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# What is the number of different committees of 4 people that

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Manager
Joined: 12 Jun 2006
Posts: 55
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What is the number of different committees of 4 people that [#permalink]  11 Aug 2007, 19:48
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Question Stats:

100% (00:00) correct 0% (00:00) wrong based on 4 sessions
What is the number of different committees of 4 people that can be selected from a group of 10 people

a) 20
b) 40
c) 80
d) 120
e) 210

Thanks for the help!
Director
Joined: 03 May 2007
Posts: 894
Schools: University of Chicago, Wharton School
Followers: 6

Kudos [?]: 63 [0], given: 7

Re: GMATPrep PS [#permalink]  11 Aug 2007, 20:08
lonesome307 wrote:
What is the number of different committees of 4 people that can be selected from a group of 10 people

a) 20
b) 40
c) 80
d) 120
e) 210

Thanks for the help!

= (10x9x8x7x6!)/(6!x4x3x2)
= 1,260
Director
Joined: 11 Jun 2007
Posts: 932
Followers: 1

Kudos [?]: 70 [0], given: 0

Re: GMATPrep PS [#permalink]  11 Aug 2007, 20:13
lonesome307 wrote:
What is the number of different committees of 4 people that can be selected from a group of 10 people

a) 20
b) 40
c) 80
d) 120
e) 210

Thanks for the help!

this is an easy one with no restrictions

10C4 = 10! /6!*4!

10*9*8*7 / 4*3*2*1 = 210 (E)
Manager
Joined: 18 Sep 2006
Posts: 58
Followers: 1

Kudos [?]: 1 [0], given: 0

10C4 = 10!/((10! - 4!)*4!)
= 10*9*8*7/4*3*2*1
= 210 (E)
Manager
Joined: 12 Jun 2006
Posts: 55
Followers: 1

Kudos [?]: 7 [0], given: 0

Thanks guys for the quick response. How would I know when I need to use these permutation?? formula?

saurster wrote:
10C4 = 10!/((10! - 4!)*4!)
= 10*9*8*7/4*3*2*1
= 210 (E)
Director
Joined: 11 Jun 2007
Posts: 932
Followers: 1

Kudos [?]: 70 [0], given: 0

lonesome307 wrote:
Thanks guys for the quick response. How would I know when I need to use these permutation?? formula?

saurster wrote:
10C4 = 10!/((10! - 4!)*4!)
= 10*9*8*7/4*3*2*1
= 210 (E)

This is actually a combination problem because order does not matter. One of the ways I differentiate between the two is if the question is asking for an arrangement, order will matter so I use permutation. If the question is asking about selection or number of ways of listing something, we apply combinatorial principle. The reason why is because with combination problems order does not matter so we will be double counting. Hope that helps and good luck!
Director
Joined: 03 May 2007
Posts: 894
Schools: University of Chicago, Wharton School
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Re: GMATPrep PS [#permalink]  12 Aug 2007, 05:59
Fistail wrote:
lonesome307 wrote:
What is the number of different committees of 4 people that can be selected from a group of 10 people

a) 20
b) 40
c) 80
d) 120
e) 210

Thanks for the help!

= (10x9x8x7x6!)/(6!x4x3x2)
= 1,260

silly me!!!
wat a stupid calculation i had???????????????

= (10x9x8x7x6!)/(6!x4x3x2)
= 210
Re: GMATPrep PS   [#permalink] 12 Aug 2007, 05:59
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