
Kemal G. answered 05/26/17
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Patient and Knowledgeable Math and Science Tutor with PhD
Hi Myra,
a) if he wants to give one of each, the possible ways of doing that are
4*3*5 = 60
b) without restrictions, it is the 3 combinations of 12.
C(12, 3) = 12!/((12-3)!*3!)
= 12*11*10*9! / (9!*3*2*1)
= 220