VeritasKarishma
Bunuel
Runners V, W, X, Y, and Z are competing in the Bayville local triathlon. If V finishes before W and W finishes before Z, how many ways can X finish before Y??
A. 5
B. 10
C. 30
D. 60
E. 120
Check this post:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/1 ... s-part-ii/We extend the same concept to this question. VWZ can be arranged in 3! ways but only 1 way is acceptable. So we divide 120 by 3! to get 20.
X finishes before Y in half of these cases so answer is 10.
Hi
VeritasKarishma /
Bunuel,
I ttotally understand the solution that you have provided. But I went wrong when I tried to solve the problem myself. Could you please help me understand where did I go wrong? This is what I did.
Total arrangements possible = 5! = 120
In order to find arrangements where v>w, we can simply divide 120 by 2 i.e. 60.
Now these 60 will have only those arrangments where v> w. Also they will have arrangements where w>z and w<z. So if we divide 60 by 2, we get only those arrangements where w> z. similarly, 30/ 2 = 15 gives me only those arrangements where x>y and hence, all conditions are taken care of.
Please explain where am i going wrong in the concept.
The 3 variables are interlinked. The only allowed sequence for them is V W Z.
Now W is before Z in only one of them so dividing by 2 is not correct.