Permutations and Combinations — Practice

Question 1

In how many ways can a team of 3 students be selected from a group of 8?

Accepted Answer

Order does not matter (a team is a team regardless of selection order). Use ⁸C₃ = 8!/(3!×5!) = (8×7×6)/(3×2×1) = 336/6 = 56.

Question 2

How many 4-digit numbers can be formed using digits 1, 2, 3, 4, 5 without repetition?

Accepted Answer

We need to arrange 4 distinct digits selected from 5 distinct digits. Order matters (4321 ≠ 1234). Answer = ⁵P₄ = 5!/(5−4)! = 5!/1! = 120. Alternatively: 5×4×3×2 = 120 choices by multiplication principle.

Question 3

A committee of 5 is to be formed from 6 men and 4 women such that there are at least 2 women. Find the number of ways.

Accepted Answer

Cases: (i) Exactly 2 women: ⁴C₂×⁶C₃ = 6×20 = 120. (ii) Exactly 3 women: ⁴C₃×⁶C₂ = 4×15 = 60. (iii) Exactly 4 women: ⁴C₄×⁶C₁ = 1×6 = 6. Total = 120+60+6 = 186.

Practice — Permutations and Combinations

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