how many different teams of 9 can be made from 22 football players?
$\displaystyle C^{22}_9=\frac{22!}{9!(22-9)!}
$
To see this consider the players choosen in turn, the first can be any of the 22, the second any of the remaininr 21, and so on.
So the number of ways the team can be choosen in sequence is:
$\displaystyle 22 \times 21 \times ... \times (22-8) = \frac{22!}{(22-9)!}$
but this overcounts the number or ways as every permutation of every
team of 9 appears in this process, so we have to divide by the number
of permutations of a team of 9 players (which is $\displaystyle 9!$), which gives the result.
RonL