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

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

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New post 16 Nov 2005, 02:01
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Of the 12 temporary employees in a certain company, 4 will be hired as permanent employees. If 5 of the 12 temporary employees are women, how many of the possible groups of 4 employees consist of 3 women and one man?

A. 22
B. 35
C. 56
D. 70
E. 105
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Re: GMAT Prep: Probabilities  [#permalink]

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New post 27 Dec 2009, 16:54
5
1
GMAT TIGER wrote:
dk94588 wrote:
This is a section I have had a lot of trouble with, and I was hoping someone could show me how to do this.

Of the 12 temporary employees in a certain company, 4 will be hired as permanent employees. If 5 of the 12 temporary employees are women, how many of the possible groups of 4 employees consist of 3 women and one man?

a) 22
b) 35
c) 56
d) 70
e) 105


Select 3 out of 5 = 5c3 = 5
Select 1 out of 7 = 7c1 = 7
Total = 5c3 x 7c1 = 5x7 = 35


Just a little correction 5C3=10 not 5. So, 5C3*7C1=10*7=70. (5C3 # of selections of 3 women out of 5 and 7C1 # of selections of 1 man out of 7)

Answer: D
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Re: GMAT Prep: Probabilities  [#permalink]

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New post 21 Feb 2010, 18:51
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dk94588 wrote:
This is a section I have had a lot of trouble with, and I was hoping someone could show me how to do this.

Of the 12 temporary employees in a certain company, 4 will be hired as permanent employees. If 5 of the 12 temporary employees are women, how many of the possible groups of 4 employees consist of 3 women and one man?

a) 22
b) 35
c) 56
d) 70
e) 105



The wording of this question is pure evil. First it gives you information regarding 4 permanent employees, then it asks you concerning groups of temporary employees. The information of the permanent employees is irrelevant to this problem. Make sure to understand exactly what the question is asking before attacking it.
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  [#permalink]

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New post 16 Nov 2005, 02:56
1
Select three women out of five in 5C3=10 ways.Select one man in 7 ways or 10x7=70 different groups of 3 women and 1 man.
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Re: PS: employees  [#permalink]

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New post 16 Nov 2005, 11:13
1
themagiccarpet wrote:
Of the 12 temporary employees in a certain company, 4 will be hired as permanent emploees. If 5 of the 12 temporary employees are women, how many of the possible groups of 4 temporary employees consist of 3 women and 1 man?

(A)22
(B)35
(C)56
(D)70
(E)105


Possible groups = Possible groups of 3 women * possible groups of 1 men

Since order does not matter, its a combination problem.

=[5!/(3!2!)} * [7!/(6!1!)]
=70
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Re: GMAT Prep: Probabilities  [#permalink]

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New post 27 Dec 2009, 12:56
2
dk94588 wrote:
This is a section I have had a lot of trouble with, and I was hoping someone could show me how to do this.

Of the 12 temporary employees in a certain company, 4 will be hired as permanent employees. If 5 of the 12 temporary employees are women, how many of the possible groups of 4 employees consist of 3 women and one man?

a) 22
b) 35
c) 56
d) 70
e) 105


Select 3 out of 5 = 5c3 = 5
Select 1 out of 7 = 7c1 = 7
Total = 5c3 x 7c1 = 5x7 = 35
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Re: GMAT Prep: Probabilities  [#permalink]

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New post 27 Dec 2009, 23:02
5c3*7c1=70

but come on 75% jobs for females? where is the right to equality :P
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Re: GMAT Prep: Probabilities  [#permalink]

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New post 29 Dec 2009, 04:41
5C3 * 7C1

= [5!/(5-3)!* 3!] = 7!/(7-1)! * 1!
=70

In general, the formula for nCr is n!/[(n-r)!] * r!

Hope this helps
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Re: GMAT Prep: Probabilities  [#permalink]

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New post 08 Jun 2012, 00:10
1
another approach :

one scenario : [WWWM]

now we have 5 ways for the first women , 4 ways for the 2nd women and , 3 ways for the third women.

and 7 ways to select one man

so 5*4*3*7 = 420

now the three women are same so combinations with win them are irrelevant the 3 women can we arranged in 3! ways

hence 420/3!= 420/6= 70
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Re: Of the 12 temporary employees in a certain company, 4 will  [#permalink]

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New post 24 Sep 2012, 06:13
Total number of Women = 5
Total Number of Men = 4

No of employers to be hired - 04
No. of Women employees to be hired - 03 (Given in the question)
No. of Men to be hired - 1 (given in the question)

Therefore number of Combinations = C(5,3) x C (7,1) = 10 x 7 = 70 (D)
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Re: Of the 12 temporary employees in a certain company,  [#permalink]

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New post 02 Jan 2013, 20:44
2
kiyo0610 wrote:
Of the 12 temporary employees in a certain company, 4 will be hired as permanent employees. If 5 of the 12 temporary employees are women, how many of the possible groups of 4 temporary employees consist of 3 women and 1 man?

(A) 22
(B) 35
(C) 56
(D) 70
(E) 105


Number of ways to select 3 women out of 5 is \(5C3\) and the number of ways to select 1 man out of 7 is \(7C1\).
Therefore number of possible groups of 4 temporary employees consist of 3 women and 1 man are \(7C1 * 5C3 = 70\).
There are 70 possible groups.
+1D
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Re: Of the 12 temporary employees in a certain company, 4 will  [#permalink]

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New post 04 Jan 2013, 04:14
I understand the question completely, however can somebody right the answer down when the order is important in both groups. I understand how to right it down with permutations like 5 P 3 and 7 P 1, however how do you right it down in faculty form?
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Re: Of the 12 temporary employees in a certain company, 4 will  [#permalink]

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New post 27 Oct 2018, 07:12
5C3=10
7C1=7
Total combinations=10*7=70
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Re: Of the 12 temporary employees in a certain company, 4 will  [#permalink]

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New post 30 Oct 2018, 17:33
Hi Bunuel, I was wondering could you please explain where I went wrong in my approach? I got E instead of D.

WWWM can be arranged in (4!/3!*1!) or 4 ways total

We have 5 ways to select the first women, 4 ways for the 2nd women and, 3 ways for the third women. Similarly, we have 7 ways to select one man

5*4*3*7 = 420

420/4 = 105
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Re: Of the 12 temporary employees in a certain company, 4 will &nbs [#permalink] 30 Oct 2018, 17:33
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