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female $= 30$
male $= 40$
committee requirement$ = 11$

In how many ways can a committee be formed such that we have at least $5$ from each gender?

my approach:
1. $30C5 \cdot 40C5 \cdot 60C1$

2.(second approach)
$[30C5 \cdot 40C6] + [30C6 \cdot 40C5]$

Which one is correct? They give different answers...

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  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Sep 19 '16 at 2:19
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The second approach is correct. If there are at least five persons of each gender on the committee, there are only two possibilities:

  1. five women and six men
  2. six women and five men

There are $\binom{30}{5}\binom{40}{6}$ ways to select five of the thirty women and six of the forty men, and $\binom{30}{6}\binom{40}{5}$ ways to select six of the thirty women and five of the forty men. Since the cases are disjoint, the total number of committees that can be formed is the sum of these.

What is wrong with the first approach?

You are counting the same committees more than once. You are counting each committee with six women six times, once for each of the six ways you could count one of the six women as one of the other sixty people. You are also counting each committee with six men six times for the same reason. Notice that $$6\binom{30}{6}\binom{40}{5} + 6\binom{30}{5}\binom{40}{6} = \binom{30}{5}\binom{40}{5}\binom{60}{1}$$

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