From a group of 8 people, how many ways can a COMMITTEE of 3 be selected?

1. The order in a committee DOESN'T matter — this is a COMBINATION problem. 2. $C(8, 3) = \binom{8}{3} = \dfrac{8!}{3!\,5!} = \dfrac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = \dfrac{336}{6} = 56$. 3. So 56 distinct committees. 4. Option A is the PERMUTATION $P(8,3) = 336$ (counts each committee 6 times — once per ordering). Option C is $P(8,3)$ also. Option D is a wild guess. 5. Common gotcha: when problems involve 'select a leader, deputy, and treasurer', that IS ordered, so $P(n,r)$ applies. Here 'committee of 3' is unordered. _Source: NCERT Class 11 Mathematics, Ch 6, §6.4 + Examples, p. 10–12._