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Combo problems. Why is one method right and the other wrong

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by Jinglander » Wed Sep 29, 2010 4:01 pm
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?


Answer is 100 i got it wrong

this is what I did

choose one prof first 4 way to do this. Then choose 2 from the remain 9 so 2C9 thus 4*2C9

I could also see doing this the way that generates the right answer which is 3C10 - 6C3. They both seem right
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by diebeatsthegmat » Wed Sep 29, 2010 5:07 pm
Jinglander wrote: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?


Answer is 100 i got it wrong

this is what I did

choose one prof first 4 way to do this. Then choose 2 from the remain 9 so 2C9 thus 4*2C9

I could also see doing this the way that generates the right answer which is 3C10 - 6C3. They both seem right
because it asks you to find at least 1 professor thus you can have 3 cases
the frist is select 1 professor and 2 students = 4*6!/2!/4!=60
the second is select 2 professors and 1 student=4!/2!2!*6!/5!*1!=36
the last is select all 3 professors=4!/3!1!=4
total = 36+4+60 =100
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by Ian Stewart » Thu Sep 30, 2010 12:27 pm
Jinglander wrote: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?

this is what I did

choose one prof first 4 way to do this. Then choose 2 from the remain 9 so 2C9 thus 4*2C9
Say your professors are A, B, C and D, and we choose a committee of three professors. If we choose A first, then choose B and C, we get the same committee as if we choose B first, then choose A and C; the order of the three professors does not matter. In your solution, you are assuming that the order of the professors does, at least partly, matter - you're picking one professor as the 'first professor', and then you're selecting the rest of the committee. Because of that, you're overcounting.

The answer you give would be correct if the question assigned a position to one of the committee members - for example, if it asked 'If a three person committee will be chosen, and one professor must be chosen as the chairperson of the committee, how many committees can be chosen?" We're then assigning a position to one professor, and your solution would be correct.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by sanabk » Thu Sep 30, 2010 1:52 pm
If the committee has to include at least 1 professor, then the number of ways = Total ways-(never case / selecting all 3 students)

Total ways = 10c3=120
Never = 6c3=20

The no. of ways the committee can be formed is = 120-20=100
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