A committee of (6) members is to be formed from (15) members with (2) fixed members included. How many ways are possible?

Correct answer: (715)

The direct answer is option A, 715. Two members are compulsory, so they already occupy two places in the six-member committee. The remaining number of places is \(6-2=4\), and the remaining available members number \(15-2=13\). Therefore choose 4 from 13: \(\binom{13}{4}=\frac{13\times12\times11\times10}{4\times3\times2\times1}=715\). Option A is correct. Option B, 1001, is \(\binom{14}{4}\), so it uses the wrong number of available members. Option C, 1365, does not equal \(\binom{13}{4}\) and is an overcount for this problem. Option D, 3003, is also far too large and does not account correctly for the two fixed members. Since this is a committee, different orders of the same four chosen members do not create new committees. Memory cue: total members minus fixed members gives the pool; committee size minus fixed members gives the number to choose.