Hi..

There are Two restrictions..
1) \(V>W>Z\)
2) \(X>Y\)

Let's take just the 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

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

Thanks chetan2u for highlighting . yup I completely ignored that condition

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

Final Answer:

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

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.
_________________

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
_________________

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.

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



_________________

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
_________________

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
_________________

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

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.
_________________

karishma ,
Thanks a lot for clearing this. I am was missing a very simple point.

Why are we dividing it by 6!. I cannot understand . Shouldnt it be subtraction

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
_________________

Help I am drawing a tree diagram and am only getting 10 different ways to arrange VWZ! Each with two blank spots. I guess because each blank spot could be x or y then there are 20 ways.

is there a faster way to do it besides draw a tree?

All solutions here (and the OA itself) is assuming that V-W-Z will finish in that order. The question asks how many ways X can finish before Y, so shouldn't we also account for the cases where X and Y finish between V-W-Z e.g. V-X-W-Z-Y? Or am I missing something here?

Hi VeritasKarishma / Bunuel,

I ttotally understand the solution that you have provided. But I went wrong when I tried to solve the problem myself. Could you please help me understand where did I go wrong? This is what I did.

Total arrangements possible = 5! = 120
In order to find arrangements where v>w, we can simply divide 120 by 2 i.e. 60.
Now these 60 will have only those arrangments where v> w. Also they will have arrangements where w>z and w<z. So if we divide 60 by 2, we get only those arrangements where w> z. similarly, 30/ 2 = 15 gives me only those arrangements where x>y and hence, all conditions are taken care of.

Please explain where am i going wrong in the concept.