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# limiting combination question -- sounds easy but is it?

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Intern
Joined: 02 Jan 2003
Posts: 3
limiting combination question -- sounds easy but is it? [#permalink]

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20 Nov 2003, 09:15
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

here is the question :
20 people are at the firm. 5 to be sent away on business. how many groups can be formed providing that Pres, VP and Accountant cant go altogether among 5.

Here is the way i approached it:
20-3(that cant go) = 17 than a simple combination pick 17!/(5!*12!)

here is the way Gmat Club solved it:
total number of combinations =20!/5!15!
restricting number of combinations with 3 that cant go 17!/)(2!15!)

than subctracting total minus restricted..

why cant the first method work i thought its the same?
whats the reasoning? Thanks in advance!!
Intern
Joined: 20 Oct 2003
Posts: 13
Location: Moscow

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20 Nov 2003, 12:03
With this kind of combinatoric problems with limitations the best & the shortest way to solve is to calculate the total number of possible combinations and substract all the cases that fall under the limiting condition.

Specificaly:
a. Total number of combinations is 5C20 = 20!/ (5!15!)
b. Now imagin that you have 5 chairs, 1st, 2nd & 3rd are taken by Pres, VP and accountant respectively. You have only 4th & 5th to be taken by someone else. i.e for the 4th place we have 17 managers, for 5th place we have 16 => C(17,2) = 17!/(2!15!)
c. Total - limits = result
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
Re: limiting combination question -- sounds easy but is it? [#permalink]

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22 Nov 2003, 05:32
Leon_v wrote:
here is the question :
20 people are at the firm. 5 to be sent away on business. how many groups can be formed providing that Pres, VP and Accountant cant go altogether among 5.

Here is the way i approached it:
20-3(that cant go) = 17 than a simple combination pick 17!/(5!*12!)

here is the way Gmat Club solved it:
total number of combinations =20!/5!15!
restricting number of combinations with 3 that cant go 17!/)(2!15!)

than subctracting total minus restricted..

why cant the first method work i thought its the same?
whats the reasoning? Thanks in advance!!

This question is poorly worded. Does this mean that NONE of the 3 can go? or does this mean that any none, one, or two of the 3 can go so long as all 3 don't go?
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

SVP
Joined: 30 Oct 2003
Posts: 1790
Location: NewJersey USA

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19 Dec 2003, 17:38
I believe we need to choose 5 people to form a commitee from 20 people with the restriction that President and VP cannot be together.
This is basically
( total combinations ) - ( combinations where VP , A & P go together )

Desired combinations = 20C5 - 18C2
Director
Joined: 13 Nov 2003
Posts: 789
Location: BULGARIA

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20 Dec 2003, 06:53
I agree with AkamaiBrah that it is not clear if 2 people can go together or one alone.
20 Dec 2003, 06:53
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