Pariksha SaathiQuantitative Aptitude
Permutation and Combination: Practice Questions with Answers
15 practice questions on Permutation and Combination, with answers; most also have an explanation or a worked solution. Try each one before you open its answer.
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Q.1
A committee of 3 people is to be formed from 5 men and 4 women, with exactly 2 men and 1 woman. In how many ways can this be done?
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Answer: 40
Solution:
- Choosing 2 men from 5: ⁵C₂ = 10. Choosing 1 woman from 4: ⁴C₁ = 4.
- These two choices are independent, so multiply them: 10 × 4 = 40.
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Q.2
In how many ways can 3 books be selected from a shelf of 8 different books?
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Answer: 56
Solution:
- Selecting books (the group chosen, not an order among them) is a combination.
- ⁸C₃ = 8! / (3! × 5!) = (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56.
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Q.3
In how many ways can a President and a Secretary be chosen from a group of 6 people, assuming one person can’t hold both positions?
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Answer: 30
Solution:
- The two roles are distinct (President ≠ Secretary), so who gets which role matters — this is a permutation.
- ⁶P₂ = 6! / 4! = 6 × 5 = 30.
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Q.4
In how many ways can 5 people be seated in a row such that 2 particular people always sit together?
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Answer: 48
Solution:
- Glue the 2 particular people into one block. This leaves 4 units to arrange: the block plus the other 3 people.
- These 4 units can be arranged in 4! = 24 ways.
- Within the block, the 2 people can switch places in 2! = 2 ways.
- Total = 24 × 2 = 48.
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Q.5
A password consists of 2 letters (A–Z) followed by 3 digits (0–9), with repetition allowed in both parts. How many different passwords are possible?
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Answer: 6,76,000
Solution:
- This is the fundamental counting principle applied step by step, not a selection of a subset.
- Letters: 26 choices for each of 2 positions = 26² = 676. Digits: 10 choices for each of 3 positions = 10³ = 1,000.
- Total passwords = 676 × 1,000 = 6,76,000.
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Q.6
In how many ways can the letters of the word "MANGO" be arranged?
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Answer: 120
Solution:
- MANGO has 5 distinct letters, and arranging them is a permutation of all 5.
- Number of arrangements = 5! = 5 × 4 × 3 × 2 × 1 = 120.
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Q.7
In how many ways can the letters of the word "BANANA" be arranged?
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Answer: 60
Solution:
- BANANA has 6 letters in total: B occurs once, A occurs 3 times, N occurs 2 times.
- Arrangements = 6! / (3! × 2!) = 720 / (6 × 2) = 720 / 12
- = 60
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Q.8
A committee of 4 is to be formed from 6 men and 4 women, with at least 2 women. In how many ways can this be done?
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Answer: 115
Solution:
- Split into cases by the exact number of women: 2, 3, or 4.
- Exactly 2 women, 2 men: ⁴C₂ × ⁶C₂ = 6 × 15 = 90.
- Exactly 3 women, 1 man: ⁴C₃ × ⁶C₁ = 4 × 6 = 24.
- Exactly 4 women, 0 men: ⁴C₄ × ⁶C₀ = 1 × 1 = 1.
- Total = 90 + 24 + 1 = 115.
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Q.9
In how many ways can a committee of 2 men and 2 women be formed from 6 men and 5 women?
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Q.10
In how many ways can 4 people be seated in a row such that 2 particular people always sit together?
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Answer: 12. Glue the pair into 1 block (3 units total): 3! × 2! = 6 × 2.
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Q.11
A code consists of 3 digits (0–9), with repetition allowed. How many different codes are possible?
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Q.12
A committee of 3 is to be formed from 5 men and 3 women, with at least 1 woman. Find the number of ways.
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Answer: 46. Total ⁸C₃ = 56; all-men (no woman) = ⁵C₃ = 10; 56 − 10.
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Q.13
In how many ways can the letters of the word "CHAIR" be arranged?
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Q.14
In how many ways can 4 students be selected from a class of 10 for a quiz team?
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Q.15
In how many ways can the letters of the word "LEVEL" be arranged?
How these questions were made
Written by Mahesh Srivastava, Founder of Pariksha Saathi. Published: 2026-10-03.