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M25 #3

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Affiliations: Beta Gamma Sigma
Joined: 14 Aug 2008
Posts: 211
Schools: Harvard, Penn, Maryland
Followers: 4

Kudos [?]: 27 [0], given: 3

M25 #3 [#permalink] New post 23 May 2009, 21:33
not that I don't agree with the answer, but could someone please explain the formulas, I see that it may help me in the future to know them, and I dont have a stats textbook handy..

4 professors and 6 students are being considered for membership on a supervisory committee which must consist of 3 people. If the committee has to include at least 1 professor, how many ways can this committee be formed?

4 professors and 6 students are being considered for membership on a supervisory committee which must consist of 3 people. If the committee has to include at least 1 professor, how many ways can this committee be formed?
* 36
* 60
* 72
* 80
* 100

First, consider an unconstrained version of the question: how many committees of 3 are possible? The answer is C_{10}^3 = \frac{10!}{(7!3!)} = 120 . Now subtract the number of committees that consist entirely of students i.e. C_{6}^3 = \frac{6!}{(3!3!)} = 20 . The final answer is C_{10}^3 - C_6^3 = 120 - 20 = 100 .
The correct answer is E.
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WE: Information Technology (Hospitality and Tourism)
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Re: M25 #3 [#permalink] New post 24 May 2009, 22:47
Expert's post
Please consider these two links for the details:

Combinations: http://en.wikipedia.org/wiki/Combination

Permutations: http://en.wikipedia.org/wiki/Permutation

Both of these come from the Combinatorics field of Math and answer a specific question of how many combinations are possible in any given situation based on the total number of members and any limitations.

The main difference between C and P is that in Combinations order does not matter and Permutations order matters - see this article for the easy explanation of that difference: http://betterexplained.com/articles/eas ... binations/
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Expert Post
Founder
Founder
User avatar
Affiliations: UA-1K, SPG-G, HH-D
Joined: 04 Dec 2002
Posts: 12340
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
WE: Information Technology (Hospitality and Tourism)
Followers: 2323

Kudos [?]: 9415 [0], given: 3699

GMAT ToolKit User Premium Member CAT Tests
Re: M25 #3 [#permalink] New post 24 May 2009, 22:55
Expert's post
This question has already been discussed. Please check this thread gmat-club-tests-master-threads-all-tests-78599.html

Continue discussions in the appropriate thread.
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Re: M25 #3   [#permalink] 24 May 2009, 22:55
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M25 #3

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