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# Combi theory and concept... pls help...

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Manager
Joined: 18 Oct 2005
Posts: 79
Location: Hong Kong
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Kudos [?]: 2 [0], given: 0

Combi theory and concept... pls help... [#permalink]  06 Dec 2005, 23:00
I took this from challenge 25, question 3.

4 professors and 6 students are being considered for membership on a supervisory committee which must consist of 3 people. In how many ways can this committe be formed if it has to include at least one professor?

I post the answer immediately since it is anyway available on the challenge site.
The best way to approach this problem is to consider an unconstrained version of the question first: how many committees of 3 are possible? The answer is 10C3 = 10!/(7!*3!) = 120. From this figure we have to subtract the number of committees that consist entirely of students i.e. 6C3 = 6!/(3!*3!) = 20. The final answer is 10C3 - 6C3 = 120 - 20 = 100.

My question is this....Why can't I do this...
I need one professor in the team. So 4C1=4.
Then I need 2 guys from the remaining 3 professors and 6 students. Therefore i use 9C2=36.
Total nos of combinations= 36*4=144? Why is this wrong?
SVP
Joined: 28 May 2005
Posts: 1743
Location: Dhaka
Followers: 6

Kudos [?]: 72 [0], given: 0

Re: Combi theory and concept... pls help... [#permalink]  07 Dec 2005, 02:03
vic wrote:
I took this from challenge 25, question 3.

4 professors and 6 students are being considered for membership on a supervisory committee which must consist of 3 people. In how many ways can this committe be formed if it has to include at least one professor?

I post the answer immediately since it is anyway available on the challenge site.
The best way to approach this problem is to consider an unconstrained version of the question first: how many committees of 3 are possible? The answer is 10C3 = 10!/(7!*3!) = 120. From this figure we have to subtract the number of committees that consist entirely of students i.e. 6C3 = 6!/(3!*3!) = 20. The final answer is 10C3 - 6C3 = 120 - 20 = 100.

My question is this....Why can't I do this...
I need one professor in the team. So 4C1=4.
Then I need 2 guys from the remaining 3 professors and 6 students. Therefore i use 9C2=36.
Total nos of combinations= 36*4=144? Why is this wrong?

The correct way to do this prombem is

3 member can be 3 professors or (2 professors, 1 student) or (1 professor, 2 student) or 3 student

=4C3 + 4C2*6C1+ 4C1*6C2 + 6C3
= 4 + 6*6 +4*15+20
= 120
_________________

hey ya......

Manager
Joined: 18 Oct 2005
Posts: 79
Location: Hong Kong
Followers: 1

Kudos [?]: 2 [0], given: 0

Thanks a lot Nakib.

you confirm my understanding.
One can really get lost with all these combi and permutation stuff...
SVP
Joined: 28 May 2005
Posts: 1743
Location: Dhaka
Followers: 6

Kudos [?]: 72 [0], given: 0

vic wrote:
Thanks a lot Nakib.

you confirm my understanding.
One can really get lost with all these combi and permutation stuff...

you are welcome vic.
_________________

hey ya......

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