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
Will
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*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

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