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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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20 Dec 2017, 02:17
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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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23 Dec 2017, 04:39
2
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  [#permalink]

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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 $$=\frac{20}{2} = 10$$

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

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

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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

GMAT assassins aren't born, they're made,
Rich
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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ 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. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 9230 Location: Pune, India Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink] Show Tags 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. _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > Target Test Prep Representative Status: Head GMAT Instructor Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2823 Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink] Show Tags 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 _________________ Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 122 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Retired Moderator Joined: 25 Feb 2013 Posts: 1214 Location: India GPA: 3.82 Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink] Show Tags 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 Final Answer: GMAT assassins aren't born, they're made, Rich 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. CEO Joined: 12 Sep 2015 Posts: 3722 Location: Canada Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink] Show Tags 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. RELATED VIDEOS _________________ Test confidently with gmatprepnow.com CEO Joined: 12 Sep 2015 Posts: 3722 Location: Canada Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink] Show Tags 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 Answer: B Cheers, Brent _________________ Test confidently with gmatprepnow.com EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 14188 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo [#permalink] Show Tags 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! GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com *****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

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

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22 Oct 2018, 02:39
Hi karishma,
I am not able to understand why have we divide 120/ 6 ?
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Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

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23 Oct 2018, 02:12
1
Piggu18 wrote:
Hi karishma,
I am not able to understand why have we divide 120/ 6 ?

5 runners - V, W, X, Y, Z

They can complete the race in 5! ways = 120 ways.

VWXYZ
WXVYZ
ZXVWY
...
etc

If we focus on only 3 of these V, W and Z, there are different arrangements possible.
VWZ
WVZ
ZWV
etc...

These 3 can be arranged in 3! ways = 3*2*1 = 6 ways
But of these 6, only 1 way is acceptable to us: VWZ

So out of every 6 ways in the 120 ways, only 1 way is acceptable to us. So we divide 120 by 6 to get 20 acceptable ways.
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Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

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23 Oct 2018, 02:26
karishma ,
Thanks a lot for clearing this. I am was missing a very simple point.
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Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

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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

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

Why are we dividing it by 6!. I cannot understand . Shouldnt it be subtraction
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Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo  [#permalink]

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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.

GMAT assassins aren't born, they're made,
Rich
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Re: Runners V, W, X, Y, and Z are competing in the Bayville local triathlo   [#permalink] 24 Apr 2019, 12:56
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