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

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Manager
Joined: 17 Jan 2006
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Of the 12 temporary employees in a certain company, 4 will [#permalink]

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30 May 2006, 17:16
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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VP
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Re: GMATPREP : probability [#permalink]

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30 May 2006, 17:58
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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Senior Manager
Joined: 08 Jun 2004
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Location: Europe

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30 May 2006, 19:55
Agree with Prof, 70 it is.

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Manager
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30 May 2006, 21:41
I'm sorry, how do you get from 5c3 to 10?

I'm terrible at these types of questions!

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SVP
Joined: 30 Mar 2006
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30 May 2006, 22:06
Ans - 70

5C3*7C1 = 10*7 = 70

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Director
Joined: 10 Oct 2005
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30 May 2006, 22:08
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
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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SVP
Joined: 05 Apr 2005
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30 May 2006, 22:21
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
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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Manager
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31 May 2006, 20:57
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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Director
Joined: 10 Oct 2005
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31 May 2006, 22:25
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
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
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Manager
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06 Jun 2006, 13:02
A little late - 5C3*7C1 = 70

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06 Jun 2006, 13:02
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# Of the 12 temporary employees in a certain company, 4 will

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