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Of the 12 temporary employees in a certain company, 4 will [#permalink]
 30 May 2006, 17:16
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]
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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Director
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Agree with Prof, 70 it is.
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I'm sorry, how do you get from 5c3 to 10?
I'm terrible at these types of questions!
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Ans - 70
5C3*7C1 = 10*7 = 70
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Director
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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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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=1510C3=10*9*8/1*2*3=1205C3 = (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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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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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=1510C3=10*9*8/1*2*3=1205C3 = (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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A little late - 5C3*7C1 = 70
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