Question 1

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 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.

Question 2

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: This is a permutation with repetition problem. Step 3: Number of arrangements = 7!/(4! × 3!) = (7 × 6 × 5 × 4!)/(4! × 3 × 2 × 1) = (7 × 6 × 5)/6 = 35.

Question 3

A student must answer 5 questions out of 8 questions in an exam. In how many ways can the student choose the questions?

Accepted Answer

Step 1: The student must select 5 questions from 8 available questions. Step 2: The order of selection does not matter, so this is a combination problem. Step 3: C(8,5) = 8!/(5! × 3!) = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56.

Question 4

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. Step 3: Number of arrangements = 5! = 5 × 4 × 3 × 2 × 1 = 120.

Question 5

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

Question 6

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, we use: n!/(n₁! × n₂! × ...). Step 3: Number of arrangements = 4!/2! = (4 × 3 × 2 × 1)/2 = 24/2 = 12.

Question 7

In how many ways can 5 different books be arranged on a shelf such that 2 specific books are always together?

Accepted Answer

Step 1: Treat the 2 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: Within the combined unit, the 2 books can be arranged in 2! = 2 ways. Step 4: Total arrangements = 24 × 2 = 48.

Question 8

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 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 0 or 5. Since only 5 is available, the last digit must be 5. Step 2: The first three positions must be filled with 3 digits chosen from {1, 2, 3, 4, 6} without repetition. Step 3: Number of ways = P(5,3) = 5 × 4 × 3 = 60.

Question 10

In a group of 8 people, how many ways can we select a president, a vice-president, and a treasurer, with no person holding more than one position?

Accepted Answer

Step 1: This is a permutation problem because the positions are distinct (order matters). Step 2: President can be selected in 8 ways. Step 3: Vice-president can be selected from remaining 7 people in 7 ways. Step 4: Treasurer can be selected from remaining 6 people in 6 ways. Step 5: Total = 8 × 7 × 6 = 336 or P(8,3) = 8!/(8−3)! = 8!/5! = 336.

Question 11

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: A number is divisible by 5 if and only if its last digit is 0 or 5. Step 2: Since we can only use digits {1,2,3,4,5,6}, the last digit must be 5. Step 3: The last digit is fixed as 5 (1 way). Step 4: First digit: 6 choices (any of 1,2,3,4,5,6). Step 5: Second digit: 6 choices (any of 1,2,3,4,5,6). Step 6: Third digit: 6 choices (any of 1,2,3,4,5,6). Step 7: Total = 6 × 6 × 6 × 1 = 216.

Question 12

A password consists of 3 distinct digits followed by 2 distinct letters (from A–Z). How many such passwords are possible?

Accepted Answer

Step 1: First 3 positions: distinct digits from {0,1,2,...,9}. Step 2: First digit: 10 choices. Step 3: Second digit: 9 choices (must differ from first). Step 4: Third digit: 8 choices (must differ from first two). Step 5: Ways to fill first 3 positions: 10 × 9 × 8 = 720. Step 6: Next 2 positions: distinct letters from {A,B,...,Z} (26 letters). Step 7: Fourth position (first letter): 26 choices. Step 8: Fifth position (second letter): 25 choices (must differ from fourth). Step 9: Ways to fill last 2 positions: 26 × 25 = 650. Step 10: Total = 720 × 650 = 468,000.

Question 13

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 14

In a circular arrangement, 8 people are to be seated around a table. If two specific people (A and B) must not sit next to each other, in how many ways can this be done?

Accepted Answer

Step 1: Total circular arrangements of 8 people: (8-1)! = 7! = 5,040. Step 2: Arrangements where A and B sit together: treat A and B as a single block. Step 3: Now we have 7 units (the A-B block plus 6 others) to arrange in a circle: (7-1)! = 6! = 720. Step 4: Within the A-B block, A and B can be arranged in 2! = 2 ways. Step 5: Arrangements with A and B together: 720 × 2 = 1,440. Step 6: Arrangements where A and B do NOT sit together: 5,040 - 1,440 = 3,600.

Question 15

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

Accepted Answer

Step 1: Word MISSISSIPPI has 11 letters: M(1), I(4), S(4), P(2). Step 2: Treat all 4 I's as one block [IIII] and all 4 S's as one block [SSSS]. Step 3: Now we have units: [IIII], [SSSS], M, P, P. That is 5 units total. Step 4: Arrange these 5 units: 5! = 120 ways. Step 5: But P appears twice, so divide by 2!: 120 ÷ 2 = 60 ways. Step 6: Within [IIII] block, all I's are identical: 1 way. Within [SSSS] block, all S's are identical: 1 way. Step 7: Total = 60.

Question 16

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

Accepted Answer

Step 1: Total committee size = 6. Constraint: at least 2 engineers AND at least 2 doctors. Step 2: Possible compositions: (2E, 4D), (3E, 3D), (4E, 2D). Step 3: Case 1 (2E, 4D): C(8,2) × C(6,4) = 28 × 15 = 420. Step 4: Case 2 (3E, 3D): C(8,3) × C(6,3) = 56 × 20 = 1,120. Step 5: Case 3 (4E, 2D): C(8,4) × C(6,2) = 70 × 15 = 1,050. Step 6: Total = 420 + 1,120 + 1,050 = 2,590.

SSC CPO Permutation & Combination

SSC CPO Permutation & Combination — Past Exam Questions

Permutation & Combination— Rules & Concept

Key Points to Remember

Exam-Specific Tips

60-Second Revision — Permutation & Combination