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Alicia lives in a town whose streets are on a grid system

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sdas
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Alicia lives in a town whose streets are on a grid system [#permalink] New post   01 Feb 2012, 11:05
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Alicia lives in a town whose streets are on a grid system, with all streets running east-west or north-south without breaks. Her school, located on a corner, lies three blocks south and three blocks east of her home, also located on a corner. If Alicia s equally likely to choose any possible path from home to school, and if she only walks south or east, what is the probability she will walk south for the first two blocks?

A. 1/10
B. 1/5
C. 1/4
D. 1/3
E. 1/2
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Bunuel
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Re: Alicia lives in a town whose streets are on a grid system [#permalink] New post   01 Feb 2012, 11:27
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sdas wrote:
Alicia lives in a town whose streets are on a grid system, with all streets running east-west or north-south without breaks. Her school, located on a corner, lies three blocks south and three blocks east of her home, also located on a corner. If Alicia s equally likely to choose any possible path from home to school, and if she only walks south or east, what is the probability she will walk south for the first two blocks?


To get to the school Alicia should walk 3 times south and 3 times east: SSSEEE. Total # of routs to the school is # of permutation of SSSEEE, which is 6!/(3!3!)=20 (# of permutations of 6 letters out of which 3 S's and 3 E's are identical);

Now, we wan to count all the routs which start with {SS}. So, {SS} is fixed and then there can be any combination of the rest 4 letters SEEE. So, all possible routs which start with {SS} equal to # of permutation of SEEE, which is 4!/3!=4 (# of permutations of 4 letters out of which 3 E's).

P=4/20=1/5.
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Re: Alicia lives in a town whose streets are on a grid system [#permalink] New post   03 Jul 2013, 14:38
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nintso wrote:
Hello,

I have a question to Bunuel. the formula you are using is Combination's formula as it is 6!/(6-3)!*3!, and since Alice has to choose 3 Ss and 3Es, where order does not matter, it has to be Combination. but you are saying that "Total # of routs to the school is # of permutation of SSSEEE, which is 6!/(3!3!)=20 (# of permutations of 6 letters out of which 3 S's and 3 E's are identical)"[u][/u]. can you please explain whether it should be a combination or permutation formula?

I am having hard time understanding this problem. Manhattan guide explains it with anagram grid, but I am not grasping that model. I tried to solve it using Slot Method, but could not do it. I can't really get it with combination/permutation formulas either. can you please explain it in an easier way with more details?


I have two keywords for Permutations and combinations - Arrangement and Selection.


SELECTION means - CHOOSING 1, more or nothing. (Combinations)

ARRANGEMENT means - RE-ORGANIZING or ORDER(Permutations)

In this given problem, Alice should definitely take 3 souths and 3 Easts to reach her school. But in any 'ORDER' of her choice....
==> I need to use Permutations and not combinations formula as I hear the word 'ORDER'

so calculating the total number of permutations = \(6!/3!3!\)= 20

If Alice needs to take 2 Souths first, then the remaining 4 steps - 1 South and 3 Easts can be re-arranged (user Permutations) in = \(4!/3!\)= 4

Probability = \(4/20 = 1/5\)
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Re: Alicia lives in a town whose streets are on a grid system [#permalink] New post   01 Feb 2012, 11:46
Thanks Bunuel: the only part I was making mistake was (SS) to be considered 1 and then SEEE as 4. I was considering it as (SS) = 1 + SEEE (4) total 5. Thats why I was going wrong.
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Re: Alicia lives in a town whose streets are on a grid system [#permalink] New post   15 Jan 2013, 05:20
Hello,

I have a question to Bunuel. the formula you are using is Combination's formula as it is 6!/(6-3)!*3!, and since Alice has to choose 3 Ss and 3Es, where order does not matter, it has to be Combination. but you are saying that "Total # of routs to the school is # of permutation of SSSEEE, which is 6!/(3!3!)=20 (# of permutations of 6 letters out of which 3 S's and 3 E's are identical)"[u][/u]. can you please explain whether it should be a combination or permutation formula?

