# 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?

Jul 25, 2016

$1884$

#### Explanation:

in general you can have $8$ choose $6$ for the men and
$10$ chose $5$ 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 $1$ from $8$ and add his combinations to the total of $7$ choose $1$ ways at the end. So lets start with the other guys

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

$7 \cdot 252 = 1764$

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

(9!)/((9-5)!5!) = 126

$1764 + 126 = 1884$