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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