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What is the number of different committees of 4 people that

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Manager
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What is the number of different committees of 4 people that [#permalink] New post 11 Aug 2007, 19:48
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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
Director
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Joined: 03 May 2007
Posts: 899
Schools: University of Chicago, Wharton School
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Re: GMATPrep PS [#permalink] New post 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 :roll:
Director
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Re: GMATPrep PS [#permalink] New post 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
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 [#permalink] New post 11 Aug 2007, 22:20
10C4 = 10!/((10! - 4!)*4!)
= 10*9*8*7/4*3*2*1
= 210 (E)
Manager
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 [#permalink] New post 11 Aug 2007, 23:58
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
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 [#permalink] New post 12 Aug 2007, 00:08
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
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Re: GMATPrep PS [#permalink] New post 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 :roll:


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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