A committee of (5) members is to be formed from (12) members with (2) fixed members included. How many ways are possible?

Correct answer: (120)

The direct answer is option A, 120. Two particular members must be included, so they are already fixed in the five-member committee. We need only choose the remaining \(5-2=3\) members from the other \(12-2=10\) members. Since a committee has no order, use a combination: \(\binom{10}{3}=\frac{10\times9\times8}{3\times2\times1}=120\). Option A is correct. Option B, 210, is \(\binom{10}{4}\), so it chooses four additional members and makes a committee of six, not five. Option C, 252, is \(\binom{10}{5}\), so it chooses five additional members and is too large. Option D, 792, corresponds to choosing five from 12 and ignores the requirement that the two fixed members must be included. The key step is to remove the compulsory members and count only the free choices. Memory cue: required members are already selected; subtract them from both the total committee size and the available people.