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

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16 Nov 2005, 03:01

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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?

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.

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

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27 Dec 2009, 12:42

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

Last edited by Bunuel on 08 Jun 2012, 03:06, edited 1 time in total.

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
_________________

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)

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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Re: Of the 12 temporary employees in a certain company, [#permalink]

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02 Jan 2013, 21:44

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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
_________________

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

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04 Jan 2013, 05:14

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?

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

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25 Nov 2014, 02:24

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

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

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11 Apr 2017, 13:45

Hello from the GMAT Club BumpBot!

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