Question 1

How many 4-digit numbers can be formed using the digits 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 (only digit divisible by 5 in the set). Step 2: Fix 5 at the units place. Step 3: Fill the remaining 3 positions with 3 digits chosen from {2, 3, 4, 6}. This is P(4,3) = 4 × 3 × 2 = 24.

Question 2

A group of 10 people consists of 6 men and 4 women. In how many ways can a team of 5 people be formed such that it has more men than women?

Accepted Answer

Step 1: 'More men than women' in a 5-person team means: (4M, 1W) or (5M, 0W). Step 2: Case 1 (4M, 1W): C(6,4) × C(4,1) = 15 × 4 = 60. Step 3: Case 2 (5M, 0W): C(6,5) × C(4,0) = 6 × 1 = 6. Step 4: Total = 60 + 6 = 66.

Question 3

How many ways can 3 prizes be distributed among 8 students if each student can receive at most one prize?

Accepted Answer

Step 1: We need to select 3 students from 8 to receive the 3 distinct prizes. Step 2: Since prizes are distinct, order matters. Step 3: This is a permutation problem: P(8,3) = 8!/(8-3)! = 8!/5! = 8 × 7 × 6 = 336.

Question 4

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

Accepted Answer

Step 1: Count total letters: M(1), I(4), S(4), P(2) = 11 letters total. Step 2: Use formula for permutations with repetition: n! / (n₁! × n₂! × ... × nₖ!). Step 3: Arrangements = 11! / (1! × 4! × 4! × 2!) = 39916800 / (1 × 24 × 24 × 2) = 39916800 / 1152 = 34650.

Question 5

A committee of 4 people 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 (2W, 2M): C(5,2) × C(6,2) = 10 × 15 = 150. Step 3: Case 2 (3W, 1M): C(5,3) × C(6,1) = 10 × 6 = 60. Step 4: Case 3 (4W, 0M): C(5,4) × C(6,0) = 5 × 1 = 5. Step 5: Total = 150 + 60 + 5 = 215.

Question 6

From a deck of 52 playing cards, in how many ways can we select 5 cards such that we have exactly 2 cards of one suit and exactly 3 cards of another suit?

Accepted Answer

Step 1: Choose 2 suits from 4 suits for the distribution (2 cards from one, 3 from another): this is an ordered selection, so we have 4 × 3 = 12 ways (first suit gets 2 cards, second suit gets 3 cards). Step 2: From the first chosen suit, select 2 cards from 13: C(13,2) = 78. Step 3: From the second chosen suit, select 3 cards from 13: C(13,3) = 286. Step 4: Total = 12 × 78 × 286 = 267,696.

Question 7

In a group of 10 people, how many ways can we select a president, a vice-president, and a secretary such that no person holds more than one position?

Accepted Answer

Step 1: We need to select 3 people for 3 distinct positions from 10 people. Step 2: This is a permutation problem (order matters). Step 3: President can be chosen in 10 ways. Step 4: Vice-president can be chosen from remaining 9 people in 9 ways. Step 5: Secretary can be chosen from remaining 8 people in 8 ways. Step 6: Total = 10 × 9 × 8 = 720. Alternatively, P(10,3) = 10!/(10-3)! = 10!/7! = 720.

Question 8

How many 4-digit numbers can be formed using digits 1, 2, 3, 4, 5, 6 (with repetition allowed) such that the number is divisible by 5?

Accepted Answer

Step 1: For divisibility by 5, the last digit must be 0 or 5. Since we only have digits 1–6, the last digit must be 5. Step 2: Last digit is fixed as 5 (1 way). Step 3: First digit can be any of 1, 2, 3, 4, 5, 6 (6 choices). Step 4: Second digit can be any of 1, 2, 3, 4, 5, 6 (6 choices). Step 5: Third digit can be any of 1, 2, 3, 4, 5, 6 (6 choices). Step 6: Total = 6 × 6 × 6 × 1 = 216.

Question 9

A committee of 6 members is to be formed from 8 engineers and 5 doctors such that the committee has at least 2 doctors and at least 2 engineers. How many ways can this be done?

Accepted Answer

Step 1: Total members needed = 6. Constraint: at least 2 doctors and at least 2 engineers. Step 2: Possible compositions: (2D, 4E), (3D, 3E), (4D, 2E), (5D, 1E) — but (5D, 1E) violates 'at least 2 engineers', so exclude it. Step 3: Case 1 (2D, 4E): C(5,2) × C(8,4) = 10 × 70 = 700. Step 4: Case 2 (3D, 3E): C(5,3) × C(8,3) = 10 × 56 = 560. Step 5: Case 3 (4D, 2E): C(5,4) × C(8,2) = 5 × 28 = 140. Step 6: Total = 700 + 560 + 140 = 1,400.

Question 10

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 first man, between each pair of men, and after last man): _M_M_M_M_M_. Step 3: Choose 4 of these 6 positions for women: C(6,4) = 15 ways. Step 4: Arrange 4 women in chosen positions: 4! = 24 ways. Step 5: Total = 5! × C(6,4) × 4! = 120 × 15 × 24 = 43,200.

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

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 the 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 13

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: Use the formula for permutations with repetition: n! / (n₁! × n₂! × ... × nₖ!), where n₁, n₂, etc. are frequencies of repeated letters. Step 3: Number of arrangements = 5! / 2! = 120 / 2 = 60. Therefore, the answer is 60.

Question 14

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

Accepted Answer

Step 1: We need to form 3-digit numbers using 5 distinct digits without repetition. Step 2: For the first position (hundreds), we have 5 choices. For the second position (tens), we have 4 remaining choices. For the third position (units), we have 3 remaining choices. Step 3: Total numbers = 5 × 4 × 3 = 60. Therefore, the answer is 60.

Question 15

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

Accepted Answer

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

Question 16

A password consists of 2 letters followed by 3 digits. How many different passwords can be formed if letters can be repeated but digits cannot?

Accepted Answer

Step 1: For the 2 letters (with repetition allowed), each position has 26 choices. Number of ways = 26 × 26 = 676. Step 2: For the 3 digits (without repetition), the first digit has 10 choices, second has 9 choices, third has 8 choices. Number of ways = 10 × 9 × 8 = 720. Step 3: Total passwords = 676 × 720 = 486,720. Therefore, the answer is 486,720.

Question 17

In how many ways can 5 different books be arranged on a shelf such that a specific book is always at one end?

Accepted Answer

Step 1: Fix the specific book at one end (either left or right). This can be done in 2 ways. Step 2: Arrange the remaining 4 books in the remaining 4 positions. This can be done in 4! = 24 ways. Step 3: Total arrangements = 2 × 24 = 48.

SSC GD Constable Permutation & Combination

SSC GD Constable 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