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A committee of five is to be chosen from five men and four women. How many different committees can be formed if the number of men on the committee is to be greater than the number of women?

Attempt

If the number of men on the committee is to be greater than the number of women. Lets go ... $5$ Men and $4$ Women If five committee is to be chosen if number of men must be greater than the number of women 4 men and 1 woman Or 3 men and 2 women = $(5C4 and 4C1)$ Or $(5C3 and 4C2)$ = $(5C4 × 4C1)$ + $(5C3 × 4C2)$ = $(5 × 4)$ + $(10 × 6)$ = $20$ + $60$ = $80$ ways.

Or

$5C5$ + $5C4$ X $4C1$ + $5C3$ x $4C2$ = $81$

Since the question said if the number of men on the committee is to be greater than the number of women. Can part of it be $5$men and $0$ women like I did in the second attempt or not?

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  • $\begingroup$ I would say, 5 > 0. $\endgroup$ – georg Sep 27 '15 at 7:14
  • $\begingroup$ Alright, thanks. So the answer would be 81 right? $\endgroup$ – user274246 Sep 27 '15 at 7:22
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Yes, $5$ men are more than $0$ women, and the answer $81$ is correct. Weird way to form a committee, though.

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  • $\begingroup$ Alright sir. Thank you. Just wanted to be sure cause combinatorics can really tricky at times. $\endgroup$ – user274246 Sep 27 '15 at 7:38

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