Peyton Velez

Answered

2022-06-24

In my textbook MP is strictly reserved to counting lists. Does what I do below to count subsets work?

Consider 3 men(Ace, Bob, Corry) and 3 women(Ann, Beth, Candace). Suppose we need to choose a team with 2 men and 2 women in it. An example team is $\{\text{Ace, Ann, Beth, Bob}\}$ which is the union of $\{\text{Ace, Bob}\}\{\text{Beth, Ann}\}.$ So we simply count 2−lists whose first element is a set of two men and whose second element is a set of two women. We get the same number of 2−lists if their first element is a set of two women. There are $(}\genfrac{}{}{0ex}{}{3}{2}{\textstyle )}=8$ two-men subsets and that many two-women subsets. So there are 8 choices for the first element and 8 choices for the second one. In all there are ${8}^{2}={{\textstyle (}\genfrac{}{}{0ex}{}{3}{2}{\textstyle )}}^{2}=64$ two element lists = teams with two men and two women in each.

Consider 3 men(Ace, Bob, Corry) and 3 women(Ann, Beth, Candace). Suppose we need to choose a team with 2 men and 2 women in it. An example team is $\{\text{Ace, Ann, Beth, Bob}\}$ which is the union of $\{\text{Ace, Bob}\}\{\text{Beth, Ann}\}.$ So we simply count 2−lists whose first element is a set of two men and whose second element is a set of two women. We get the same number of 2−lists if their first element is a set of two women. There are $(}\genfrac{}{}{0ex}{}{3}{2}{\textstyle )}=8$ two-men subsets and that many two-women subsets. So there are 8 choices for the first element and 8 choices for the second one. In all there are ${8}^{2}={{\textstyle (}\genfrac{}{}{0ex}{}{3}{2}{\textstyle )}}^{2}=64$ two element lists = teams with two men and two women in each.

Answer & Explanation

Marlee Guerra

Expert

2022-06-25Added 25 answers

it's all good except $(}\genfrac{}{}{0ex}{}{3}{2}{\textstyle )}=3$ (not 8) so the true answer should be 9.

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