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Of the 12 temporary employees in a certain company, 4 will

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Manager
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Of the 12 temporary employees in a certain company, 4 will [#permalink] New post 30 May 2006, 16:16
Of the 12 temporary employees in a certain company, 4 will be hired as permanent employees. If 5 of the 12 temporary are women. how many possible groups of 4 temporary employees consist of 3 women an 1 man

22
35
56
70
105
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Re: GMATPREP : probability [#permalink] New post 30 May 2006, 16:58
dinesh8 wrote:
Of the 12 temporary employees in a certain company, 4 will be hired as permanent employees. If 5 of the 12 temporary are women. how many possible groups of 4 temporary employees consist of 3 women an 1 man?

22
35
56
70
105


=5c3x7c1
= 10x7
=70
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 [#permalink] New post 30 May 2006, 18:55
Agree with Prof, 70 it is.
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 [#permalink] New post 30 May 2006, 20:41
I'm sorry, how do you get from 5c3 to 10?

I'm terrible at these types of questions!
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 [#permalink] New post 30 May 2006, 21:06
Ans - 70

5C3*7C1 = 10*7 = 70
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 [#permalink] New post 30 May 2006, 21:08
bz9 wrote:
I'm sorry, how do you get from 5c3 to 10?

I'm terrible at these types of questions!

5C3=5*4*3/3*2*1=10 hope this helps :wink:
And for example
6C4=6*5*4*3/4*3*2*1=15
10C3=10*9*8/1*2*3=120
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 [#permalink] New post 30 May 2006, 21:21
Yurik79 wrote:
bz9 wrote:
I'm sorry, how do you get from 5c3 to 10?

I'm terrible at these types of questions!

5C3=5*4*3/3*2*1=10 hope this helps :wink:
And for example
6C4=6*5*4*3/4*3*2*1=15
10C3=10*9*8/1*2*3=120


5C3 = (5x4x3!)/3!2! = (5x4x3x2x1)/[(3x2x1)(2x1)] = 10
6C4=(6*5*4!)/(4!2!) = 15
10C3=(10*9*8x7!)/(7!x3x2x1) = 120
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 [#permalink] New post 31 May 2006, 19:57
Thanks Everyone,

Just so I'm 100% clear, what is your approach if there are 7 women and 5 men in this question?

Thanks,
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 [#permalink] New post 31 May 2006, 21:25
HIMALAYA wrote:
Yurik79 wrote:
bz9 wrote:
I'm sorry, how do you get from 5c3 to 10?

I'm terrible at these types of questions!

5C3=5*4*3/3*2*1=10 hope this helps :wink:
And for example
6C4=6*5*4*3/4*3*2*1=15
10C3=10*9*8/1*2*3=120


5C3 = (5x4x3!)/3!2! = (5x4x3x2x1)/[(3x2x1)(2x1)] = 10
6C4=(6*5*4!)/(4!2!) = 15
10C3=(10*9*8x7!)/(7!x3x2x1) = 120

Yes I know the formula ))Just simplified a little
in 6C4=(6*5*4!)/(4!2!) = 15 4! is cancelled anyway
5C3 = (5x4x3!)/3!2! = (5x4x3x2x1)/[(3x2x1)(2x1)] = 10 same here 3! is canceled :wink:
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 [#permalink] New post 06 Jun 2006, 12:02
A little late - 5C3*7C1 = 70
  [#permalink] 06 Jun 2006, 12:02
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