A three-person committee must be chosen from a group of 7 : Quant Question Archive [LOCKED]
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# A three-person committee must be chosen from a group of 7

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

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04 Dec 2006, 15:44
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

The normal way I'd do this would be 17C3 - 10C3 but when I was tried it in a different way:

7*16*15/3*2*1 = 230 What is wrong with this method?
VP
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04 Dec 2006, 16:52
7C1*10C2 + 7C2*10C1 + 7C3 = 560
Director
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04 Dec 2006, 16:59
This is the way I did it. I am not clear about what your logical reasoning was, to explain what you might have done wrong.

since 1 prof is always required, the way to select 3 people is:

1 Prof, 2 students (7C1*10C2)
2 Profs, 1 student (7C2*10C1)
3 Profs (7C3)

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04 Dec 2006, 17:05
But why does it not work this way:

We have to have at least 1 professor so we have 7 choices: 7

Next we can choose any one member from a group of 16, so 16 choices: 16

Finally we can choose any one member from a group of 15 - so 15 choices: 15

7*16*15 = 1680

Order not important so 1680/3*2*1 = 280

What is wrong with the method?
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04 Dec 2006, 19:29
trivikram wrote:
7C1*10C2 + 7C2*10C1 + 7C3 = 560

Getting same result..
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05 Dec 2006, 03:48
But why does it not work this way:

We have to have at least 1 professor so we have 7 choices: 7

Next we can choose any one member from a group of 16, so 16 choices: 16

Finally we can choose any one member from a group of 15 - so 15 choices: 15

7*16*15 = 1680

Order not important so 1680/3*2*1 = 280

What is wrong with the method?

Sorry I still don't get it. I get the numerator, don't get how you are taking care of the ordering.
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05 Dec 2006, 11:54
anindyat wrote:
trivikram wrote:
7C1*10C2 + 7C2*10C1 + 7C3 = 560

Getting same result..
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05 Dec 2006, 12:25
7*16*15 is the number of ways to permute 1 choice of 7 professors and then two further choices from a group of 16.

Within this 1680 we'll have repeats ABC, CBA right?

So divided 1680 by 3*2*1 (no of ways you can permute 3 choices) to just give number of unique possibilites. What did I do wrong?

1680/6 = 280
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06 Dec 2006, 03:08
I have a mind block with the way I did it now. It looks like someone else will have to explain why yours is wrong.
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06 Dec 2006, 10:29
I understand that the normal way to solve it would be 17C3 - 10C3 but can anyone shed light on why the other way doesn't work??
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06 Dec 2006, 16:16
I understand that the normal way to solve it would be 17C3 - 10C3 but can anyone shed light on why the other way doesn't work??

OK you are assuming 17C3 is picking total 3 people from a pack of 17
Who knows if the selected 3 are graduate students from the total profs and graduate students.

Then your 10C3 is worng 'cos the number of graduate students to be removed are now not 10 but 7

So 17C3-10C3 is wrong

Like wise of the 3 selected from 17 , 2 are graduate student and 1 prof you have only 8 graduate students out of whom you have to select 3 which is again wrong.
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06 Dec 2006, 16:50
trivikram wrote:
I understand that the normal way to solve it would be 17C3 - 10C3 but can anyone shed light on why the other way doesn't work??

OK you are assuming 17C3 is picking total 3 people from a pack of 17
Who knows if the selected 3 are graduate students from the total profs and graduate students. It doesn't matter because we are simply looking at the number of total combinations - number of combinations we are not interested in.
Then your 10C3 is worng 'cos the number of graduate students to be removed are now not 10 but 7

So 17C3-10C3 is wrong (This actually gives the correct answer and is the suggested method)

Like wise of the 3 selected from 17 , 2 are graduate student and 1 prof you have only 8 graduate students out of whom you have to select 3 which is again wrong.

This is getting worse!

Firstly 17C3 - 10C3 = 560 (The correct answer) and is also the suggested method from the combinations course on this website --- see:

http://www.gmatclub.com/content/courses ... ations.php

This method makes sense - you are finding the total number of ways of choosing 3 people from 17 and then subtracting the total number of ways of choosing a committee of 3 from 10 students alone.

However - this is fine and is NOT what the post was querying at all!!!!!!!!

The question is - what is wrong with the following method?:

We have to have at least 1 professor so we have 7 choices: 7

Next we can choose any one member from a group of 16, so 16 choices: 16

Finally we can choose any one member from a group of 15 - so 15 choices: 15

7*16*15 = 1680

Order not important so 1680/3*2*1 = 280 (because no. of ways to permute 3 people is 3*2*1)

Why doesn't this yield the correct answer?

Last edited by MBAlad on 06 Dec 2006, 16:59, edited 3 times in total.
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06 Dec 2006, 18:35
17C3 - 10C3 is definitely right here.
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06 Dec 2006, 19:11
Come on tennis - what's wrong with the other method?
06 Dec 2006, 19:11
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