The number of possible combinations is given by
... C(18, 3) = 18!/(3!(18-3)!) = 18·17·16/(3·2·1) = 816 . . . . possible combinations
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There are 18 ways to choose the first one; 17 ways to choose the second one, and 16 ways to choose the 3rd one. The same 3 students can be chosen in any of 3! = 6 different orders, so the product 18·17·16 must be divided by 6 to get the number of possible combinations in which order doesn't matter.
You need the Least Common Multiple of 12 and 9.
That's 36.
3 x 12 = 36
4 x 9 = 36
Answer:
Please see the attached picture for the full solution.