranged in a row?

Accepted Answer

Step 1: We have 7 balls total: 4 red (identical) and 3 blue (identical). Step 2: Use the formula for permutations with repetition: n! / (n₁! × n₂!), where n = total objects, n₁ = frequency of red, n₂ = frequency of blue. Step 3: Arrangements = 7! / (4! × 3!) = 5040 / (24 × 6) = 5040 / 144 = 35. Therefore, the answer is 35.

Question 13

How many 2-digit numbers can be formed using the digits 3, 5, 7, and 9 without repetition?

Accepted Answer

Step 1: We need to form 2-digit numbers using 4 different digits without repetition. Step 2: For the first digit (tens place), we have 4 choices. Step 3: For the second digit (units place), we have 3 remaining choices (since repetition is not allowed). Step 4: Total numbers = 4 × 3 = 12. Alternatively, use P(4,2) = 4! / (4-2)! = 4! / 2! = 12. Therefore, the answer is 12.

Question 14

In how many ways can the letters of the word 'LETTER' be arranged?

Accepted Answer

Step 1: The word 'LETTER' has 6 letters total. Step 2: Count repeated letters: L appears 1 time, E appears 2 times, T appears 2 times, R appears 1 time. Step 3: Use the formula for permutations with repetition: n! / (n₁! × n₂! × ... × nₖ!) = 6! / (2! × 2!) = 720 / 4 = 180. Therefore, the answer is 180.

Question 15

A student must answer 4 questions out of 7 questions in an exam. In how many ways can the student select the questions?

Accepted Answer

Step 1: The student must choose 4 questions from 7 available questions. Step 2: Order of selection does not matter, so this is a combination problem. Step 3: C(7,4) = 7! / [4! × 3!] = (7 × 6 × 5) / (3 × 2 × 1) = 210 / 6 = 35.

Question 16

How many ways can a committee of 3 members be selected from a group of 8 people?

Accepted Answer

Step 1: We need to select 3 people from 8 where order does not matter (committee selection). Step 2: Use combination formula: C(n,r) = n! / [r!(n-r)!]. Step 3: C(8,3) = 8! / [3! × 5!] = (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56.

Question 17

How many ways can 6 students be divided into two groups of 3 each?

Accepted Answer

Step 1: We need to divide 6 students into two groups of 3 each. Step 2: If we select 3 students for the first group, the remaining 3 automatically form the second group. Step 3: The number of ways to select 3 from 6 is C(6,3) = 6! / (3! × 3!) = 720 / (6 × 6) = 720 / 36 = 20. Step 4: However, since the two groups are indistinguishable (dividing into groups, not assigning to Group A and Group B), we divide by 2!: 20 / 2 = 10. Therefore, the answer is 10.

Question 18

How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5 without repetition?

Accepted Answer

Step 1: We need to form 3-digit numbers using 5 different digits, where no digit repeats. Step 2: For the first position (hundreds place), we have 5 choices. For the second position (tens place), we have 4 remaining choices. For the third position (units place), we have 3 remaining choices. Step 3: Total = 5 × 4 × 3 = 60. Alternatively, P(5,3) = 5! / (5-3)! = 5! / 2! = 120 / 2 = 60. Therefore, the answer is 60.

Question 19

How many ways can 3 students be selected from a group of 7 students?

Accepted Answer

Step 1: We need to select 3 students from 7, where order does not matter. Step 2: Use the combination formula: C(n,r) = n! / [r!(n-r)!] = C(7,3) = 7! / (3! × 4!) = (7 × 6 × 5) / (3 × 2 × 1) = 210 / 6 = 35. Therefore, the answer is 35.

Question 20

A committee of 3 members is to be formed from 6 men and 4 women. In how many ways can this be done if the committee must have at least 1 woman?

Accepted Answer

Step 1: Total ways to select 3 from 10 people = C(10,3) = 120. Step 2: Ways to select 3 with NO women (all men) = C(6,3) = 20. Step 3: Ways with at least 1 woman = Total ways - Ways with no women = 120 - 20 = 100. Therefore, the answer is 100.

Question 21

In how many ways can 5 different books be arranged on a shelf?

