Question 1150200
<i>a) In how many ways can a committee of 4 people be selected?</i>
<pr>
12C4 = {{{12!/(4!*8!)}}} = <b>495</b>
<pr>
<i>b) In how many ways can this committee be selected if there must be exactly 2 men and 2 women on the committee?</i>
<pr>
7C2 * 5C2 = {{{(7!/(2!*5!))*(5!/(2!*3!))}}} = <b>210</b>
<pr>
<i>c) In how many ways can this committee be selected if there must be at least 2 women on the committee?<i>
<pr>
Ways to choose 4 women and 0 men: 7C4 = {{{(7!/(4!*3!))}}} = 35
Ways to choose 3 women and 1 man: 7C3 * 5C1 = {{{(7!/(3!*4!))*(5!/(1!*4!))}}} = 175
Ways to choose 2 women and 2 men: 7C2 * 5C2 = {{{(7!/(2!*5!))*(5!/(2!*3!))}}} = 210
35 + 175 + 210 = <b>420</b>