How many ways can a committee of six be made from 4 students and 8 teachers if the committee contains at least three students?

The obvious answer would be to select 3 students and 3 teachers or 4 students and 2 teachers, which is $${4 \choose 3}{8 \choose 3} + {4 \choose 4}{8 \choose 2} = 252$$ which corresponds to the given answer.

However, my first approach at the problem was a bit different and I don't see why it is wrong: First select 3 students, so that the condition of at least 3 students is satisfied, then choose any 3 out of the remaining 9 people (8 teachers + 1 student). $${4 \choose 3}{9 \choose 3} = 336 \neq 252$$

So why is this method incorrect?

  • 3
    $\begingroup$ You counted, for example, choosing students A, B, C "first" and D later as different from choosing A, B, D and C later. $\endgroup$ – André Nicolas Jun 22 '16 at 3:21

In your fist approach, you are counting same things twice.

Say students are $A,B,C,D$. Now first choosing $A,B,C$ and then among the remaining nine choosing $D$ and two teachers is same as choosing $A,B,D$ and then choosing $C$ and the same two teachers but these two are counted twice in your approach.

Did I make it clear?

  • $\begingroup$ Ahh, i get it. Thank you both! $\endgroup$ – Tony Tarng Jun 22 '16 at 3:24
  • 2
    $\begingroup$ Notice that Tony's first approach counts each selection of all four students four times, once each for each way of selecting three of them. Therefore, $$\binom{4}{3}\binom{9}{3} = \binom{4}{3}\binom{8}{3} + \binom{4}{3}\binom{4}{4}\binom{8}{2}$$ $\endgroup$ – N. F. Taussig Jun 22 '16 at 9:42
  • $\begingroup$ @N.F.TaussigYes...that is good of you to notice...I missed that. $\endgroup$ – User Not Found Jun 23 '16 at 3:41

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