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A three-person committee must be chosen from a group of 7

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A three-person committee must be chosen from a group of 7 [#permalink] New post 18 Jan 2006, 19:20
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A three-person committee must be chosen from a group of 7 professors and 10 graduate students. If at least one of the people on the committee must be a professor, how many different groups of people could be chosen for the committee?

A. 70
B. 560
C. 630
D. 1,260
E. 1,980

Plz, help me out with this, guys.

My solution: C(7;1)*C(16;2)=7*120=840
What's wrong with my reasoning here.
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 [#permalink] New post 18 Jan 2006, 19:34
total number of groups if the three people are chosen randomly= 17C3=680

total number of groups if no professor is chosen = 10C3 = 120

total number of groups if at least a professor is chosen = 680-120 =560
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 [#permalink] New post 18 Jan 2006, 19:35
I think the answer is B 560.

You could do it in two ways:

Total combinations - Combinations without any professor = 17C3 - 10C3 =560

OR

Exactly One professor + Exactly 2 professors + Exactly 3 professors
= 7C1*10C2 + 7C2*10C1+ 7C3 = 315 + 210 + 35 = 560
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 [#permalink] New post 18 Jan 2006, 21:33
Option are

1 Prof + 2 Student C(7,1) * C(10, 2) +
2 Prof + 1 Student C(7,2) * C(10, 1) +
3 Prof C(7,3) +
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 [#permalink] New post 19 Jan 2006, 03:08
Atleast 1 prof = All - No Prof
= 17C3 - 10C3 = 680-120 = 560
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 [#permalink] New post 19 Jan 2006, 08:45
Out of three places, 1 must be a professor. So let's consider the remaining two places.

Of those two places, we can either have: (professor, grad), (grad, grad) or (professor, professor)

Case 1: (professor, grad) -> Team will comprise 2 professors, 1 grad
# of ways = 7C2*10C1

Case 2: (grad, grad) -> Team will comprise 1 professor, 2 grads
# of ways = 7C1*10C2

Case 3: (professor, professor) -> Team will comprise 3 professors
# of ways = 10C3

Total # of ways = Case 1 + Case2 + Case 3
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Re: combination problem PS [#permalink] New post 30 Aug 2007, 13:19
rlevochkin wrote:
A three-person committee must be chosen from a group of 7 professors and 10 graduate students. If at least one of the people on the committee must be a professor, how many different groups of people could be chosen for the committee?

A. 70
B. 560
C. 630
D. 1,260
E. 1,980

Plz, help me out with this, guys.

My solution: C(7;1)*C(16;2)=7*120=840
What's wrong with my reasoning here.


many posted the right solution here, which is 560. but could some permutation / combination expert please explain why the solution proposed above: C(7;1)*C(16;2)=7*120=840 is not applicable here?? i cant find an explanation...

thanks a lot

EDIT:

with the help of my little sister i found the flaw myself... :-) the problem is when you select 2 out of 16, i.e. C(16;2) then these two guys might be two professors or one professor and you cannot multiply them with 7 (the #professors) since only 5 or 6 professors would be left. therefore, as the correct solution suggest, you have differentiate between the scenarios 1prof/2studs, 2profs/1stud, 3profs.

Last edited by skkingdom on 30 Aug 2007, 13:49, edited 1 time in total.
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Re: combination problem PS [#permalink] New post 30 Aug 2007, 13:47
skkingdom wrote:
many posted the right solution here, which is 560. but could some permutation / combination expert please explain why the solution proposed above: C(7;1)*C(16;2)=7*120=840 is not applicable here?? i cant find an explanation...

thanks a lot


not so sure, but this how i understand it. you are trying to multiply the probabilities of choosing from the two overlaping sets. so you have many duplicates.
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Re: combination problem PS [#permalink] New post 30 Aug 2007, 13:51
ankita wrote:
skkingdom wrote:
many posted the right solution here, which is 560. but could some permutation / combination expert please explain why the solution proposed above: C(7;1)*C(16;2)=7*120=840 is not applicable here?? i cant find an explanation...

thanks a lot


not so sure, but this how i understand it. you are trying to multiply the probabilities of choosing from the two overlaping sets. so you have many duplicates.


i think i just edited above at the same time you answered ;-) yes you are right, there is a considerable overlap, that makes the 840 possibilities incorrect
Re: combination problem PS   [#permalink] 30 Aug 2007, 13:51
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