Question 1088182
I'm pretty sure this was answered already, but here it is again.

1. Order doesn't matter so we use the combination formula C(n,k) = n!/((n-k)!k!).   The outcomes in this case are also mutually exclusive, so we simply add them:
     C(3,3) + C(4,3) + C(5,3) = 1 + 4 + (5*4/2) = <font color=red><b>15</b></font>    

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2.  Order doesn't matter, so another combination scenario.  Use counting principle (independent events multiply the number of possible outcomes):
     C(3,1)*C(4,1)*C(5,1) = 3*4*5 = <font color=red><b>60</b></font>