Question 1

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!). Step 3: C(8,3) = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56.

Question 2

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: Number of arrangements = 5!/2! = (5 × 4 × 3 × 2 × 1)/2 = 120/2 = 60.

Question 3

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

Accepted Answer

Step 1: We have 7 balls total: 4 identical red and 3 identical blue. Step 2: When arranging objects where some are identical, use: n!/(n₁! × n₂!). Step 3: Number of arrangements = 7!/(4! × 3!) = 5040/(24 × 6) = 5040/144 = 35.

Question 4

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 different digits without repetition. Step 2: This is a permutation problem: P(5,3) = 5!/(5-3)! = 5!/2!. Step 3: P(5,3) = 5 × 4 × 3 = 60.

Question 5

A team of 4 players is to be selected from 6 batsmen and 5 bowlers such that the team has at least 2 batsmen. How many ways can this be done?

Accepted Answer

Step 1: At least 2 batsmen means: (2 batsmen, 2 bowlers) OR (3 batsmen, 1 bowler) OR (4 batsmen, 0 bowlers). Step 2: Case 1: C(6,2) × C(5,2) = 15 × 10 = 150. Step 3: Case 2: C(6,3) × C(5,1) = 20 × 5 = 100. Step 4: Case 3: C(6,4) × C(5,0) = 15 × 1 = 15. Step 5: Total = 150 + 100 + 15 = 265.

Question 6

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 7

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

Accepted Answer

Step 1: Total ways to form committee with at least 2 women = (2 women, 3 men) + (3 women, 2 men) + (4 women, 1 man) + (5 women, 0 men). Step 2: C(6,2)×C(8,3) = 15×56 = 840. Step 3: C(6,3)×C(8,2) = 20×28 = 560. Step 4: C(6,4)×C(8,1) = 15×8 = 120. Step 5: C(6,5)×C(8,0) = 6×1 = 6. Step 6: Total = 840 + 560 + 120 + 6 = 1526.

Question 8

In how many ways can 7 different books be arranged on a shelf such that 3 specific books are always together?

Accepted Answer

Step 1: Treat the 3 specific books as a single unit. Step 2: Now we have 5 units to arrange (the block of 3 books + 4 other books). Step 3: These 5 units can be arranged in 5! = 120 ways. Step 4: Within the block, the 3 books can be arranged among themselves in 3! = 6 ways. Step 5: Total arrangements = 5! × 3! = 120 × 6 = 720.

Question 9

In a group of 10 people, how many ways can we select a president, a vice-president, and a treasurer (all different people)?

Accepted Answer

Step 1: This is a permutation problem because the order matters (different roles). Step 2: For president: 10 choices. Step 3: For vice-president: 9 remaining choices (one person already selected). Step 4: For treasurer: 8 remaining choices. Step 5: Total ways = 10 × 9 × 8 = 720. Alternatively, P(10,3) = 10!/(10−3)! = 10!/7! = 10 × 9 × 8 = 720.

Question 10

A team of 6 players is to be selected from 5 forwards, 4 midfielders, and 3 defenders. How many ways can the team be formed if it must contain exactly 2 forwards, 2 midfielders, and 2 defenders?

Accepted Answer

Step 1: Select 2 forwards from 5: C(5,2) = 10. Step 2: Select 2 midfielders from 4: C(4,2) = 6. Step 3: Select 2 defenders from 3: C(3,2) = 3. Step 4: Apply multiplication principle: Total = C(5,2) × C(4,2) × C(3,2) = 10 × 6 × 3 = 180.

Question 11

How many 6-digit numbers can be formed using the digits {1,2,2,3,3,3} such that the two 2's are not adjacent and the three 3's are not all adjacent?

Accepted Answer

Step 1: Total arrangements of {1,2,2,3,3,3} = 6!/(2!×3!) = 60. Step 2: Arrangements where the two 2's are adjacent: treat '22' as a single unit. Arrange {1,22,3,3,3} = 5!/3! = 20. Step 3: Arrangements where three 3's are all adjacent: treat '333' as a single unit. Arrange {1,2,2,333} = 4!/2! = 12. Step 4: Arrangements where both conditions occur (2's adjacent AND 3's all adjacent): arrange {1,22,333} = 3! = 6. Step 5: By inclusion-exclusion, arrangements with 2's adjacent OR 3's all adjacent = 20 + 12 - 6 = 26. Step 6: Arrangements with 2's NOT adjacent AND 3's NOT all adjacent = 60 - 26 = 34.

Question 12

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

Accepted Answer

Step 1: Total ways to form a 5-member committee from 14 people = C(14,5) = 2002. Step 2: Ways with 0 women (all men) = C(8,5) = 56. Step 3: Ways with 1 woman = C(6,1) × C(8,4) = 6 × 70 = 420. Step 4: Ways with at least 2 women = 2002 - 56 - 420 = 1526.

Question 13

In how many ways can 10 identical red balls and 8 identical blue balls be distributed into 4 distinct boxes such that each box contains at least one ball?

Accepted Answer

Step 1: This is a stars and bars problem with constraints. We need to distribute 18 balls into 4 boxes with each box non-empty. Step 2: First, place 1 ball in each box (4 balls used), leaving 14 balls to distribute freely. Step 3: We need to find the number of non-negative integer solutions to r₁ + r₂ + r₃ + r₄ = 10 and b₁ + b₂ + b₃ + b₄ = 8 where rᵢ, bᵢ ≥ 0. Step 4: This equals C(10+4-1, 4-1) × C(8+4-1, 4-1) = C(13,3) × C(11,3) = 286 × 165 = 47190.

SBI PO Permutation & Combination

SBI PO Permutation & Combination — Past Exam Questions

Permutation & Combination— Rules & Concept

Key Points to Remember

Exam-Specific Tips

60-Second Revision — Permutation & Combination