Probability Question

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antimatter1: Newbie | Next Rank: 10 Posts; Posts: 9; Joined: Wed May 13, 2009 4:46 am

by antimatter1 » Sun May 17, 2009 4:00 am

From GMATPrep Exam 1 (not verbatim):

If there are 12 temporary employees, 5 of which are women, and 4 of the 12 will be hired as permanent employees, how many possible combinations of 3 women and 1 man could be hired?

I understand that it came down to [(5*4*3*7)/2] = 70. Can anyone please explain the logic behind this?

Thanks in advance for the help!

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Source: Beat The GMAT — Problem Solving | Sun May 17, 2009 4:00 am

Jose Ferreira: GMAT Instructor; Posts: 99; Joined: Thu Mar 05, 2009 5:14 pmFollowed by: 9 members; Location: NYC; Thanked: 43 times; GMAT Score:800

by Jose Ferreira » Sun May 17, 2009 9:13 am

So, this is a collection probability. Let's say the five women are Annie, Barbara, Catherine, Denise and Elizabeth. Clearly we need to choose three from this bunch, so the tendency is to say 5*4*3. However, this misses the point that (Annie Barbara Catherine) is the same group as (Catherine Barbara Annie) or (Barbara Catherine Annie).

The formula for collections in this case is:
(5!)/(3!*2!)=10
or you could think of it as:
(Total!)/[(Number selected!)((Total-Number Selected)!)]

And think about this logically. To get a collection of three women from a group of five, we simply need to see how many combinations of two women we can exclude.

Here they are:
AB, AC, AD, AE, BC, BD, BE, CD, CE, DE
A total of ten.

Of course, for each of these 10 women groupings, there are 7 possible men that could be matched.

10*7=70

Jose Ferreira
Founder and CEO, Knewton, Inc.
https://www.knewton.com/gmat

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