A three-person committee must be chosen from a group of 7 : GMAT Problem Solving (PS)
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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]

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26 May 2012, 10:58
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Hi All,

Need help to understand what is wrong with my approach.

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

My approach :-
There are 3 seats. One has to be a professor. The remaining can be occupied by remaining professors or students.

Thus ways of filling seat 1 is 7 (You have to pick one professor out of 7)
Ways of filling remaining 2 seats will be to pick 2 out of 16 (10 S and 6 P as one P has already been picked. ) i.e. C(16,2) = 16*15/2 = 120

Total 120* 7 = 840

I know something is wrong with this approach but what exactly am I missing.
[Reveal] Spoiler: OA

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

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27 May 2012, 22:36
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@anujkch,

When you do 7C1*16C2, you make a critical error. Consider the 7 professors to be A,B,C,D,E,F,G. Supposing you select A to be separate (i.e. part of the 7C1 selection) and the other two professors to be selected from B through G. You will then get a case A(BC). Now consider the case where you select B as part of the separate case, and choose the other two professors from A,C,D,E,F,G,H. Again you will get a case B(AC), which is the same as ABC in terms of selection. Therefore your method causes double counting at places and gives you an answer which is greater than the actual one.

@ankitbansal85,
The error you make is different. In the PPP case, 7*6*5 is the number of ways you can arrange, and not just select three professors from 7 professors (i.e. 7*6*5 = 7P3). The number of ways to select 3 professors from 7 professors is 7C3. Similarly, in the PSS case, you must use 7*10C2, not 7*10*9, the latter is the number of ways to arrange 1 professor and 2 students in a row, not just select them. Correct this approach and you will get the right answer.

As for the right answer, it can be obtained in multiple ways,
Method 1:
7C1 * 10C2 + 7C2 * 10C1 + 7C3 = 315 + 210 + 35 = 560

Method 2:
Total number of ways to select at least one professor = Number of ways to select 3 people from 10 students and 7 professors - Number of ways to select 3 people from 10 student (i.e. without including any professor)
= 17C3 - 10C3 = 680 - 120 = 560
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Re: A three-person committee must be chosen from a group of 7 [#permalink]

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19 May 2013, 10:55
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The question stem states that at least 1 professor should be present. So all combinations of 3 people out of 17 people minus the combinations that no professors are present will give the answer:

All combinations: 17! / (3! 14!) = 680.
No professors present: 10! / 3! 7! = 120.

680-120 = 560.
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Re: Help with permutation n combination pblm [#permalink]

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26 May 2012, 14:16
The committee has to be formed from 2 different groups of people (professors, graduates):
So, 3 cases will be formed:
1P,2G 2P,1G 3P,0G
7C1*10C2
+
7C2*10C1
+
7C3*10C0
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Re: Help with permutation n combination pblm [#permalink]

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26 May 2012, 18:45
The missing part in this approach is that you cannot merge both the groups together as when you try to select out of 16 persons the next 2 people then you have included both the professors and the graduates.So if you try to select 2 people out of the 16 from the lot then you can again select professors also.

This question asks you to give the combinations of the committee wherein the committee could be differentiated on the basis of the number of the people selected from two groups (professors/graduates).
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Re: A three-person committee must be chosen from a group of 7 [#permalink]

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27 May 2012, 21:02
I agree that if I try to select 2 people out of the 16 from the lot then you can again select professors also. But thats what the requirement is i.e. One should be professor, that we have already selected, and for remaining two seats we don't have a specific condition i.e. whether they have to Professors or Graduates. Even your set 3P,0G points to the condition where no Grad is selected.
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Re: A three-person committee must be chosen from a group of 7 [#permalink]

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27 May 2012, 21:14
Hi ,

I took the following approach and got an incorrect answer.please let me know what is wrong with my approach

Let p = Professor
s= Student.

we have 3 cases ppp,pps,pss

PPP = 7*6*5 = as there are a total of 7 professor to choose from we can fill the first slot in 7 ways, next in 6 ways and last in 5 ways = 210

PPS = 7*6*10 = same first slot 7 profs, second 6 profs and lat we have 10 student we can fill it in 10 ways = 420

PSS - 7*10*9 = same first slot 7 profs to choose from., second slot 10 students to choose from and third slot 9 stuidents to choose from = 630

wiith this my answer is 1260 ( Option D) which is incorrect. can some one please let me know what incorrect in my approach.

Regards,
Ankit
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Re: A three-person committee must be chosen from a group of 7 [#permalink]

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27 May 2012, 22:44
Thanks @GyanOne ......a very basic mistake indeed....

so we can only use the factorial thing when we are arranging or other words permutation....and not while doing selection in other words combination

If possible can you give me some theory behind it....some example....
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Re: A three-person committee must be chosen from a group of 7 [#permalink]

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27 May 2012, 23:42
@ankitbansal85,

Suppose you have three people A,B, and C. How many ways can you:
1) Arrange them in a row?
2) Select two from these three?

Ans: 1) The number of ways to arrange them in a row are ABC,BCA,CAB,ACB,BAC,and CBA for a total of 6 ways. 3! = 6, so the number of ways to arrange n people in a row is simply n!
2) Two can be selected from these three in just three ways: AB, BC, or CA. Remember that the selection AB is the same as the selection BA because in selections order does not matter. Therefore number of ways to select = 3C2 = 3 ways. Therefore the number of ways to select r objects from n different object is simply nCr.

Hope this helps.
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Re: A three-person committee must be chosen from a group of 7 [#permalink]

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24 Sep 2012, 06:48
Number of Professors - 7

Number of Students - 10

Committee size - 3

CONDITION - At least ONE professor per committee

With these conditions we can formulate the following scenarios :

Scenario I) 1 Professor and 2 Students
Scenario II) 2 professors and 1 student
Scenario III) 3 professors and ZERO students (this is the only tricky part of the question, WE HAVE to assume that the committee can consist of only professors since it is not explicitly mentioned in the question that their must be a "student representative" in the committee )

Solving I --> C(7,1) x C(10,2) = 7 x 45 = 315
Solving II --> C(7,2) x C (10,1) = 21 x 10 = 210
Solving III --> C(7,3) x C (10,0) = 35 x 1 = 35

Adding I , II and III we get 560 (B)
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Re: A three-person committee must be chosen from a group of 7 [#permalink]

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24 Sep 2012, 07:32
This is a common combinatorics question - first trick is to immediately recognize the "at least one" phrase. This is very common on the GMAT.

Think of it as all possibilities - possibility for ZERO professors.
P(All) - P(0 professors)

Choose 3 from 17 - Chooose 3 from 10 (all 10 coming from the graduate students)
17C3 - 10C3

17! / (3!*14!) - 10! - (3! *7!)

15*16*17 / (3*2) - 8*9*10 / (3*2)

5*8*17 - 4*3*10
40*17 - 40*3
40(17 - 3)
40*14
400 + 160
560
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24 Oct 2014, 06:04
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Re: A three-person committee must be chosen from a group of 7 [#permalink]

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03 Oct 2016, 02:24
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Re: A three-person committee must be chosen from a group of 7 [#permalink]

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17 Nov 2016, 18:05
Prob(at least 1 prof) = Total - prob(all grad) --> 17C3 - 10C3 = 560

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

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24 Nov 2016, 05:24
I solved it this way:

7C1 * 10C2 = 315
7C2 * 10C1 = 210
7C3 = 35

315+210+35= 560 Option B.
Re: A three-person committee must be chosen from a group of 7   [#permalink] 24 Nov 2016, 05:24
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