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How many ways to assemble a team of 5 out of 15 girls and 10 boys, if the team must contain at least two boys and two girls?

Why is it wrong to count the following way(?):

  1. Choose 2 boys $ = {10 \choose 2}$
  2. Choose 2 girls $ = {15 \choose 2}$
  3. Choose one more boy/girl from the rest $ = 21$

$$ {10 \choose 2}{15 \choose 2}21 = 99225$$

I quiet sure that this count has duplicates, if so, how do I eliminate them?


The correct answer is: $${10 \choose 2}{15 \choose 3}+{10 \choose 3}{15 \choose 2}=33075 $$

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1 Answer

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The problem is that for the gender of which there are $3$, there are three ways of picking $2$ of the $3$ at first and then one later, so you're counting each team three times. You can either divide by that factor of three, or take the approach in the answer you gave: there must be either $2$ boys and $3$ girls or vice versa, so add the numbers of ways you can make those two selections.

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thanks for the quick answer! Is there something that can help me identify duplicate counts quickly? –  MichaelS Oct 19 '11 at 10:51
    
@MichaelS: I don't know of anything more general, but I think this is quite a typical overcounting problem: Fixing things and not taking into account that different items could have been fixed. You'll probably get better at spotting these kinds of problems as you gain more experience. –  joriki Oct 19 '11 at 11:02
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The following suggestion is not necessarily quick, but should be used when feasible. Do a very careful explicit "list and count" for the same basic problem, but with numbers small enough to make the explicit listing feasible. Then compute the answer that your proposed method gives for the "smaller" problem. If the proposed method is flawed, it is likely that the two numbers won't match. A comparison of the two numbers may even suggest a fix. –  André Nicolas Oct 19 '11 at 17:05
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