
Arnold F. answered 10/10/15
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The total number of ways of choosing 3 of the 20 players is N=C(20,3) where C(n,r)=(n!)/[(r!)(n-r)!] i.e the number of combinations of choosing 3 out of 20.
Assuming five players are designated as starters then all three were chosen from the remaining 15. That would be NA=C(15,3).
The probability would be NA/N.
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