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Permutations

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Permutations [#permalink] New post 07 Jun 2006, 08:58
Please explain your answer.
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 [#permalink] New post 07 Jun 2006, 09:18
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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 [#permalink] New post 07 Jun 2006, 09:29
nope, four more tries... :lol:
There is a similar question in OG, though much more easier...
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 [#permalink] New post 07 Jun 2006, 09:47
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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Last edited by giddi77 on 07 Jun 2006, 16:20, edited 3 times in total.
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Re: Permutations [#permalink] New post 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? :roll: ). So it is A.
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 [#permalink] New post 07 Jun 2006, 14:06
Its definitely C

will explain if im right
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Re: Permutations [#permalink] New post 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? :roll: ). So it is A.


Prof, question asks what the prob person will NOT travel through C
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Re: Permutations [#permalink] New post 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? :roll: ). 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.
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Re: Permutations [#permalink] New post 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? :roll: ). 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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Re: Permutations [#permalink] New post 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? :roll: ). 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 :drunk cuz i do not drink but my condition is more than i am :drunk
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 [#permalink] New post 07 Jun 2006, 19:59
The OA is B.

The approach of giddi77 is correct
You can use the same approach with Q195 from OG11.
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 [#permalink] New post 07 Jun 2006, 23:09
good giddi ! i am definitely out of it :?
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 [#permalink] New post 07 Jun 2006, 23:38
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!
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 [#permalink] New post 08 Jun 2006, 07:21
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.
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 [#permalink] New post 08 Jun 2006, 08:02
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 :thanks . Also reminded me the formula for permutation of n objects wherein a, b.. objects are identicle.
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 [#permalink] New post 08 Jun 2006, 08:28
Great question and explanation!....totally undestand this question now.
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Re: Permutations [#permalink] New post 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? :roll: ). 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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 [#permalink] New post 10 Jun 2006, 17:54
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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  [#permalink] 10 Jun 2006, 17:54
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