A three-person committee must be chosen from a group of 7 : PS Archive
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# A three-person committee must be chosen from a group of 7

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Founder
Affiliations: AS - Gold, HH-Diamond
Joined: 04 Dec 2002
Posts: 14457
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
Followers: 3727

Kudos [?]: 23047 [0], given: 4515

A three-person committee must be chosen from a group of 7 [#permalink]

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26 Feb 2003, 22:48
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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
Consultant
Joined: 03 Feb 2003
Posts: 5
Location: Moscow, Russia
Followers: 1

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

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26 Feb 2003, 23:47
We should consider three scenarios:
1 prof - 7*2C10
2 prof - 2C7*1C10
3 fprof - 3C7

what do you think?
Founder
Affiliations: AS - Gold, HH-Diamond
Joined: 04 Dec 2002
Posts: 14457
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
Followers: 3727

Kudos [?]: 23047 [0], given: 4515

### Show Tags

26 Feb 2003, 23:50
Hakob wrote:
We should consider three scenarios:
1 prof - 7*2C10
2 prof - 2C7*1C10
3 fprof - 3C7

what do you think?

YES. I took this one from my Permutations/Combinations guide available from this site. I think i will add many of hte problems you published here guys. Thanks.... Here is my formal explanation from back in the days.

Scenario 1. We get C(17, 3) = = 680 total possible committees. Now, the number of teams with students only is, using the same formula C(10,3) = = 120.
Now, 680-120=560.

Scenario 2. For a committee with 1 professor and 2 students, we will get 7xC(10,2) = = 45x7=315. (we multiply by 7 because For 2 professors and 1 student, we will get C(7, 2)x10 = 21x10=210, and for 3 professors, we will get C(7, 3)=35. Adding up the combinations we will get: 315+210+35=560.
Princeton Review suggests using Scenario 2 because it is supposedly simpler to understand, however, taking into consideration that you can make a mistake in the endless calculations and that it requires 3 complex operations in contrast to 2 in the first case, it has a weaker standing. In any case, both ways get the correct answer, but you need to choose the one that appeals more to youтАФthe one easier to use.
Manager
Joined: 24 Jun 2003
Posts: 54
Followers: 0

Kudos [?]: 1 [0], given: 0

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09 Sep 2003, 14:41
I completely understand these approaches, however I am curious as to why the following does not work:

7 ways to pick the professor * 2 C 16

Any thoughts?

Thanks
Htown
Manager
Joined: 13 Aug 2009
Posts: 203
Schools: Sloan '14 (S)
Followers: 3

Kudos [?]: 103 [0], given: 16

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14 Nov 2009, 05:54
Another easy way to solve this problem is to take:

Total number of combinations - Number of combinations WITHOUT any professors

17C3 - 10C3

Re: Combin/Perm   [#permalink] 14 Nov 2009, 05:54
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