Asked by Benjamin
out of 7 men and 5 women, 5 members of a committee are selected. in how many ways can this be done if (a) there must be exactly 3 men (b) there must be more women than men?
Answers
Answered by
Reiny
exactly 3 men ---> need 2 women
number of choices
= C(7,3) x C(5,2)
= 35 x 10
= 350
more women than men
---> 5 women, 4W-1M, 3W-2M
= C(5,5) + C(5,4)xC(7,1) + C(5,3)xC(7,2)
= 1 + 35 + 10(21)
= 246
number of choices
= C(7,3) x C(5,2)
= 35 x 10
= 350
more women than men
---> 5 women, 4W-1M, 3W-2M
= C(5,5) + C(5,4)xC(7,1) + C(5,3)xC(7,2)
= 1 + 35 + 10(21)
= 246
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