Runners V, W, X, Y, and Z are competing in the Bayville loca

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by BTGmoderatorAT » Sat Dec 23, 2017 5:40 am

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

OA: B

I'm confused how to set up the formulas here. Can any experts help?

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Source: Beat The GMAT — Problem Solving | Sat Dec 23, 2017 5:40 am

elias.latour.apex: Master | Next Rank: 500 Posts; Posts: 228; Joined: Thu Apr 20, 2017 1:02 amFollowed by: 3 members; Location: Global; Thanked: 32 times; GMAT Score:770

by elias.latour.apex » Tue Dec 26, 2017 9:33 am

We can tackle this problem by imagining five chairs in a row. Three people will sit in the other chairs leaving two chairs open for X and Y. Since X will always sit to the left of Y (representing having finished first) all we need to do is to calculate how many different two-vacant chair combinations there are.

If X sits in the far left chair, Y will have 4 possible places to sit.
If X sits in the second chair, Y will have 3 possible places to sit.
If X sits in the middle chair, Y will have 2 possible places to sit.
If X sits in the fourth chair, Y will have 1 possible place to sit.

So the answer is: 4+3+2+1 = 10

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by Jeff@TargetTestPrep » Thu Feb 22, 2018 8:31 am

ardz24 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

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

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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by GMATGuruNY » Thu Feb 22, 2018 9:17 am

ardz24 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

Of the 5 positions in the race, 2 must be occupied by X-Y, in that order.
From the 5 positions, the number of ways to choose 2 for X-Y = 5C2 = (5*4)/(2*1) = 10.
The remaining 3 positions must be occupied by V-W-Z, in that order.
Number of options for V-W-Z = 1. (They must occupy the remaining 3 positions, with V to the left of W and W to the left of Z.)
To combine the options above, we multiply:
10*1 = 10.

The correct answer is B.

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by Brent@GMATPrepNow » Thu Feb 22, 2018 9:54 am

ardz24 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

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
--------------------------

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Brent Hanneson - Creator of GMATPrepNow.com

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by Brent@GMATPrepNow » Thu Feb 22, 2018 10:01 am

ardz24 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

Another option is to systematically list and count the possible outcomes
To begin, Z, W and V must have the following order: Z-W-V

Now insert an X and a Y so that X is ahead of Y.
We get:
1) YZWVX
2) ZYWVX
3) ZWYVX
4) ZWVYX
5) YZWXV
6) ZYWXV
7) ZWYXV
8) YZXWV
9) ZYXWV
10) YXZWV

Answer: B

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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