Question 1

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

Accepted Answer

Step 1: Identify vowels and consonants in 'GARDEN'. Vowels: A, E (2 vowels). Consonants: G, R, D, N (4 consonants). Step 2: The word has 6 letters. Odd positions are 1, 3, 5 — that's 3 odd positions. We need to place 2 vowels in 3 odd positions. Ways = P(3,2) = 3×2 = 6. Step 3: The remaining 4 consonants occupy the remaining 4 positions. Ways = 4! = 24. Step 4: Total arrangements = 6 × 24 = 144. Therefore, answer is 144.

Question 2

A student must choose 2 subjects from 5 available subjects. In how many ways can this be done?

Accepted Answer

Step 1: We need to select 2 subjects from 5 subjects where order does not matter. Step 2: This is a combination: C(5,2) = 5!/(2! × 3!) = (5 × 4)/(2 × 1) = 20/2 = 10. Therefore, the answer is 10.

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: The number of arrangements of n distinct objects is n!. Step 3: Therefore, arrangements = 5! = 5 × 4 × 3 × 2 × 1 = 120.

Question 4

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). 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 5

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 from 4 distinct 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 numbers = 4 × 3 = 12. Alternatively, P(4,2) = 4! / (4-2)! = 4! / 2! = 12.

Question 6

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

Accepted Answer

Step 1: Total balls = 4 + 3 = 7. Step 2: We have 4 identical red balls and 3 identical blue balls. Step 3: Use formula for permutations with repetition: n! / (n₁! × n₂!) = 7! / (4! × 3!) = 5040 / (24 × 6) = 5040 / 144 = 35.

Question 7

A student needs to select 2 subjects from 5 available subjects for a project. In how many ways can this be done?

Accepted Answer

Step 1: The student is selecting 2 subjects from 5, where the order of selection does not matter. Step 2: Use combination formula: C(5,2) = 5! / [2! × (5-2)!] = 5! / (2! × 3!) = (5 × 4) / (2 × 1) = 20 / 2 = 10.

Question 8

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: When objects are repeated, use formula: n! / (n₁! × n₂! × ... × nₖ!). Step 3: Number of arrangements = 4! / 2! = 24 / 2 = 12.

Question 9

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

Accepted Answer

Step 1: The specific book can be placed at either the left end or right end = 2 ways. Step 2: The remaining 4 books can be arranged in the remaining 4 positions = 4! = 24 ways. Step 3: Total arrangements = 2 × 24 = 48.

Question 10

How many 4-digit numbers can be formed using the digits 2, 3, 4, 5, 6, 7 without repetition, such that the number is even?

Accepted Answer

Step 1: For a number to be even, the last digit must be 2, 4, or 6 (3 choices). Step 2: Once the last digit is fixed, 3 remaining positions must be filled from the remaining 5 digits. Step 3: Number of ways to fill first 3 positions = P(5,3) = 5 × 4 × 3 = 60. Step 4: Total = 3 × 60 = 180.

Question 11

A password consists of 3 letters followed by 2 digits. The letters must be distinct and chosen from {A, B, C, D, E}, and the digits must be distinct and chosen from {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. How many such passwords are possible?

Accepted Answer

Step 1: Choose and arrange 3 distinct letters from 5 available = P(5,3) = 5 × 4 × 3 = 60. Step 2: Choose and arrange 2 distinct digits from 10 available = P(10,2) = 10 × 9 = 90. Step 3: Total passwords = 60 × 90 = 5400.

Question 12

A committee of 4 people is to be selected from a group of 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 13

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 0 or 5. Since we only have digits 1–6, the last digit must be 5. Step 2: Fix 5 in the units place. Step 3: For the remaining 3 positions (thousands, hundreds, tens), we select and arrange 3 digits from the remaining 5 digits {1, 2, 3, 4, 6}. Step 4: Number of ways = P(5,3) = 5!/(5−3)! = 5!/2! = 120/2 = 60.

Question 14

In a group of 8 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: This is a permutation problem where order matters (different positions are distinct). Step 2: Select a president from 8 people: 8 choices. Step 3: Select a vice-president from the remaining 7 people: 7 choices. Step 4: Select a secretary from the remaining 6 people: 6 choices. Step 5: Total ways = 8 × 7 × 6 = 336. Alternatively, P(8,3) = 8!/(8−3)! = 8!/5! = 336.

Question 15

In how many ways can 8 identical red balls and 5 identical blue balls be arranged in a row?

Accepted Answer

Step 1: Total balls = 8 + 5 = 13. Step 2: Since balls of the same colour are identical, this is a permutation with repetition problem. Step 3: Number of arrangements = 13! / (8! × 5!) = C(13,5) = (13 × 12 × 11 × 10 × 9) / (5 × 4 × 3 × 2 × 1) = 1287.

Question 16

From a group of 7 men and 4 women, a team of 5 people is to be selected. In how many ways can this be done if there must be more men than women on the team?

Accepted Answer

Step 1: 'More men than women' with 5 people means: (4M, 1W) or (5M, 0W). Step 2: Case 1 (4M, 1W): C(7,4) × C(4,1) = 35 × 4 = 140. Step 3: Case 2 (5M, 0W): C(7,5) × C(4,0) = 21 × 1 = 21. Step 4: Total = 140 + 21 = 161.

Question 17

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

Accepted Answer

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

Question 18

A committee of 4 members is to be formed 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 + 2 men, OR 3 women + 1 man, OR 4 women + 0 men. 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 19

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 4? (A number is divisible by 4 if its last two digits form a number divisible by 4.)

Accepted Answer

Step 1: For divisibility by 4, the last two digits must form a number divisible by 4. Step 2: From digits {1,2,3,4,5,6}, find all 2-digit endings divisible by 4: 12, 16, 24, 32, 36, 52, 56, 64 (8 valid endings). Step 3: For each valid 2-digit ending, we have 4 remaining digits to fill the first two positions. Step 4: Number of ways to fill first two positions = P(4,2) = 4 × 3 = 12. Step 5: Total = 8 × 12 = 96.

Question 20

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 21

From a group of 7 men and 6 women, a team of 5 is to be selected. In how many ways can this be done if the team must have more women than men?

Accepted Answer

Step 1: Total team size = 5. Constraint: more women than men means women > men. Step 2: Possible distributions (Men, Women): (0,5), (1,4), (2,3). Step 3: Case 1 (0M, 5W): C(7,0) × C(6,5) = 1 × 6 = 6. Step 4: Case 2 (1M, 4W): C(7,1) × C(6,4) = 7 × 15 = 105. Step 5: Case 3 (2M, 3W): C(7,2) × C(6,3) = 21 × 20 = 420. Step 6: Total = 6 + 105 + 420 = 531.

SSC CHSL Permutation & Combination

SSC CHSL 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