Data Sufficiency

Runners V, W, X, Y and Z are competing in the Bayville local triathlon. How many different ways are there for X to complete the race ahead of Y?
(A) 5
(B) 10
(C) 30
(D) 60
(E) 120

this one was tough for me...here was my thinking

the only ways for x to complete race ahead of y

1*2*3*y*x=6
1*2*y*3*x=6
1*y*2*3*x=6
y*1*2*3*x=6
1*2*y*x*3=6
1*y*2*x*3=6
y*1*2*x*3=6
1*y*x*2*3=6
y*1*x*2*3=6
y*x*1*2*3=6

6*10=60 but answer is 10. what am i missing? is it that it is a combination instead of a permutation? and is there faster way to solve?

thanks.

I am also weak on probability. But I solved this question in the following way:

1st way - V, W, Z, X, Y - 1 way over here
2nd way - V, W, X, Y, Z - 2 ways over here
3rd way - V, X, W, Z, Y - 3 ways over here
4th way - X, W, Z, V, Y - 4 ways over here

Total ways = 10

i think the answer should be 60 not 10

The OA of 10 is wrong unless there are conditions in the problem that you've omitted. If X is first, for example, there are 4 * 3 * 2 * 1 = 24 ways for the other racers to finish, so there are 24 arrangements there alone.

An easy way to do a problem like this: there are 5 * 4 * 3 * 2 * 1 = 120 ways to run the race. Half the time X beats Y (and half the time Y beats X) so 1/2 of 120 = 60. We're done!

The nitty gritty casework, if you want it:

X finishes first. The other four finish in any way: 4 * 3 * 2 * 1 = 24 ways.
X finishes second. Any of (V, W, Z) can finish first, then we have 3 * 2 * 1 ways left for the other three races. So 3 * 6 = 18 ways.
X finishes third. Any two of (V, W, Z) finish first and second, so 3 * 2. Then we have 2 ways left for the remaining racers, so 3 * 2 * 2 = 12 ways.
X finishes fourth. Y must finish fifth, and we have 3 * 2 * 1 = 6 ways for the first three runners.

So we have 24 + 18 + 12 + 6 = 60 ways.

Now if the question means "Ways" in the sense that we're only comparing the placement of X and Y and the other three runners are irrelevant, the answer is 10. Here's why:

If X is 1st, Y could be 2nd, 3rd, 4th, or 5th: 4 ways
If X is 2nd, Y could be 3rd, 4th, or 5th: 3 ways
If X is 3rd, Y could be 4th or 5th: 2 ways
If X is 4th, Y could be 5th: 1 way

So 10 total ways. But the question isn't worded correctly: it would have to say "How many different ways are there for X to finish ahead of Y, IRRESPECTIVE of the other three runners", or something like that.