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

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Manager
Joined: 12 Oct 2011
Posts: 105
GMAT 1: 700 Q48 V37
GMAT 2: 720 Q48 V40
A committee of 3 people is to be chosen from the president  [#permalink]

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08 Apr 2012, 05:43
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Difficulty:

55% (hard)

Question Stats:

63% (01:55) correct 38% (02:03) wrong based on 144 sessions

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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
Math Expert
Joined: 02 Sep 2009
Posts: 62289
Re: A committee of 3 people is to be chosen from the president  [#permalink]

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08 Apr 2012, 08:21
1
2
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.

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

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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Location: United States
Concentration: General Management, Entrepreneurship
GMAT 1: 750 Q49 V44
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WE: Engineering (Computer Software)
Re: A committee of 3 people is to be chosen from the president  [#permalink]

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Updated on: 08 Apr 2012, 11:26
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, 10:16.
Last edited by pinchharmonic on 08 Apr 2012, 11:26, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 62289
Re: A committee of 3 people is to be chosen from the president  [#permalink]

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08 Apr 2012, 10:20
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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Joined: 06 Aug 2011
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Concentration: General Management, Sustainability
WE: Engineering (Telecommunications)
Re: A committee of 3 people is to be chosen from the president  [#permalink]

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02 Jul 2015, 04:05
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.
Math Expert
Joined: 02 Sep 2009
Posts: 62289
Re: A committee of 3 people is to be chosen from the president  [#permalink]

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02 Jul 2015, 04:40
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]

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16 Feb 2019, 07:59
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

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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If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.
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Joined: 13 Jun 2018
Posts: 39
GMAT 1: 700 Q49 V36
Re: A committee of 3 people is to be chosen from the president  [#permalink]

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16 Feb 2019, 08:52
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?

Bunuel
Re: A committee of 3 people is to be chosen from the president   [#permalink] 16 Feb 2019, 08:52
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