# Counting principles questions

There are six boys and five girls in a school tennis club. A team of two boys and two girls will be selected to represent the school in a tennis competition.

(a) In how many different ways can the team be selected?

(b) Tim is the youngest boy in the club and Anna is the youngest girl. In how many different ways can the team be selected if it must include both of them?

(c) What is the probability that the team includes both Tim and Anna?

Those are the questions. The markscheme says:

I need some clarification:

For (a), you need to divide by two because the order doesn't matter?

(b), it's $(6-1)\times(5-1)$ because you are excluding one person from the boys and one person from the girls; only 5 of the 6 are eligible for the boys, and 4 of the 5 are eligible for the girls?

(c), divide the number of ways Tim and Anna can be chosen by the total.

Is my thinking correct? Is there a better way to phrase my thinking?

Your question about (a) is vague. You need to count the number of teams. If you count something different, then sometimes you will need to divide in order to determine the number of teams. The book's solution is correct; ${6\choose 2}$ counts the number of ways to pick a set of two boys, and ${5\choose 2}$ counts the number of ways to pick a set of two girls. You then multiply these, as this counts the number of ways to pick a team consisting of a set of two boys and a set of two girls. You do not need to divide by 2 because you are not overcounting. If it were possible to have $\{boy A, boy B\}\cup \{girl X, girl Y\}$ and also $\{girl X, girl Y\}\cup \{boy A, boy B\}$ then you would divide by $2$. But that's impossible, since you never get $\{girl X, girl Y\}$ when choosing two boys.
Say the boys are Alfred, Ben, Charles, David, Errol, Frank (ABCDEF for short). The pairs available are AB,AC,AD,AE,AF,BC,BD,BE,BF,CD,CE,CF,DE,DF,EF, $15$ in total. We are counting AB as the same as BA for example.