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# M25 #3

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Manager
Affiliations: Beta Gamma Sigma
Joined: 14 Aug 2008
Posts: 209
Schools: Harvard, Penn, Maryland

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23 May 2009, 22:33
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 $$C_{10}^3 - C_6^3 = 120 - 20 = 100$$ .
Founder
Joined: 04 Dec 2002
Posts: 15148
Location: United States (WA)
GMAT 1: 750 Q49 V42

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24 May 2009, 23:47

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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Founder
Joined: 04 Dec 2002
Posts: 15148
Location: United States (WA)
GMAT 1: 750 Q49 V42

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24 May 2009, 23:55

Continue discussions in the appropriate thread.
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Re: M25 #3   [#permalink] 24 May 2009, 23:55
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# M25 #3

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