There are 7,315 different combinations, thus, the correct option is c
If we have a set of N elements, the number of different combinations of K elements (such that K ≤ N) of these N elements that we can make is:
[tex]C(N, K) = \frac{N!}{(N - K)!K!}[/tex]
In this case we have K = 4 and N = 22, replacing that, we get:
[tex]C(22, 4) = \frac{22!}{(22 - 4)!*4!} = \frac{22*21*20*19}{4*3*2*1} = 7,315[/tex]
So there are 7,315 different combinations, thus, the correct option is c.
If you want to learn more about combinations:
https://brainly.com/question/11732255
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