I am having hard time understanding this problem. Manhattan guide explains it with anagram grid, but I am not grasping that model. I tried to solve it using Slot Method, but could not do it. I can't really get it with combination/permutation formulas either. can you please explain it in an easier way with more details?
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Re: Alicia lives in a town whose streets are on a grid system [#permalink] New post   03 Jul 2013, 06:54
A slightly lengthier method would be:

Total number of ways to go home, considering that the order order matters =
Options: SSSEEE
Slots : ------
total permutations of 6 options in 6 slots: 6P6 = 6! = 6*5*4*3*2

Total ways to select "South" in the first 2 positions and anything else in the subsequent 4 positions =
Options: SS ????
Slots: -- ----

Permutations of 3 "S" in 2 slots AND Permutations of 4 Choices in 4 slots =
3P2 * 4P4 = 3! * 4! = 3*2*4*3*2

The probability is therefore:

3P2 * 4P4
--------------- =
6P6

3*2*4*3*2
-------------- = 1/5
6*5*4*3*2
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Re: Alicia lives in a town whose streets are on a grid system [#permalink] New post   03 Jul 2013, 07:50
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sdas wrote:
Alicia lives in a town whose streets are on a grid system, with all streets running east-west or north-south without breaks. Her school, located on a corner, lies three blocks south and three blocks east of her home, also located on a corner. If Alicia s equally likely to choose any possible path from home to school, and if she only walks south or east, what is the probability she will walk south for the first two blocks?


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Re: Alicia lives in a town whose streets are on a grid system [#permalink] New post   03 Jul 2013, 11:23
Expert Reply
1. Alicia can choose only the following as the first two in the path: S1 S2, E1 E2, S1 E1 and E1 S1
2. The total number of paths starting with each of the above are 4,4,6 and 6 respectively
3. Therefore the probability that Alicia chooses two south as the first two is 4/20=1/5.
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Re: Alicia lives in a town whose streets are on a grid system [#permalink] New post   02 Dec 2013, 12:16
if the order does not matter, the solution with combinatorics is correct. however, the question itself is open to argument. route has a totally different meaning - it is more identical to a decision tree rather than a problem where the 3 Ss and 3 Es are identical SSSEEE-. Therefore, Manhattan GMAT should reconsider this question before using it as an example.


nintso wrote:
Hello,

I have a question to Bunuel. the formula you are using is Combination's formula as it is 6!/(6-3)!*3!, and since Alice has to choose 3 Ss and 3Es, where order does not matter, it has to be Combination. but you are saying that "Total # of routs to the school is # of permutation of SSSEEE, which is 6!/(3!3!)=20 (# of permutations of 6 letters out of which 3 S's and 3 E's are identical)"[u][/u]. can you please explain whether it should be a combination or permutation formula?

I am having hard time understanding this problem. Manhattan guide explains it with anagram grid, but I am not grasping that model. I tried to solve it using Slot Method, but could not do it. I can't really get it with combination/permutation formulas either. can you please explain it in an easier way with more details?
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Re: Alicia lives in a town whose streets are on a grid system [#permalink] New post   13 Aug 2017, 01:59
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I'm rusty on combinatorics, so I solves this in a way I consider easier:
1st step: the girl is leaving her home. She can either walk south or east; 3/6 = 1/2 for both ways. Since she went south, now she has two more ways to go south, or three more ways to go east. But she still has to go south, so: (3/6) * (2/5) = 6/30 = 2/10 = 1/5.

Don't know if this methodology is right, or if I was lucky to get the correct answer even though going through a wrong method, but let me know.
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Re: Alicia lives in a town whose streets are on a grid system [#permalink] New post   21 May 2018, 03:13
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Alicia's trip involves 3 movements south (SSS) and 3 movements east (EEE). We want to know the probability that her first 2 movements are SS. This question is no different from the following:

A bag contains three marbles labeled S and three marbles labeled E. If two marbles are randomly selected from the bag, what is the probability that both are labeled S?

P(SS) = 3/6 * 2/5 = 1/5.
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Re: Alicia lives in a town whose streets are on a grid system [#permalink] New post   23 Jul 2021, 15:39
Came here to recommend a method that Plalud has already covered. Don't get lost in permutations and combinations! It is a simple probability question. Adding a more detailed, simpler explanation below:

South = S1, S2, S3 = 3
East = E1, E2, E3 = 3

Probability of Alicia choosing south as her first path = S1 or S2 or S3/ all choices = 3/6 = 1/2 (lets say she chose S1 here)
Probability of Alicia choosing south as her second path = S2 or S3/ remaining choices = 2/5

Total probability = first case (AND) second case = 1/2*2/5 = 1/5
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Re: Alicia lives in a town whose streets are on a grid system [#permalink] New post   06 Aug 2023, 03:09
Bunuel can you plz help me with the confusion. My doubt is because Alicia is standing at her home and she can only move east or south at a time the probability to move south first is 1/2, again at the second block the probability is 1/2 so why not the answer is 1/2 * 1/2 = 1/4.
I am sure I am missing something conceptually but I am not able to figure it out.
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Re: Alicia lives in a town whose streets are on a grid system [#permalink]
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