A team of (4) students is to be formed from (9) students. Exactly (2) of (3) special students must be included. How many ways are there?

Correct answer: (45)

The direct answer is option B: 45 teams. There are 3 special students, and exactly 2 of them must be included. First choose those 2 special students: \(\binom{3}{2}=3\) ways. There are 9 total students, so 6 are not special. The team has 4 members, and after selecting 2 special students we must choose 2 ordinary students from those 6: \(\binom{6}{2}=15\) ways. By the multiplication rule, the total is \(\binom{3}{2}\binom{6}{2}=3\times15=45\). Option B is correct. Option A, 36, does not use the required two-stage selection correctly. Option C, 54, overcounts the valid choices. Option D, 63, also includes invalid teams or uses an incorrect combination. We choose exactly two special students, not one, three, or an unspecified number. Memory cue: “exactly” means choose the required number from the special group, then fill the remaining places from everyone else.