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A committee of 3 people is to be chosen from the president

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BN1989
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A committee of 3 people is to be chosen from the president [#permalink] New post   08 Apr 2012, 06:43
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A committee of 3 people is to be chosen from the president and vice president of four different companies. What is the number of different committees that can be chosen if two people who work for the same company cannot both serve on the committee?

A) 16
B) 24
C) 28
D) 32
E) 40
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D
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Bunuel
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Re: A committee of 3 people is to be chosen from the president [#permalink] New post   08 Apr 2012, 09:21
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BN1989 wrote:
A committee of 3 people is to be chosen from the president and vice president of four different companies. What is the number of different committees that can be chosen if two people who work for the same company cannot both serve on the committee?

A) 16
B) 24
C) 28
D) 32
E) 40


Each company can send only one "representative" to the committee. Let's see in how many ways we can choose 3 companies (as there should be 3 members) to send only one "representatives" to the committee: 4C3=4.

But these 3 chosen companies can send two persons (either president or vice president ): 2*2*2=2^3=8.

Total # of ways: 4C3*2^3=32.

Answer: D.

Or: 8*6*4/3!=32, we are dividing by 3! because the order in the committee is not important.

Answer: D.

Similar problems with different approaches:
combination-permutation-problem-couples-98533.html
ps-combinations-94068.html
committee-of-88772.html

Hope it helps.
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pinchharmonic
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Re: A committee of 3 people is to be chosen from the president [#permalink] New post   Updated on: 08 Apr 2012, 12:26
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man bunuel always has such a quick and elegant way. Can you comment if this approach is correct?

i just did the total number of possibilities first:

ab cd ef gh are the 4 companies for example

8c3. this will include the problem counts of a pres/vp of the same company

how many ways did we over count = How many ways can a president and VP of the same company be part of this group?

choose the group, 4c1, then take both the president and vp, 2c2, then a remaining member from one of the other 3, 6c1

4c1 * 2c2 * 6c1
4 * 1 * 6 = 24

8c3 - 24 = 56 - 24 = 32

Originally posted by pinchharmonic on 08 Apr 2012, 11:16.
Last edited by pinchharmonic on 08 Apr 2012, 12:26, edited 1 time in total.
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Re: A committee of 3 people is to be chosen from the president [#permalink] New post   08 Apr 2012, 11:20
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pinchharmonic wrote:
man bunuel always has such a quck and elegant way. Can you comment if this approach is correct?

i just did the total number of possibilities first:

ab cd ef gh are the 4 companies for example

8c3. this will include the problem counts of a pres/vp of the same company

how many ways did we over count = How many ways can a president and VP of the same company be part of this group?

choose the group, 4c1, then take both the president and vp, 2c2, then a remaining member from one of the other 3, 6c1

4c1 * 2c2 * 6c1
4 * 1 * 6 = 24

8c3 - 24 = 56 - 24 = 32


Yes, that's also a correct approach.
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mygmatsuccess
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Re: A committee of 3 people is to be chosen from the president [#permalink] New post   02 Jul 2015, 05:05
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Sorry to bump into an old thread.

Can we approach this question the following way?

We need 3 people.

1st can be anyone from the 8.

2nd one cannot be the one from same company. So there will be 6 possibilities.

3rd cannot be from the 1st and 2nd guys' company. So there will be 4 possibilities.

And order does not matter and so we should divide by 3!.

(8*6*4)/3!

This will give 32.
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Re: A committee of 3 people is to be chosen from the president [#permalink] New post   16 Feb 2019, 08:59
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BN1989 wrote:
A committee of 3 people is to be chosen from the president and vice president of four different companies. What is the number of different committees that can be chosen if two people who work for the same company cannot both serve on the committee?

A) 16
B) 24
C) 28
D) 32
E) 40


Now this is quite amusing, i would consider the P & VP to be a couple :-D

So there will be 4 couples, out which we need a committee which wont have one

Total number of ways, without any restriction =\(5C_3\) = 56 ways

Now if we keep a restriction we can calculate the ways in which one couple is present = \(4C_1 * 6C_1\)= 24 ways

Now the difference will give us 32 ways
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Re: A committee of 3 people is to be chosen from the president [#permalink] New post   02 Jul 2015, 05:40
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pptkrishanth wrote:
Sorry to bump into an old thread.

Can we approach this question the following way?

We need 3 people.

1st can be anyone from the 8.

2nd one cannot be the one from same company. So there will be 6 possibilities.

3rd cannot be from the 1st and 2nd guys' company. So there will be 4 possibilities.

And order does not matter and so we should divide by 3!.

(8*6*4)/3!

This will give 32.


Yes. This is the same exact approach given in my first reply above.
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Re: A committee of 3 people is to be chosen from the president [#permalink] New post   16 Feb 2019, 09:52
mygmatsuccess wrote:
Sorry to bump into an old thread.

Can we approach this question the following way?

We need 3 people.

1st can be anyone from the 8.

2nd one cannot be the one from same company. So there will be 6 possibilities.

3rd cannot be from the 1st and 2nd guys' company. So there will be 4 possibilities.

And order does not matter and so we should divide by 3!.

(8*6*4)/3!

This will give 32.



I understand that we are dividing by 3! because the order does not matter. Can someone please explain the math behind it?

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Re: A committee of 3 people is to be chosen from the president [#permalink] New post   22 Oct 2022, 07:34
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Re: A committee of 3 people is to be chosen from the president [#permalink]
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