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Runners V, W, X, Y, and Z are competing in the Bayville local triathlo
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20 Dec 2017, 02:17

14

00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

55% (02:16) correct 45% (01:57) wrong based on 141 sessions

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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??

Runners V, W, X, Y, and Z are competing in the Bayville local triathlo
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23 Dec 2017, 04:39

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1

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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Runners V, W, X, Y, and Z are competing in the Bayville local triathlo
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23 Dec 2017, 03:11

4

4

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 \(=\frac{20}{2} = 10\)

Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo
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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 . yup I completely ignored that condition

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

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

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
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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.

Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo
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22 Feb 2018, 09:23

1

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

Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo
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22 Feb 2018, 10:55

Top Contributor

1

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.

Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo
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22 Feb 2018, 11:03

1

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

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

Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo
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20 Oct 2018, 21:17

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 \(\frac{120}{6}=20\) 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

For my understanding, which principle has been used to calculate Restriction 1 (5!/3!)? Mississipi Formula? Can we express it in terms of combinations?

Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo
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24 Apr 2019, 05:33

EMPOWERgmatRichC wrote:

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

Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo
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24 Apr 2019, 12:56

Hi AlN,

You could certainly calculate ALL of the ways that "don't fit" what we're looking for - and then subtract that number from 120 - but that would be a LOT of work.

Here's another way to consider the given information:

Let's focus on just the 3 runners: V, W and Z. If there were no 'restrictions' on how they finished (and there were no "ties"), then there would be 6 possible outcomes:

VWZ VZW WVZ WZV ZWV ZVW

This prompt DOES give us the restriction that V has to finish before W and W has to finish before Z. Thus, of those 6 possibilities, 5 of them have to be removed. Here, we can do that by dividing 6 by 3!, since 6/3! = 1... the 1 option that "fits" what we were told. Including a 4th or 5th runner does not change that math.