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# Permutations

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Senior Manager
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Permutations [#permalink]  07 Jun 2006, 08:58
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VP
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D) !

ALL possible combinations are 36C11

ALL possible combinations w/o one point (respectively point c) are 35C11

so it is 35C11/36C11, which is 25/36 !
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Senior Manager
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nope, four more tries...
There is a similar question in OG, though much more easier...
VP
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It is B! Made a mistake..

5 to the right and 5 down. If we arrange all the possible routes RRRRRDDDDD, all the possible unique shortest routes are 10!/(5!*5!) = 252

Routes that constitute A->C->B = (5!/4!)*(5!/4!) = 25

Probability that the person will pass through the Point C on his way to B through the shortest path = 25/252

Required probability = 1-25/252 = 227/252
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"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

Last edited by giddi77 on 07 Jun 2006, 16:20, edited 3 times in total.
VP
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Re: Permutations [#permalink]  07 Jun 2006, 13:58
it is definitely not B (227/225), D(25/36) and E(10/21). it is between A and C. but i am not sure which one is correct..

it should be A (1/6) because a person can reach B from A through C in 25 ways. so total ways to reach B from A is more than 125 (how do we count the paths here? ). So it is A.
Director
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Its definitely C

will explain if im right
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Director
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Re: Permutations [#permalink]  07 Jun 2006, 14:27
Professor wrote:
it is definitely not B (227/225), D(25/36) and E(10/21). it is between A and C. but i am not sure which one is correct..

it should be A (1/6) because a person can reach B from A through C in 25 ways. so total ways to reach B from A is more than 125 (how do we count the paths here? ). So it is A.

Prof, question asks what the prob person will NOT travel through C
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Re: Permutations [#permalink]  07 Jun 2006, 16:19
gmatmba wrote:
Professor wrote:
it is definitely not B (227/225), D(25/36) and E(10/21). it is between A and C. but i am not sure which one is correct..

it should be A (1/6) because a person can reach B from A through C in 25 ways. so total ways to reach B from A is more than 125 (how do we count the paths here? ). So it is A.

Prof, question asks what the prob person will NOT travel through C

oh yes, its E.

the probability cannot be significantly less than 1/2. so A(1/6), B (227/225 = absurd), and C(1/5) are clearly out. the choice is between D and E. if D is not OA, then only E remains.
Director
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Re: Permutations [#permalink]  07 Jun 2006, 16:37
Professor wrote:
gmatmba wrote:
Professor wrote:
it is definitely not B (227/225), D(25/36) and E(10/21). it is between A and C. but i am not sure which one is correct..

it should be A (1/6) because a person can reach B from A through C in 25 ways. so total ways to reach B from A is more than 125 (how do we count the paths here? ). So it is A.

Prof, question asks what the prob person will NOT travel through C

oh yes, its E.

the probability cannot be significantly less than 1/2. so A(1/6), B (227/225 = absurd), and C(1/5) are clearly out. the choice is between D and E. if D is not OA, then only E remains.

are the answer choices changing every time. I see that I picked C above, but I wanted to pick B.

Prof - B is 227/252 ... not sure where you got 225 from...maybe it changed from the time you looked.
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VP
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Re: Permutations [#permalink]  07 Jun 2006, 16:47
gmatmba wrote:
Professor wrote:
gmatmba wrote:
Professor wrote:
it is definitely not B (227/225), D(25/36) and E(10/21). it is between A and C. but i am not sure which one is correct..

it should be A (1/6) because a person can reach B from A through C in 25 ways. so total ways to reach B from A is more than 125 (how do we count the paths here? ). So it is A.

Prof, question asks what the prob person will NOT travel through C

oh yes, its E.

the probability cannot be significantly less than 1/2. so A(1/6), B (227/225 = absurd), and C(1/5) are clearly out. the choice is between D and E. if D is not OA, then only E remains.

are the answer choices changing every time. I see that I picked C above, but I wanted to pick B.

Prof - B is 227/252 ... not sure where you got 225 from...maybe it changed from the time you looked.

i am not cuz i do not drink but my condition is more than i am
Senior Manager
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The OA is B.

The approach of giddi77 is correct
You can use the same approach with Q195 from OG11.
VP
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good giddi ! i am definitely out of it
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Manager
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giddi77 wrote:
It is B! Made a mistake..

5 to the right and 5 down. If we arrange all the possible routes RRRRRDDDDD, all the possible unique shortest routes are 10!/(5!*5!) = 252

Routes that constitute A->C->B = (5!/4!)*(5!/4!) = 25

Probability that the person will pass through the Point C on his way to B through the shortest path = 25/252

Required probability = 1-25/252 = 227/252

Giddi - Please elaborate on the logic behind the lines in bold. Thanks!
VP
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shobhitb wrote:
giddi77 wrote:
It is B! Made a mistake..

