I think answer should be 72.

9C1*8C1*1

Above scenario holds true when order matters. But in given case, order of players doesn't matter. hence the answer is 36.

9C2*1 = 36

Since a team should consist of John, we only have to find out the ways in which we can select the remaining players out of \(9\), since John, who is one among the \(10\) is already selected.

And since the order does not matter, it is

\(2C9 = \dfrac{9!}{2!(9-2)!} = \dfrac{9 \times 8 \times 7!}{2! \times 7!} = \dfrac{9 \times 8}{2} = 36\)

There are \(36\) ways of selecting a team of \(3\) players out of \(10\) players so that John, who is one among the \(10\) players, is selected in the team.