A team of (5) students is to be formed from (11) students. Exactly (2) of (4) special students must be included. How many ways are there?

Correct answer: (210)

Direct answer: Option B, 210 ways. A team has no order, so use combinations. Choose exactly 2 from the 4 special students: \(\binom{4}{2}=6\). The team needs 5 students, so choose the remaining 3 from the non-special students. Their number is \(11-4=7\), and the choices are \(\binom{7}{3}=35\). Therefore the total is \(6\times35=210\). Option A, 175, is not the product of the required choices and undercounts. Option B, 210, is correct because every selection contains exactly two special and three non-special students. Option C, 245, gives an incorrect total. Option D, 280, is too large and overcounts. The key idea is that “exactly 2” excludes selections with 0, 1, 3, or 4 special students. Memory cue: required special choices first, leftover team places from the ordinary group second, then multiply.