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Standing on the origin of an xy-coordinate plane, John takes [#permalink]

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13 Feb 2012, 21:36

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Difficulty:

85% (hard)

Question Stats:

51% (03:14) correct
49% (02:23) wrong based on 103 sessions

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Standing on the origin of an xy-coordinate plane, John takes a 1-unit step at random in one of the following 4 directions: up, down, left, or right. If he takes 3 more steps under the same random conditions, what is the probability that he winds up at the origin again?

This is the Manhattan Gmat Problem of the week this week. Extremely difficult to explain this without drawing. But lets suppose he takes a step in any one direction. Since he can take 4 different directions from there on in and he does this 3 times the total number of possibilities is \(4*4*4=64\)

Now if you start drawing on a piece of paper, you will realise that there are \(9\) such possibilities where he can end up back on the origin so answer should be \(\frac{9}{64}\). I am going to attach an image of all these possibilities along-with this post as well.

Attachment:

routes.jpg [ 90.8 KiB | Viewed 2486 times ]

Hence B _________________

"Nowadays, people know the price of everything, and the value of nothing."Oscar Wilde

Standing on the origin of an xy-coordinate plane, John takes a 1-unit step at random in one of the following 4 directions: up, down, left, or right. If he takes 3 more steps under the same random conditions, what is the probability that he winds up at the origin again?

(A) 7/64 (B) 9/64 (C) 11/64 (D) 13/64 (E) 15/64

Find the total number of possibilities first. He takes total 4 steps . He can take each step in any direction so there are a total of 4*4*4*4 possibilities (this includes UUUU, UDLR, DDLR etc etc)

He needs to be at the origin after 4 steps. So if he takes a step up, he needs to take a step down at some time. If he takes a step to the left, he needs to take one to the right at some time. Say if he takes two steps in this way - UL, his next two steps are defined - DR/RD. If instead, he takes two steps in this way - UU, his next two steps have to be DD. There are two possibilities: 1. He goes only Up and Down or only Left and Right. UUDD can be arranged in 4!/(2!*2!) ways (includes UDUD, DUDU, DDUU etc). LLRR can also be arranged in 4!/(2!*2!) ways. 2. He goes Up/Down as well as Left/Right. UDLR can be arranged in 4! ways.

Total possible arrangements = 2*4!/(2!*2!) + 4!

Probability he comes back to the origin = (2*4!/(2!*2!) + 4!)/4*4*4*4 = 9/64
_________________

Re: Standing on the origin of an xy-coordinate plane, John takes [#permalink]

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26 Sep 2013, 22:31

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Standing on the origin of an xy-coordinate plane, John takes [#permalink]

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20 May 2014, 23:57

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This post received KUDOS

It took me about 3.5 mins to solve...I forgot some cases initially

Total no of cases=4^4 Way 1- 1 each of L,R,U,D- They can be arranged in 4! ways..4! Way 2- 2 each of R & L..RRLL- They can be arranged in 4!/2!*2!= 6 Way 2- 2 each of D & U..DDUU- They can be arranged in 4!/2!*2!= 6

36 ways possible/4*4*4*4 =9/64

I think it helps to think directions as numbers with opposite signs...In this case the sum of the 4 numbers should be 0
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Appreciate the efforts...KUDOS for all Don't let an extra chromosome get you down..

Re: Standing on the origin of an xy-coordinate plane, John takes [#permalink]

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25 Jan 2016, 22:23

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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