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

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Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink]

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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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Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink]

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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


No of ways 5 runners can run the race \(= 5!=120\)

Now order of finishing the race for V, W & Z is V...W...Z. VWZ among themselves can have \(3!=6\) positions but only \(1\) position is allowed.

So number of ways runners can run a race \(= \frac{120}{6}=20\)

Half the time X beats Y (and half the time Y beats X)

so no of ways X finishes before Y \(=\frac{20}{2} = 10\)

Option B
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Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink]

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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



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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Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink]

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New post 23 Dec 2017, 04:49
chetan2u wrote:
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



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 120/6=20 ways..
2) now half of 20 will have x before y and half y before x..
So 20/2=10
B...

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


Thanks chetan2u for highlighting :thumbup: . yup I completely ignored that condition :(
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Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink]

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New post Updated on: 22 Feb 2018, 13:21
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

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Originally posted by EMPOWERgmatRichC on 19 Feb 2018, 13:45.
Last edited by EMPOWERgmatRichC on 22 Feb 2018, 13:21, edited 1 time in total.
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Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink]

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New post 20 Feb 2018, 04:09
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


Check this post:

https://www.veritasprep.com/blog/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.
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Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink]

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New post 22 Feb 2018, 09:15
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



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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Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink]

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EMPOWERgmatRichC wrote:
Hi All,

Assuming that there are no "ties", the 5 runners would have 5! = 120 different outcomes. Since Runner X will finish ahead of Runner Y half the time, the number of options is 120/2 = 60

Final Answer:

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Hi EMPOWERgmatRichC

Your solution ignores one essential condition that VWZ also has to maintain a particular order, hence you are getting answer as option D.
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Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink]

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New post 22 Feb 2018, 10:55
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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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Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink]

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New post 22 Feb 2018, 11:03
Expert's post
Top Contributor
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


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

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Brent
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Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink]

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New post 22 Feb 2018, 13:23
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!

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Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo   [#permalink] 22 Feb 2018, 13:23
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