Bunuel wrote:
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
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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