Question 1150272
<i>a) All the members are to be of the same sex.</i>
<pr>
Ways to choose 4 men: 6C4 = {{{6!/(4!*2!)}}} = 15
Ways to choose 4 women: 4C4 = {{{4!/(4!*0!)}}} = 1
<pr>
Ways to choose all members of  the same sex: 15 + 1 = <b>16</b>
<pr>
<i>b) There must be an equal number of men and women. (Without sister provision.)</i>
<pr>
Ways to choose 2 men and 2 women: 6C2 * 4C2 = {{{(6!/(2!*4!))*(4!/(2!*2!))}}} = 15 * 6 = <b>90</b>
<pr>
<i>Given that the 4 women include 2 sisters, find the total number of ways in which the team can be selected if either of the sisters, but not both, must be included.</i>
<pr>
Let's re-categorize the people as such:
<pr>
6 men
2 non-sisters
1 sister (A)
1 sister (B)
<pr>
Ways to choose 2 men, 1 non-sister, and sister (A): {{{(6!/(2!*4!))*(2!/(1!*1!)) * (1!/(1!*0!))}}} = 15 * 2 * 1 = 30
Ways to choose 2 men, 1 non-sister, and sister (B): {{{(6!/(2!*4!))*(2!/(1!*1!)) * (1!/(1!*0!))}}} = 15 * 2 * 1 = 30
<pr>
Ways that the team can choose 2 men and 2 women, where one of the women must be a "sister," but the other woman cannot be a "sister":
<pr>
30 + 30 = <b>60</b>