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A company has 20 employees, 12 males, and 8 females. Suppose we need to form a committee of 5 employees.

  1. How many ways are there to form this committee if we need 3 males and 2 females? I believe this is a permutation problem and calculate it as one. 12 * 11 * 10 * 8 * 7 = 73,920

  2. How many ways are there to form this committee if we need at least 4 females? Same as the above except we need four females so, 8 * 7 * 6 * 5 * 12 = 20,169

  3. How many ways are there to form this committee if we need at least 2 males and at least 2 female? 12 * 11 * 8 * 7 * 16 (because that's how many people are left) = 118,272.

People got different answers, am I looking at this wrong?

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    $\begingroup$ This seems like a homework problem; if so, it should have the self-study tag. See this help page $\endgroup$ – Peter Flom - Reinstate Monica Feb 3 '14 at 10:38
  • $\begingroup$ It's a practice midterm. Should I still put the self-study tag? $\endgroup$ – itsSLO Feb 3 '14 at 10:45
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    $\begingroup$ Yes, you should. If you look at the link, the self-study tag is for "routine questions" - just of this sort. In fact, it was changed from "homework" some time ago, in order to deal with this sort of question. $\endgroup$ – Peter Flom - Reinstate Monica Feb 3 '14 at 11:01
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True you must use combinations.

1) chossing 3 from 12 AND choosing 2 from 8
\begin{equation} {12 \choose 3} \times {8 \choose 2} = 220 \times 28 \end{equation}

2) choosing 4 females AND a man, OR, all five are from females \begin{equation} {12 \choose 1} \times {8 \choose 4} + {8 \choose 5} = 840 + 56 \end{equation}

3) either we have 3 men and two females, or 2 men and 3 females. \begin{equation} {12 \choose 3} \times {8 \choose 2} + {12 \choose 2} \times {8 \choose 3} = 6160 + 3696 \end{equation}

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  • $\begingroup$ Actually I just realized, aren't these all wrong because they should be combinations not permutations? $\endgroup$ – itsSLO Feb 3 '14 at 10:42

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