The following approach is reasonably systematic. There will be lots of words, but at the end there will be a more or less compact formula.
We count first the teams that have no couple, then, in basically the same way, the teams that have $1$ couple. A couple, viewed as an entity, will be called a family. There are $5$ families.
No couples: Maybe we choose $6$ singles. That can be done in $\binom{6}{6}$ ways. Of course this is $1$, but we call it by the complicated name $\binom{6}{6}\binom{5}{0}2^0$. Soon that will look reasonable!
Or else we pick $5$ singles, $1$ family, and a representative of the family. This can be done in $\binom{6}{5}\binom{5}{1}2^1$ ways.
Or else we pick $4$ singles, $2$ families, and a representative of each family. This can be done in $\binom{6}{4}\binom{5}{2}2^2$ ways.
Or else we pick $3$ singles, $3$ families, and a representative of each family. This can be done in $\binom{6}{3}\binom{5}{3}2^3$ ways.
Or else we pick $2$ singles, $4$ families, and a representative of each family. This can be done in $\binom{6}{2}\binom{5}{4}2^4$ ways.
Or else, finally, we pick $1$ singles, $5$ families, and a representative of each family. This can be done in $\binom{6}{1}\binom{5}{5}2^5$ ways.
Add up. A number of cases, but only one idea.
One couple: The idea is the same. There are $\binom{5}{1}$ ways o pick the couple. We will count the number of ways to pick the remaining $4$ people, add them up, and multiply by $\binom{5}{1}$. But rom here on, we only count the ways of picking the $4$.
We could pick $4$ singles. This can be done in $\binom{6}{4}$ ways, but we call the number $\binom{6}{4}\binom{5}{0}2^0$.
Or else we pick $3$ singles, $1$ family, and a representative of the family. This can be done in $\binom{6}{3}\binom{5}{1}2^1$ ways.
Or else we pick $2$ singles, $2$ families, and a representative of each family. This can be done in $\binom{6}{2}\binom{5}{2}2^2$ ways.
Or else we pick $1$ single, $3$ families, and a representative of each family. This can be done in $\binom{6}{1}\binom{5}{3}2^3$ ways.
Or else, finally, we pick $0$ singles, $4$ families, and a representative of each family. This can be done in $\binom{6}{0}\binom{5}{4}2^4$ ways.
Final answer: We gather the whole thing into a compact formula.
$$\sum_{i=0}^5 \binom{6}{6-i}\binom{5}{i}2^i +\binom{5}{1}\sum_{i=0}^4 \binom{6}{4-i}\binom{5}{i}2^i.$$
Part (b) is a case where the negative space is more easy to calculate: the probability that there are no experienced players.
The unconstrained choice has $\binom {15}{9}$ $ = \frac{\large 15!}{\large 9!6!} = \frac{\large 15\cdot14 \cdot13 \cdot12 \cdot11 \cdot10}{\large 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$ $ = 5005$ options.
The no-experienced-players choice has $\binom {12}{9} = \frac{\large 12!}{\large 9!3!} =\frac{\large 12 \cdot 11 \cdot 10}{\large 3 \cdot 2 \cdot 1} = 220$ options.
So the probability of having an experienced player on the team from a blind choice is $\frac{\large 5005-220}{\large 5005}$ $ = \frac{\large 4785}{\large 5005}$ $ = \frac{\large 87}{\large 91}$
Part (c) potentially depends on player skill mix, but absent any constraint will be $\binom {9}{2,2,5}$ $ = \binom 92\binom 72$
Best Answer
You need to choose the captian so its only a way to choose the captain, for the defence you have 3 players so you have 3 choices and for the forward players you have 5 choices so the number of choices is : $1$x$5$x$3$=$15$ way or choice