5 to the right and 5 down. If we arrange all the possible routes RRRRRDDDDD, all the possible unique shortest routes are 10!/(5!*5!) = 252

Routes that constitute A->C->B = (5!/4!)*(5!/4!) = 25

Probability that the person will pass through the Point C on his way to B through the shortest path = 25/252

Required probability = 1-25/252 = 227/252

Giddi - Please elaborate on the logic behind the lines in bold. Thanks!

If you take a simple 2 by 2 path (see picture), From X to Y the shortest path will always have only 4 LEGS (by a LEG I mean either a horizontal or vertical bar) on the picture.

If you denote R = right , D = down, then one of the shortest paths form X to Y could be
R-R-D-D
Similarly, D-D-R-R, D-R-R-D, D-R-D-R, R-D-R-D, R-D-D-R are also shortest possible unique routes. In all there are 6 routes.

Rather than counting all of them we can say that, all possible shortest routes are unique arrangements of 4 letters R, R, D, D
Hence number of routes = 4!/(2!*2!) = 6

In this problem we have 5 Right and 5 Down LEGS.. Hence total routes = 10!/(5!*5!)

Similarly from A->C we have 1-D, and 4-Rs.. Hence total routes from A-> C = 5!/4!
From C-> we again have 1-R and 4-Ds. Hence 5!/4! routes.

From A->B->C = (5!/4!)*(5!/4!)

HTH! BTW when I first got similar problem in my PP test, I couldn't solve it either. I tried to solve this problem for next 2 days and then gave up Found the solution in OG.
Attachments

Shortest_route.GIF [ 1.26 KiB | Viewed 686 times ]

_________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

Manager
Joined: 09 May 2006
Posts: 99
Location: Bangalore, India
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giddi77 wrote:
shobhitb wrote:
giddi77 wrote:
It is B! Made a mistake..

5 to the right and 5 down. If we arrange all the possible routes RRRRRDDDDD, all the possible unique shortest routes are 10!/(5!*5!) = 252

Routes that constitute A->C->B = (5!/4!)*(5!/4!) = 25

Probability that the person will pass through the Point C on his way to B through the shortest path = 25/252

Required probability = 1-25/252 = 227/252

Giddi - Please elaborate on the logic behind the lines in bold. Thanks!

If you take a simple 2 by 2 path (see picture), From X to Y the shortest path will always have only 4 LEGS (by a LEG I mean either a horizontal or vertical bar) on the picture.

If you denote R = right , D = down, then one of the shortest paths form X to Y could be
R-R-D-D
Similarly, D-D-R-R, D-R-R-D, D-R-D-R, R-D-R-D, R-D-D-R are also shortest possible unique routes. In all there are 6 routes.

Rather than counting all of them we can say that, all possible shortest routes are unique arrangements of 4 letters R, R, D, D
Hence number of routes = 4!/(2!*2!) = 6

In this problem we have 5 Right and 5 Down LEGS.. Hence total routes = 10!/(5!*5!)

Similarly from A->C we have 1-D, and 4-Rs.. Hence total routes from A-> C = 5!/4!
From C-> we again have 1-R and 4-Ds. Hence 5!/4! routes.

From A->B->C = (5!/4!)*(5!/4!)

HTH! BTW when I first got similar problem in my PP test, I couldn't solve it either. I tried to solve this problem for next 2 days and then gave up Found the solution in OG.

Thanks a ton giddi! absolutely crystal now . Also reminded me the formula for permutation of n objects wherein a, b.. objects are identicle.
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Great question and explanation!....totally undestand this question now.
VP
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Re: Permutations [#permalink]  08 Jun 2006, 08:36
Professor wrote:
it is definitely not B (227/225), D(25/36) and E(10/21). it is between A and C. but i am not sure which one is correct..

it should be A (1/6) because a person can reach B from A through C in 25 ways. so total ways to reach B from A is more than 125 (how do we count the paths here? ). So it is A.

this is how i was trying to solve this problem (however this is not an elegant approach) but i was drunk (messed up) without drink.

number of ways somebody can pass from A to B through C = 25
total number of ways somebody can pass from A to B = x
number of ways somebody can pass from A to B not through C = x - 25

The prob that somebody can pass from A to B not through C = (x - 25)/x

if we consider on choice B [(252-25)/252 = 227/252), it is exactly what (x - 25)/x is.

again this doesnot work always.
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Great solution giddi. And the absolute correct one. I cheated myself though. I approximated that from A to B it takes 10 steps and each step there are two options so total number of route is 2^10. Same thing from A to C total number of route is 2^5. So the possibility of passing C is 2^5/2^10 which is 1/2^5, and the possibility of not passing C is 1-1/2^5. I know that the 2^5 and 2^10 are not exact since you don't get two options when you reach the bottom. But I get the sense that the result should be something close to 31/32, or close to 1. And therefore I chose B.

Thought I'd share it too since sometimes this kind of "cheat" or approximation might be useful, especially if you are short on time, or at a loss of what is the correct way to solve a question.
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keep on seeking, and you will find;
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