A committee of (7) members is to be formed from (16) members with (3) fixed members included. How many ways are possible?

Correct answer: (715)

The direct answer is option A, 715. Three specified members must be included in a committee of seven, so those three are already fixed. The remaining places are \(7-3=4\). The remaining people are \(16-3=13\), and we choose four of them without regard to order: \(\binom{13}{4}=715\). Hence option A is correct. Option B, 1001, is not the value of \(\binom{13}{4}\); it corresponds to a different combination. Option C, 1365, is also not the required value and overcounts the choices. Option D, 3003, is much larger and cannot result from choosing only four people from the remaining 13. The fixed three members are not selected again and their internal order is irrelevant because this is a committee. The correct chain is: seven total places, three already occupied, four places left; sixteen people, three fixed, thirteen available; choose four from thirteen. Memory cue: subtract compulsory members first, then calculate one combination.