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 Post subject: Mystery Teammates - Combinatorics
 Post Posted: Tue Apr 07, 2009 7:44 am 
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Course Students


Posts: 5
Hello,

This question is from the Word Translations Question bank. The exact working on the question is as follows:

How many different 5-person teams can be formed from a group of x individuals?

(1) If there had been x + 2 individuals in the group, exactly 126 different 5-person teams could have been formed.

(2) If there had been x + 1 individuals in the group, exactly 56 different 3-person teams could have been formed.

I answered this question correctly but for the wrong reasons...When reading through the answer explanation- I don't understand how to get to the following equation for statement 1: 9! ÷ [(5!)(4!)] = 126; and the following equation for statement 2:8! ÷ [(5!)(3!)] = 56). Specifically I don't understand why you have to divide by 4! and 3!.

Thanks for your help in advance!


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 Post subject: Re: Mystery Teammates - Combinatorics
 Post Posted: Wed Apr 08, 2009 5:32 am 
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ManhattanGMAT Staff


Posts: 7146
karishma.bhargava wrote:
Specifically I don't understand why you have to divide by 4! and 3!.

Thanks for your help in advance!


have you read the word translations strategy guide, and/or studied some sort of reference material on combinations?

this is the standard form of the formula for combinations. if you need to select r items from a total pool of n items, then the total number of ways in which you can make that selection is
n! / (r! (n - r)!)

in our strategy guide, this formula is also the general result of our ANAGRAM GRID method. if you don't feel like memorizing the formula itself, or you have difficulty doing so, then you can apply the "anagram grid" to each of the 2 problems.

note that the two numbers whose factorials appear in the denominator will add up to the number n. this happens above: 3 + 5 is indeed 8, and 4 + 5 is indeed 9.

--

in any case, you don't have to do any actual math to solve this problem.

it should be clear that the larger the pool, the more combinations that can be chosen.
the consequence is that if i tell you that there are a specific number of ways of choosing a group of some specific size, then you MUST know the size of the overall pool. for instance, if there are 126 ways of choosing 5 things out of a pool of n, then there can only be ONE value of n satisfying this criterion. we know this is the case, because any larger pool would give more than 126 such combinations, and any smaller pool would give fewer.

remember that, since this is data sufficiency, you don't care what the actual numbers are! just being able to find them, or just knowing that you can find them, is plenty good.

so, here's the minimal thought process required to solve this problem:

(1)
oh hey, this means i have a unique value for x + 2.
therefore, i have a unique value for x.
therefore, i can find how many combinations of 5 items i'd have.

(2)
oh hey, this means i have a unique value for x + 1.
therefore, i have a unique value for x.
therefore, i can find how many combinations of 5 items i'd have.

so, ans = (d)


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 Post subject: Re: Mystery Teammates - Combinatorics
 Post Posted: Wed Apr 08, 2009 11:05 pm 
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Course Students


Posts: 5
Thanks for your help! I need to go back and refresh my memory on that chapter! :)


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 Post subject: Re: Mystery Teammates - Combinatorics
 Post Posted: Sun Apr 12, 2009 11:00 am 
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Posts: 10
Heyy I have a doubt...
here they are asking how many different teams can be formed so we need to know the value of X to determine the value and for that we need both A abd B options to calculate the value...which comes out to be 7.
so I think the ans needs to be C.

Please let me know if I am incorrect


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 Post subject: Re: Mystery Teammates - Combinatorics
 Post Posted: Mon Apr 13, 2009 9:06 pm 
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ManhattanGMAT Staff


Posts: 901
Location: St. Louis, MO
kinjal.nandy wrote:
here they are asking how many different teams can be formed so we need to know the value of X to determine the value and for that we need both A abd B options to calculate the value...which comes out to be 7.
so I think the ans needs to be C.

Please let me know if I am incorrect

Your rephrased question "What is x?" is correct.

However, you are incorrect about which statement(s) provide the answer. The correct answer is D, either statement alone provides x.

Here is the math translation of each statement:
(1) (x+2)!/5!(x+2-5)! = 126 --> A little trial and error reveals that x+2=9, so x = 7.

(2) (x+1)!/5!(x+1-5)! = 56 --> A little trial and error reveals that x+1=8, so x = 7.

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Emily Sledge
Instructor
ManhattanGMAT


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 Post subject: Re: Mystery Teammates - Combinatorics
 Post Posted: Sat Nov 05, 2011 5:05 am 
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Students


Posts: 45
Dear Esledge,

Thanx for the ans.

BR


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 Post subject: Re: Mystery Teammates - Combinatorics
 Post Posted: Wed Nov 23, 2011 2:03 pm 
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ManhattanGMAT Staff


Posts: 2242
Location: Southwest Airlines, seat 21C
On behalf of esledge, you’re welcome!

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Tim Sanders
Manhattan GMAT Instructor


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