Accepted Answer

Step 1: We need to arrange 5 different books in a line. Step 2: Use the permutation formula for n distinct objects: n! = 5! = 5 × 4 × 3 × 2 × 1 = 120. Therefore, the answer is 120.

Question 22

A committee of 4 people is to be formed from 6 men and 4 women. In how many ways can this be done if the committee must have at least 2 women?

Accepted Answer

Step 1: 'At least 2 women' means either 2 women or 3 women or 4 women. Step 2: Case 1 (2 women, 2 men): C(4,2) × C(6,2) = 6 × 15 = 90. Case 2 (3 women, 1 man): C(4,3) × C(6,1) = 4 × 6 = 24. Case 3 (4 women, 0 men): C(4,4) × C(6,0) = 1 × 1 = 1. Step 3: Total = 90 + 24 + 1 = 115. Therefore, the answer is 115.

Question 23

In how many ways can 2 red balls and 3 blue balls be arranged in a row?

Accepted Answer

Step 1: Total balls = 2 + 3 = 5. We have 2 identical red balls and 3 identical blue balls. Step 2: Use the formula for permutations with identical objects: n! / (n₁! × n₂!) = 5! / (2! × 3!) = (5 × 4 × 3 × 2 × 1) / (2 × 1 × 3 × 2 × 1) = 120 / (2 × 6) = 120 / 12 = 10. Therefore, the answer is 10.

Question 24

In how many ways can the letters of the word 'BOOK' be arranged?

Accepted Answer

Step 1: The word 'BOOK' has 4 letters, but the letter 'O' repeats twice. Step 2: Use the formula for permutations with repetition: n! / (n₁! × n₂! × ... × nₖ!) = 4! / 2! = (4 × 3 × 2 × 1) / (2 × 1) = 24 / 2 = 12. Therefore, the answer is 12.

Question 25

In how many ways can 5 different books be arranged on a shelf such that two specific books are always adjacent to each other?

Accepted Answer

Step 1: Treat the two specific books as a single unit. Now we have 4 units to arrange (the pair + 3 other books). Step 2: These 4 units can be arranged in 4! = 24 ways. Step 3: Within the pair, the 2 books can be arranged in 2! = 2 ways. Step 4: Total arrangements = 24 × 2 = 48.

Question 26

In how many ways can 5 men and 4 women be arranged in a row such that no two women sit adjacent to each other?

Accepted Answer

Step 1: First arrange 5 men in a row: 5! = 120 ways. Step 2: This creates 6 possible positions for women (before the first man, between each pair of men, and after the last man). Step 3: Choose 4 of these 6 positions for the 4 women: C(6,4) = 15 ways. Step 4: Arrange 4 women in the chosen positions: 4! = 24 ways. Step 5: Total arrangements = 5! × C(6,4) × 4! = 120 × 15 × 24 = 43,200.

Question 27

In how many ways can the letters of the word 'MISSISSIPPI' be arranged such that all the S's are together and all the I's are together?

Accepted Answer

Step 1: The word MISSISSIPPI has 11 letters: M(1), I(4), S(4), P(2). Step 2: Treat all 4 S's as one block and all 4 I's as one block. Step 3: Now we have: 1 M, 1 block of S's, 1 block of I's, 2 P's = 5 objects to arrange. Step 4: Arrange these 5 objects: 5! ÷ 2! = 120 ÷ 2 = 60 ways (divide by 2! because 2 P's are identical). Step 5: Within the S block, arrange 4 identical S's: 1 way. Step 6: Within the I block, arrange 4 identical I's: 1 way. Step 7: Total = 60 × 1 × 1 = 60.

Question 28

How many different 3-letter codes can be formed using the letters A, B, C, D, E, F such that no letter is repeated and the code must start with a vowel?

Accepted Answer

Step 1: Identify vowels from {A, B, C, D, E, F}: only A and E are vowels (2 choices for first position). Step 2: After selecting the first letter (a vowel), 5 letters remain. Step 3: Second position can be filled in 5 ways. Step 4: Third position can be filled in 4 ways (from remaining 4 letters). Step 5: Total = 2 × 5 × 4 = 40.

RBI Grade B Permutation & Combination

RBI Grade B Permutation & Combination — Past Exam Questions

Permutation & Combination— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Permutation & Combination — Practice Questions

60-Second Revision — Permutation & Combination