Out of 8 men and 10 women, a committee consisting of 6 men and 5 women is to be formed. How many such committees can be formed when one particular man A refuses to be a member of the committee in which his boss's wife is there?

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Out of 8 men and 10 women, a committee consisting of 6 men and 5 women is to be formed. How many such committees can be formed when one particular man A refuses to be a member of the committee in which his boss's wife is there?

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Answer 1 · 2016-07-25T14:48:07

$1884$ $1884$

Explanation:

in general you can have $8$ $8$ choose $6$ $6$ for the men and
$10$ $10$ chose $5$ $5$ for the women. Don't ask me why you have more women and your committee is requesting less representation but that is another story.

Okay so the catch is that 1 of these guys refuses to work with one of these girls. So this particular person cannot be used with all guys so we subtract $1$ $1$ from $8$ $8$ and add his combinations to the total of $7$ $7$ choose $1$ $1$ ways at the end. So lets start with the other guys

$\frac{7!}{(7 - 6)! 6!} = 7$ $(7!)/((7-6)!6!) = 7$ now these can be matched up with $\frac{10!}{(10 - 5)! 5!} = 252$ $(10!)/((10-5)!5!) = 252$ ways for women or

$7 \cdot 252 = 1764$ $7*252 = 1764$

now for the last guy who refused work with one girl. he can only work with $9$ $9$ choose $5$ $5$ women so

$\frac{9!}{(9 - 5)! 5!} = 126$ $(9!)/((9-5)!5!) = 126$

$1764 + 126 = 1884$ $1764+126 = 1884$

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Out of 8 men and 10 women, a committee consisting of 6 men and 5 women is to be formed. How many such committees can be formed when one particular man A refuses to be a member of the committee in which his boss's wife is there?

1 Answer

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