800guy wrote:Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?
(A) 720
(B) 360
(C) 120
(D) 24
(E) 21
Common sense and logic go a long way to solving combinatorics problems on the GMAT.
If we didn't care about the order of G and N, there would be 6! = 720 possible arrangements of horses.
However, we do care about the order of G and N; we know that N always finishes before G does.
So, let's use a tiny bit of common sense. Assuming that there are no ties (something which should have been explicitly stated in the question, by the way - otherwise the question is waaaay more complicated and the correct answer isn't even present), N will finish ahead of G in 50% of the possible arrangements (if it wasn't exactly 50%, that would mean that G finishes ahead of N more or less than half the time, which makes no sense).
Therefore, the correct answer is simply 6! * (1/2) = 720/2 = 360... choose B.
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