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# GMAC 11th Edition Review Guide PS Question

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Intern
Joined: 04 Jan 2009
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GMAC 11th Edition Review Guide PS Question [#permalink]

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08 Mar 2009, 19:57
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In the 11th Edition, there is a PS problem that I cannot figure out:

Pat will walk from intersection X to intersection Y and to get there, he must walk upwards (U) twice and rightwards (R) three times. How many routes can he take to have the minimum possible length?

The answer is Ten, but I have no clue how to do it more easily. The answer lists the possible combinations but does not provide any formula.

UURRR
URRUR
URRRU
etc...

Does anyone know a faster way to figure this problem out without listing all possibilities and counting them? Thanks!

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Senior Manager
Joined: 30 Nov 2008
Posts: 482
Schools: Fuqua
Re: GMAC 11th Edition Review Guide PS Question [#permalink]

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08 Mar 2009, 22:08
Well, this is somewhat a disguise question of P & C(Permutations and Combinations).

Patt must walk 2 ways upwards(U) and 3 ways to the right(R). In how many ways can he take the route.

This is nothing but to find out "In how ways can we arrange the letters UURRR? "

Ans - 5 ! / (2! * 3!) = 10.
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Re: GMAC 11th Edition Review Guide PS Question [#permalink]

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09 Mar 2009, 16:15
Why is it that the answer to the following question is solved in the exact same way?

-How many different groups of 3 people can be formed from a group of 5 people?

To solve this problem, don't I also use the equation: 5!/(3! x 2!) = 10 ? In this case, n!/k!(n-k)!.

Now I'm confused why in the previous question, there are 2U's and 3 R's, making a total of 5 letters. However, I am not selecting only 3 letters for my combination so why does is this problem solved with the formula n!/k!(n-k)! ? Please advise. Thanks!
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Re: GMAC 11th Edition Review Guide PS Question [#permalink]

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09 Mar 2009, 20:24
I will take a stab at explaining this.

1. The original problem you had asked is about Permutation with repeating elements. The formula to solve those is Total number of permutations = N!/(K!*L!) where N is the number of items where K of the N elements are of the same type as well as L of the N elements are of the same type.

In your example: N = 5 (U,U,R,R,R)
K = 2 (U,U)
L=3 (R,R,R)

N!/(K!*L!) = 5!/(2!*3!) = 10

2. Removing second solution as I was thinking permutation and not combination

I hope this helps.

Last edited by nookway on 09 Mar 2009, 20:35, edited 1 time in total.
Senior Manager
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Re: GMAC 11th Edition Review Guide PS Question [#permalink]

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09 Mar 2009, 20:29
jenyang5268 wrote:
Why is it that the answer to the following question is solved in the exact same way?

-How many different groups of 3 people can be formed from a group of 5 people?

To solve this problem, don't I also use the equation: 5!/(3! x 2!) = 10 ? In this case, n!/k!(n-k)!.

Now I'm confused why in the previous question, there are 2U's and 3 R's, making a total of 5 letters. However, I am not selecting only 3 letters for my combination so why does is this problem solved with the formula n!/k!(n-k)! ? Please advise. Thanks!

The two questions you are asking are different.

-How many different groups of 3 people can be formed from a group of 5 people?

This is a combination since the order in which 3 people are selected is NOT important. It is just the selection. Hence 5C3 is correct.

Applying the formula, nCk = n! / (r! * n-r!), we have 5! / (2! * 3!) = 10.

Now coming to the original question, there are 5 paths where 2 paths are upwards(U) and 3 paths are sideways(S).

To reach from source to distination, the order of the path is important. So this means it is Permutation.

Let me take an example - How many ways can we arrange PAGES. 5 distinct letters are there and order is important so it is 5!.
How many ways can we arrange FEELS - Again 5 words with a E repeated twice. It will be 5! / 2!.

Similary coming to our problem, UUEEE can be arranged in 5! / (2! * 3!) since U is repeated twice and E is repeated thrice.

http://regentsprep.org/regents/math/permut/LpermRep.htm

Google for Permutations with Repitions and you should find some information.

Last edited by mrsmarthi on 09 Mar 2009, 20:47, edited 1 time in total.
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Re: GMAC 11th Edition Review Guide PS Question [#permalink]

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09 Mar 2009, 20:39
Oh ok great! Thanks a lot =)

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: GMAC 11th Edition Review Guide PS Question   [#permalink] 09 Mar 2009, 20:39
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# GMAC 11th Edition Review Guide PS Question

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