
John M. answered 04/04/14
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Analytical assistance -- Writing, Math, and more
Dalia,
Since arrangement does not matter (i.e. A, B, C is the same as C, B, A) you have a combination problem, as the total sample space is the total number of possible combinations. In this case you have sample spaces of
15, C, 2
6, C, 4
and
18, C, 8
The formula for combination given n, C, r is
n!/[(r!)(n-r!)] (sometimes k is used instead of r)
So for 15, C, 2, you have 105 items in the sample space as
15!
-------- = 15 x 14 / 2 = 15x7 = 105
2!(13!)
Repeat for the other two.