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There is a club consisting of six distinct men and seven distinct women. How many ways can we select a committee of three men and four women?

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  • $\begingroup$ Typo! Typo! Typo! $\endgroup$ – Camron Ignazio Jul 23 '13 at 1:44
  • $\begingroup$ Typo! Typo! Typo! $\endgroup$ – Camron Ignazio Jul 23 '13 at 1:44
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    $\begingroup$ Do you know how many ways there are to select just a committee of $3$ men from the set of $6$ men? If not, you need to figure that out first. Then do the same thing for selecting $4$ of the $7$ women. Now in how many ways can you combine one of those $3$-man subcommittees with one of the $4$-woman subcommittees? $\endgroup$ – Brian M. Scott Jul 23 '13 at 1:44
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There are "$6$-choose-$3$" $=\binom{6}{3}$ ways to select the men, and "$7$-choose-$4$" =$\binom{7}{4}$ ways to select the women. That's because

$(1)$ We have a group of $6$ men, from which we need to choose $3$ for the committee.

$(2)$ We have $7$ women from which we need to choose $4$ to sit in on the committee.

Since anyone is either on the committee or not, there is no need to consider order: position or arrangement of those chosen for the committee isn't of concern here, so we don't need to consider permuting the chosen men or women.

So we just *multiply*$\;$ "ways of choosing men" $\times$ "ways of choosing women".
[Recall th Rule of the Product, also known as the multiplication principle.]

$$\binom 63 \cdot \binom 74 = \dfrac{6!}{3!3!} \cdot \dfrac{7!}{4!3!} = \dfrac{6\cdot 5\cdot 4}{3!}\cdot \dfrac{7\cdot 6\cdot 5}{3!}$$

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  • $\begingroup$ I got that much, but what is the reasoning. I guess I should have stated that in the original question. $\endgroup$ – Camron Ignazio Jul 23 '13 at 1:45
  • $\begingroup$ You're welcome. Does that make sense now? $\endgroup$ – amWhy Jul 23 '13 at 1:52
  • $\begingroup$ Yes! >>>>>>>>>> $\endgroup$ – Camron Ignazio Jul 23 '13 at 1:57
  • $\begingroup$ best answer. +1 for you. @amwhy $\endgroup$ – Software Jul 23 '13 at 2:00
  • $\begingroup$ Be careful, @Software, about over-doing votes ;-) $\endgroup$ – amWhy Jul 23 '13 at 2:45
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Hint: You can splitt the problem in three parts.

(1) in how many ways can you pick a group of 3 men out of 6.

(2) in how many ways can you pick a group of four women out of 7.

(3) in how many ways can you combine these two groups? ( it's just (1)$\times$(2))

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  • $\begingroup$ you're welcome (: $\endgroup$ – sigmatau Jul 23 '13 at 1:47

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