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.