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Permutation and Combination Questions with Solutions

The whole topic comes down to one question: does the order of arrangement matter? Get that right and the correct formula follows automatically. Here’s the concept, four fully worked examples, and a few to try yourself.

The concept, quickly

  • Permutation counts ARRANGEMENTS, where order matters (ABC is different from BCA). Combination counts SELECTIONS, where order doesn’t matter (choosing the group {A, B, C} is the same regardless of order).
  • Factorial is the building block for both: n! = n × (n−1) × (n−2) × … × 1, and by convention 0! = 1.
  • Permutation of n items taken r at a time: ⁿPᵣ = n! / (n−r)!. Combination of n items taken r at a time: ⁿCᵣ = n! / (r! × (n−r)!) — a combination is a permutation divided by the r! ways to arrange the chosen r items, since order no longer matters.
  • Watch the wording: "arrange", "order", "ranks", "different ways to stand in a line" signal permutation. "Choose", "select", "form a committee" (where only the group matters, not who does what within it) signal combination.

Worked examples

1. In how many ways can the letters of the word "MANGO" be arranged?

Solution

  1. MANGO has 5 distinct letters, and arranging them is a permutation of all 5.
  2. Number of arrangements = 5! = 5 × 4 × 3 × 2 × 1 = 120.

Answer: 120

2. In how many ways can 3 books be selected from a shelf of 8 different books?

Solution

  1. Selecting books (the group chosen, not an order among them) is a combination.
  2. ⁸C₃ = 8! / (3! × 5!) = (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56.

Answer: 56

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?

Solution

  1. The two roles are distinct (President ≠ Secretary), so who gets which role matters — this is a permutation.
  2. ⁶P₂ = 6! / 4! = 6 × 5 = 30.

Answer: 30

4. 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?

Solution

  1. Choosing 2 men from 5: ⁵C₂ = 10. Choosing 1 woman from 4: ⁴C₁ = 4.
  2. These two choices are independent, so multiply them: 10 × 4 = 40.

Answer: 40

Try these yourself

Decide permutation or combination first, then check against the answer.

1. In how many ways can the letters of the word "CHAIR" be arranged?

Answer: 120

2. In how many ways can 4 students be selected from a class of 10 for a quiz team?

Answer: 210

3. In how many ways can a committee of 2 men and 2 women be formed from 6 men and 5 women?

Answer: 150

Where this comes up

Permutation and Combination is tested directly in the Quantitative/Numerical Aptitude section of every exam Pariksha Saathi covers — SSC CGL, SSC MTS, SSC CHSL, IBPS PO, IBPS Clerk, SBI PO and SBI Clerk — and underlies Probability questions too.

Practice more Permutation and Combination questions

This page covers the concept and a handful of worked examples. Pariksha Saathi has full topic-wise practice sets for Permutation and Combination (and every other Quant/Reasoning/English topic) with instant scoring and explanations — free, no account required.

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