Question 1

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

Accepted Answer

Step 1: The word 'APPLE' has 5 letters with 'P' repeated twice. Step 2: When objects are repeated, we use: n!/(n₁! × n₂! × ...). Step 3: Arrangements = 5!/2! = (5 × 4 × 3 × 2 × 1)/2 = 120/2 = 60. Therefore, the answer is 60.

Question 2

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

Accepted Answer

Step 1: We need to form 2-digit numbers using 4 different digits without repetition. Step 2: For the first position, we have 4 choices. For the second position, we have 3 remaining choices. Step 3: Total = 4 × 3 = 12. Alternatively, P(4,2) = 4!/(4-2)! = 4!/2! = 12. Therefore, the answer is 12.

Question 3

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: This is a permutation problem where all 5 objects are used: P(5,5) = 5! = 5 × 4 × 3 × 2 × 1 = 120. Therefore, the answer is 120.

Question 4

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

Accepted Answer

Step 1: We have 7 balls total: 4 identical red balls and 3 identical blue balls. Step 2: When arranging objects with repetitions, we use: n!/(n₁! × n₂!). Step 3: Arrangements = 7!/(4! × 3!) = (7 × 6 × 5 × 4!)/(4! × 3 × 2 × 1) = (7 × 6 × 5)/6 = 35. Therefore, the answer is 35.

Question 5

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 indistinguishable 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: Initial selection = C(6,3) = 20. Step 4: Since the two groups are indistinguishable (dividing, not arranging), we divide by 2!: 20/2 = 10. Therefore, the answer is 10.

Question 6

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. Step 2: This is a combination problem: C(8,3) = 8!/(3! × 5!) = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56. Therefore, the answer is 56.

Question 7

A committee of 4 members is to be selected from 6 men and 5 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 2 women or 3 women or 4 women. Step 2: Case 1 (2 women, 2 men): C(5,2) × C(6,2) = 10 × 15 = 150. Step 3: Case 2 (3 women, 1 man): C(5,3) × C(6,1) = 10 × 6 = 60. Step 4: Case 3 (4 women, 0 men): C(5,4) × C(6,0) = 5 × 1 = 5. Step 5: Total = 150 + 60 + 5 = 215.

Question 8

In a group of 8 students, a team of 3 is to be selected for a project. Additionally, from the remaining 5 students, 2 must be chosen as alternates. In how many ways can this be done?

Accepted Answer

Step 1: Select 3 students from 8 for the main team: C(8,3) = 56. Step 2: From the remaining 5 students, select 2 as alternates: C(5,2) = 10. Step 3: Total ways = 56 × 10 = 560.

Question 9

How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 without repetition, such that the number is divisible by 5?

Accepted Answer

Step 1: For divisibility by 5, the last digit must be 5 (since 0 is not available). Step 2: Fix 5 in the units place. Step 3: Now we need to fill the first 3 positions with digits from {1, 2, 3, 4, 6} without repetition. Step 4: Number of ways = P(5,3) = 5 × 4 × 3 = 60.

Question 10

From a deck of 52 playing cards, in how many ways can 5 cards be selected such that there are exactly 2 cards of one suit and exactly 3 cards of another suit?

Accepted Answer

Step 1: Choose 2 suits from 4 suits: C(4,2) = 6 ways. Step 2: Decide which suit gets 2 cards and which gets 3 cards: 2 ways (suit A has 2 and suit B has 3, or vice versa). Step 3: Select 2 cards from the first chosen suit: C(13,2) = 78. Step 4: Select 3 cards from the second chosen suit: C(13,3) = 286. Step 5: Total = 6 × 2 × 78 × 286 = 268,656.

Question 11

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

Accepted Answer

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

Question 12

A committee of 5 members is to be formed from 8 men and 6 women. In how many ways can this be done such that the committee has at least 2 women and at least 2 men?

Accepted Answer

Step 1: Identify valid compositions: (2W, 3M), (3W, 2M), (4W, 1M) is invalid (needs ≥2 men). Step 2: Calculate C(6,2)×C(8,3) = 15×56 = 840. Step 3: Calculate C(6,3)×C(8,2) = 20×28 = 560. Step 4: Total = 840 + 560 = 1400.

Question 13

In a tournament, 12 teams are divided into 3 groups of 4 teams each. The groups are indistinguishable. In how many ways can this division be made?

Accepted Answer

Step 1: If groups were distinguishable, we would choose 4 teams for group 1, then 4 for group 2, then 4 for group 3 = C(12,4) × C(8,4) × C(4,4) = 495 × 70 × 1 = 34650. Step 2: Since the 3 groups are indistinguishable, we divide by 3! = 6 to account for overcounting. Step 3: Total = 34650 / 6 = 5775.

Question 14

A password consists of 4 digits followed by 2 letters. The digits must be distinct and chosen from {1,2,3,4,5,6,7,8,9}, and the letters must be distinct and chosen from {A,B,C,D,E,F,G,H}. How many such passwords are possible?

Accepted Answer

Step 1: Choose and arrange 4 distinct digits from 9 available digits = P(9,4) = 9!/(9−4)! = 9!/5! = 9×8×7×6 = 3024. Step 2: Choose and arrange 2 distinct letters from 8 available letters = P(8,2) = 8!/(8−2)! = 8!/6! = 8×7 = 56. Step 3: Total passwords = 3024 × 56 = 169344.

Question 15

A bookshelf has 6 distinct mathematics books and 5 distinct physics books. In how many ways can these 11 books be arranged such that no two physics books are adjacent?

Accepted Answer

Step 1: First, arrange the 6 mathematics books in a row = 6! = 720 ways. Step 2: This creates 7 gaps (before the first book, between each pair, and after the last book) where physics books can be placed: _M_M_M_M_M_M_. Step 3: Choose 5 of these 7 gaps for the 5 physics books = C(7,5) = 21 ways. Step 4: Arrange the 5 distinct physics books in the chosen gaps = 5! = 120 ways. Step 5: Total = 720 × 21 × 120 = 1814400.

Question 16

How many 5-digit numbers can be formed using the digits {1,2,3,4,5,6,7,8,9} such that the digits are in strictly increasing order?

Accepted Answer

Step 1: Since we need 5 distinct digits in strictly increasing order, once we choose 5 digits from the 9 available, there is exactly one way to arrange them (in increasing order). Step 2: Number of ways to choose 5 digits from 9 = C(9,5) = 9!/(5!×4!) = (9×8×7×6)/(4×3×2×1) = 3024/24 = 126.

IBPS Clerk Permutation & Combination

IBPS Clerk 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