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# Anthony and Michael sit on the six-member board of directors

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VP
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Anthony and Michael sit on the six-member board of directors [#permalink]  04 Oct 2008, 05:11
Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?

* 20%
* 30%
* 40%
* 50%
* 60%
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Intern
Joined: 04 Oct 2008
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Re: MGMAT - Combinatorics [#permalink]  04 Oct 2008, 08:13
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The question is not "what percent of all the possible subcommittees include both Michael and Anthony?"

Then the answer would be 20%

The question is "what percent of all the possible subcommittees that include Michael also include Anthony?"

So you divide the number of possible combinations that include Michael and Anthony by the number of possible combinations that include Michael.

So, as zoinnk said there are 4 possible committees that include both Michael and Anthony.

I think it's easier to find the number of committees on which Michael does not appear at all, which would mean 5C3=10.
20-10 = 10 committees on which Michael appears.

4/10 = 40%
Director
Joined: 12 Jul 2008
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Schools: Wharton
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Re: MGMAT - Combinatorics [#permalink]  04 Oct 2008, 06:25
amitdgr wrote:
Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?

* 20%
* 30%
* 40%
* 50%
* 60%

Total # of ways to choose = 6C3 = 20
Total number of committees with both Michael and Anthony = 4*1*1 = 4

20%
Director
Joined: 12 Jul 2008
Posts: 518
Schools: Wharton
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Re: MGMAT - Combinatorics [#permalink]  04 Oct 2008, 08:55
csvobo wrote:
The question is not "what percent of all the possible subcommittees include both Michael and Anthony?"

Then the answer would be 20%

The question is "what percent of all the possible subcommittees that include Michael also include Anthony?"

So you divide the number of possible combinations that include Michael and Anthony by the number of possible combinations that include Michael.

So, as zoinnk said there are 4 possible committees that include both Michael and Anthony.

I think it's easier to find the number of committees on which Michael does not appear at all, which would mean 5C3=10.
20-10 = 10 committees on which Michael appears.

4/10 = 40%

Sorry. Misread the question. You're right! +1
VP
Joined: 30 Jun 2008
Posts: 1048
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Re: MGMAT - Combinatorics [#permalink]  04 Oct 2008, 20:16
csvobo wrote:
The question is not "what percent of all the possible subcommittees include both Michael and Anthony?"

Then the answer would be 20%

The question is "what percent of all the possible subcommittees that include Michael also include Anthony?"

So you divide the number of possible combinations that include Michael and Anthony by the number of possible combinations that include Michael.

So, as zoinnk said there are 4 possible committees that include both Michael and Anthony.

I think it's easier to find the number of committees on which Michael does not appear at all, which would mean 5C3=10.
20-10 = 10 committees on which Michael appears.

4/10 = 40%

Even I did the same mistake as zoinnk ... Tricky words I guess ... zoinnk doesn't miss too many questions....

csvobo ... can you explain "to find the number of committees on which Michael does not appear at all, which would mean 5C3"

Thanks
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Manager
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Re: MGMAT - Combinatorics [#permalink]  04 Oct 2008, 20:57
C is also my answer.

If Michael was sit, there is 5 positions left, in which only 2 for Michael's side. So 2/5 = 40%
VP
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Re: MGMAT - Combinatorics [#permalink]  04 Oct 2008, 21:17
lylya4 wrote:
C is also my answer.

If Michael was sit, there is 5 positions left, in which only 2 for Michael's side. So 2/5 = 40%

wow!! your method seems so much easier !!
_________________

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VP
Joined: 17 Jun 2008
Posts: 1404
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Re: MGMAT - Combinatorics [#permalink]  04 Oct 2008, 23:06
amitdgr wrote:
Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?

* 20%
* 30%
* 40%
* 50%
* 60%

Say there 6 members we need to form 3 member committee out of 3 members we know one is michael hence we need to select 2 out of 5
5C2 there 10 combinations

Now consider anthony and michael !!! hence select 1 out of 4 ,4C1
4 combitions

4/10 =2/5 0.4 =40%
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Its Now Or Never

VP
Joined: 30 Jun 2008
Posts: 1048
Followers: 11

Kudos [?]: 340 [0], given: 1

Re: MGMAT - Combinatorics [#permalink]  04 Oct 2008, 23:10
spriya wrote:
amitdgr wrote:
Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?

* 20%
* 30%
* 40%
* 50%
* 60%

Say there 6 members we need to form 3 member committee out of 3 members we know one is michael hence we need to select 2 out of 5
5C2 there 10 combinations

Now consider anthony and michael !!! hence select 1 out of 4 ,4C1
4 combitions

4/10 =2/5 0.4 =40%

Thanks for your reply priya. nice technique
_________________

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Re: MGMAT - Combinatorics   [#permalink] 04 Oct 2008, 23:10
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# Anthony and Michael sit on the six-member board of directors

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