A team of (5) students is to be formed from (10) students. Exactly (2) of (4) special students must be included. How many ways are there?

Correct answer: (120)

Direct answer: Option C, 120 ways. A team is a group, so order does not matter; therefore combinations are used. Exactly 2 of the 4 special students must be chosen, so first choose them in \(\binom{4}{2}=6\) ways. The team has 5 members, leaving 3 places. There are 10-4=6 ordinary students, and 3 of them can be chosen in \(\binom{6}{3}=20\) ways. By the multiplication rule, the total is \(\binom{4}{2}\binom{6}{3}=6\times20=120\). Option A, 90, is too small and does not equal this product. Option B, 100, also comes from an incorrect count. Option C, 120, is correct because it satisfies both conditions exactly. Option D, 160, is too large and overcounts selections. Remember: for “exactly,” choose the required number from the special group and the remaining number from everyone else, then multiply.