There are a few possible areas in this problem where you can go wrong.

First off: does the solution require combinations or permutations?

For the mathematics dept., in which you are selecting 1 person from 7, it is irrelevant whether you use combinations or permutations – the answer is the same. Also, whenever you see nC1, remember that the answer is n (don’t feel you have to set up all the factorials).

With the computer science dept., you have two identical positions. Now you have to address the initial question: combinations or permutations. The order in which any two candidates are chosen (say, candidate A and candidate B) is irrelevant (AB is the same as BA) thus you should use the combinations formula. 10C2.

The quick math in this case is as follows: when you have nC2, where n is any integer greater than or equal to 4, multiply n(n-1)/2 to get the answer. In this case n = 10 so (10)(9)/2 = 45.

The second trouble spot is whether to add or multiple the 45 and the 7. Because each of the 7 math departments can be matched up with any 45 of the comp. sci. dept., you want to multiply. The 7 different possibilities for group A can be matched up with the 45 different possibilities from Group B to get: 7 x 45 = 315.

Hope that was helpful

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Christopher Lele

Magoosh Test Prep