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
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Hi..

There are Two restrictions..
1) \(V>W>Z\)
2) \(X>Y\)

Let's just take first restriction...
1) \(V>W>Z\)
Total ways for all \(5 = 5!=120\)
V,W,Z can be arranged in 3! or 6 ways AND only one of it will be V>W>Z
So \(\frac{120}{6}=20\) ways..

2) now half of 20 will have x before y and half y before x..
So \(\frac{20}{2}=10\)
B...

niks18, you have missed out on first restriction and so you are getting D as answer above
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Hi All,

Assuming that there are no "ties" - and IF there were no 'restrictions', then the 5 runners would have 5! = 120 different outcomes. However, there ARE some restrictions that we have to deal with:

1) V finishes before W and W finishes before Z. By placing these 3 runners in these three 'relative positions', we have to eliminate all of the possibilities in which this outcome does not occur. That can be done by dividing the total by 3! = 6. Thus 120 options are reduced to 120/6 = 20 options
2) Runner X will finish ahead of Runner Y HALF the time, so the number of options is further reduced by half: 20/2 = 10

Final Answer:

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Without any restrictions, we have 5! = 120 arrangements. There are 3! = 6 arrangements for V, W and Z but only one counts because we want “V > W > Z,” so we only have 1/6 x 120 = 20 arrangements possible when “V > W > Z.” Of these 20 arrangements, half of them are “X > Y” and the other half are “Y > X.” Thus there are 20/2 = 10 arrangements where “X > Y” i.e., X finishes before Y.

Answer: B
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A different approach...

Take the task of arranging the 5 runners and break it into stages.

GIVEN: V finishes before W and W finishes before Z
So, the order is: Z - W - V
Our goal is to now place the remaining 2 runners (runner X and runner Y)

NOTE: I'm going to IGNORE the restriction that says X must finish before Y
You'll see why shortly.

Stage 1: place runner X into the existing order.
Notice that if we already have the arrangement Z - W - V, then we can place spaces in the areas where runner X might go.
We have: _Z_W_V_
Since there are 4 spaces where we can place runner X, we can complete stage 1 in 4 ways

Stage 2: place runner Y into the existing order.
At this point, we have placed runners Z, W, V and X
Let's pretend for a moment, that the arrangement is ZXWV
From here, we can place spaces in the areas where runner Y might go.
We have: _Z_X_W_V_
Since there are 5 spaces where we can place runner Y, we can complete stage 2 in 5 ways

By the Fundamental Counting Principle (FCP), we can the 2 stages (and thus arrange all 5 runners) in (4)(5) ways (= 20 ways)

IMPORTANT: If we IGNORE the restriction that says X must finish before Y, then there are 20 possible arrangements.
However, in HALF of those 20 arrangements, X is ahead of Y, and in the other HALF of those 20 arrangements, Y is ahead of X.

So, the number of arrangements in which X is ahead of Y = 20/2 = 10

Answer: B

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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Hi niks18,

Nice catch! Looking over my notes, it appears that I didn't write anything down about V, W and Z. There's a big 'takeaway' there about making sure that proper note-taking is done consistently!

GMAT assassins aren't born, they're made,
Rich
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