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my gmat is in 10 hours please help!

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by houstonrockets16 » Sun Jun 28, 2009 10:31 pm
In a certain group of ten members, 4 teach french, and the rest teach spanish or german. if a group of 3 is chosen, and at least one must teach french; how many committes can be chosen?

100

please tell me how!! thanks!
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by shanmugam.d » Sun Jun 28, 2009 11:28 pm
The key here is atleast 1 french teacher:
so it is the sum of combinations of 1 french teacher + 2 french teachers + all 3 french teachers
(If 4 are french teachers then remaining 6 are either spanish or german)

-> 4C1x6C2 + 4C2x6C1 + 4C3
= 60+36+4
= 100
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Re: my gmat is in 10 hours please help!

by GMATQuantCoach » Mon Jun 29, 2009 5:58 am
houstonrockets16 wrote:In a certain group of ten members, 4 teach french, and the rest teach spanish or german. if a group of 3 is chosen, and at least one must teach french; how many committes can be chosen?

100

please tell me how!! thanks!
In combinatorics, when you see the words "at least", most of the time it is easier to think of the complement. In our case, the complement of at least one must teach french is none teaches french. If you can find the # of ways forming the committee with no one teaching french, subtract that from the # of ways of forming the committee when there is no restrictions to get your answer.

Choose 3 from 10: 10C3
Choose 3 from the 6 that don't teach french: 6C3

Therefore 10C3 - 6C3 = 120 - 20 = 100

Different approach, but same answer. :)
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by aj5105 » Mon Jun 29, 2009 7:16 am
4C1x6C2 + 4C2x6C1 + 4C3 = 100
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by VP_Jim » Mon Jun 29, 2009 8:08 am
If your GMAT is in 10 hours (actually, less now), relax! You've done all the studying that you could! Cramming for the GMAT is not a good idea!
Jim S. | GMAT Instructor | Veritas Prep
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by hurdler » Tue Jun 30, 2009 5:10 pm
Could someone tell me where my solution is faulty?

If I'm reading this problem correctly, we have 10 people, and 4 of them teach French; the other 6 do not teach French. We need to choose a committee of people, with the condition that one member of the committee teaches French.

So, 1 slot on the committee has 4 possible options (the 4 French teachers). The only condition has been satisfied.

There are 9 people remaining, with only 2 slots to fill; so, 9C2.

4 (French teachers eligible for 1 slot) * 9C2 (Remaining 9 people eligible for 2 slots) = 144

Thanks.
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by VP_Jim » Tue Jun 30, 2009 5:28 pm
They key issue on this problem - and what makes it hard - is that AT LEAST one must teach French. That means the team might have one French teacher, two French teachers, or three French teachers. We have to find the number of combinations for each of those scenarios and then add them together. The same is true for any "at least" combinatorics problem.
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by shibal » Wed Jul 01, 2009 7:20 pm
let spice up things... instead of having at least one, how many combinations could be chosen with exactly 2 french guys.....
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by hurdler » Thu Jul 02, 2009 4:34 pm
Could anyone explain where my solution above breaks down? I understand the reasoning behind how to calculate the correct answer (100), but I'm not sure why my particular solution (totaling 144) is wrong... thanks!
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by doclkk » Sun Jul 05, 2009 6:59 am
hurdler wrote:Could anyone explain where my solution above breaks down? I understand the reasoning behind how to calculate the correct answer (100), but I'm not sure why my particular solution (totaling 144) is wrong... thanks!
so how'd you do man?
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