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A committe of five is to b chosen from six men and eight women. Find how many committees are possible if a particular man must be included? The answer is715 but i don't understand. @jim_thompson5910
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"A committe of five ... a particular man must be included" so you really have 4 slots left because this particular man must be locked in.
6 men total 6-1 = 5 men left over after that person is locked in 8 women total so there are 8+5 = 13 people who aren't locked in there are 4 slots, so you need to compute C(13,4)
Formula: \[\Large C(n,r) = \frac{n!}{r!(n-r)!}\]
thanks so much
np
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