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Alicia lives in a town whose streets are on a grid system [#permalink]

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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? _________________

"When the going gets tough, the tough gets going!"

Re: Alicia lives in a town whose streets are on a grid system [#permalink]

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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).

Re: Alicia lives in a town whose streets are on a grid system [#permalink]

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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. _________________

"When the going gets tough, the tough gets going!"

Re: Alicia lives in a town whose streets are on a grid system [#permalink]

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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?

Re: Alicia lives in a town whose streets are on a grid system [#permalink]

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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

Re: Alicia lives in a town whose streets are on a grid system [#permalink]

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03 Jul 2013, 07:50

Expert's post

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?

Re: Alicia lives in a town whose streets are on a grid system [#permalink]

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03 Jul 2013, 11:23

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. _________________

Re: Alicia lives in a town whose streets are on a grid system [#permalink]

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03 Jul 2013, 14:38

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

Re: Alicia lives in a town whose streets are on a grid system [#permalink]

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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?

Re: Alicia lives in a town whose streets are on a grid system [#permalink]

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24 Feb 2015, 14:34

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Re: Alicia lives in a town whose streets are on a grid system [#permalink]

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29 Jun 2016, 05:18

Bunuel wrote:

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.

Quick question - why do we look at the entire series and not just the first two movements? I.e. For the first two blocks, she can go SE, ES, SS, EE and we want probability of one of these, so 1/4 ?

Please let me know why this is wrong?

gmatclubot

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