CA Foundation · Quantitative Aptitude · Permutations and Combinations
A committee of 3 members is to be chosen from 8 teachers at a school in Pune. In how many ways can this be done?
The committee can be chosen in 56 ways. Since the members of a committee are not ranked, order is irrelevant, so we use combinations: 8C3 equals 8×7×6 divided by 3×2×1, which gives 56. Permutations would overcount the selections.
- A24
- B56Correct
- C336
- D112
Explanation
Order does not matter, so the number of ways is 8C3 = (8×7×6)/(3×2×1) = 56. The value 336 is 8P3, which wrongly treats the selection as an arrangement.
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