Question

Suppose that a department contains 8 men and 15 women. How many ways are there to form a committee with six members if it must have more women than men?

Answer 1

Answer:

$Selection = 67249\ ways$

Step-by-step explanation:

Given

$Men = 8$

$Women = 15$

$Committee = 6$

Required

Determine the number of selections if the committee must have more women

To have more women, the selection has to be:

(4 women and 2 men) or (5 women and 1 man) or (6 women and 0 man)

Where each selection is calculated using:

$^nC_r = \frac{n!}{(n - r)!r!}$

So, the selection is calculated as thus:

$Selection = (^{15}C_4 * ^{8}C_2) + (^{15}C_5 * ^{8}C_1) + (^{15}C_6 * ^{8}C_0)$

$Selection = (\frac{15!}{(15 - 4)!4!} * \frac{8!}{(8 - 2)!2!}) + (\frac{15!}{(15 - 5)!5!} * \frac{8!}{(8 - 1)!1!}) + (\frac{15!}{(15 - 6)!6!} * \frac{8!}{(8 - 0)!0!})$

$Selection = (\frac{15!}{11!4!} * \frac{8!}{6!2!}) + (\frac{15!}{10!5!} * \frac{8!}{7!1!}) + (\frac{15!}{9!6!} * \frac{8!}{8!0!})$

$Selection = (\frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1} * \frac{8 * 7}{2 * 1}) + (\frac{15 * 14 * 13 * 12 * 11}{5*4*3*2*1} * \frac{8}{1}) + (\frac{15 * 14 * 13 * 12 * 11 * 10}{6*5*4*3*2*1} * \frac{1}{1})$

$Selection = (\frac{32760}{24} * \frac{56}{2}) + (\frac{360360}{120} * 8) + (\frac{3603600}{720})$

$Selection = (1365 * 28) + (3003 * 8) + (5005)$

$Selection = 38220 + 24024 + 5005$

$Selection = 67249\ ways$