A soccer squad contains $3$ goalkeepers, $7$ defenders, $9$ midfielders and $4$ forwards. A soccer squad contains $3$ goalkeepers, $7$ defenders, $9$ midfielders and $4$ forwards. So I understood the first part of the question: $(i)$ In how many ways can a team of $1$ goalkeeper, $4$ defenders, $4$ midfielders, and $2$ forwards be chosen from this squad? **Answer:** $ \binom31 \cdot \binom74 \cdot \binom94 \cdot \binom42$ However, I don't quite understand how to do the next bit: $(ii)$ Two of the defenders refuse to play together. In how many ways can a team be chosen that contains at most one of these two defenders?

The thing to notice is that we can easily compute the number of teams where these two defenders _do_ play together: just change the ${7\choose 4}$ to ${5\choose 2}$. Then we can subtract this from what you found in the first part to get the answer to the second.