diegocml wrote:I haven't actually seen permutation in my GMAT studies yet. Having said that:
n!/(n-k)!
The fact that Drew and Eileen cannot serve together in any capacity makes me think that n = 6, but if I plug that I get > 6!/(6-3)! = 120, which is not even in the answer choices. Alternatively, if n = 7! I have 210, which I doubt is the correct answer.
I'm clueless, but I think the catch is with Drew and Eileen. Curious to see how to hack this problem.
210 is a good start, and you can tell that it's not the right answer, because 210 is the total possible permutations of all of them, and the answer does not include some of the permutations of all of them. So it has to be less than 210.
You could get the answer this way.
Start with 210, and then subtract all the ones that have Drew and Eileen in them.
The ones that have Drew and Eileen have D, E and 5 other people.
So we fill the slots this way.
D, E, 1 of 5
D, 1 of 5, E
1 of 5, D, E
and so on.
So we have three elements, D, E, 1 of 5, going into three slots. That is a 3 x 2 x 1 = 6 permutation.
Then we have 5 ways to choose 1 of five. So there are five ways to do the 3 x 2 x 1 = 6 permutation.
So we have 6 x 5 = 30 permutations that include Drew and Eileen.
210 - 30 = 180
The correct answer is
D.
An alternative way to handle it is to add up all the permutations that don't include both Drew and Eileen.
We have the ones with just Drew and the five others.
Drew can be in one of three slots.
D - -
- D -
- - D
Then the other 5 get arranged in the other two slots.
D 5 4
5 D 4
5 4 D
So we have 3 x 5 x 4 = 60 that include Drew but not Eileen.
The same thing can be done with Eileen and not Drew, for another 60.
Then, using just the others there are 5 x 4 x 3 = 60.
60 + 60 + 60 = 180
The correct answer is
D.
