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Permutation percentage combination

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Junior | Next Rank: 30 Posts
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by deltaforce » Sun Jul 05, 2009 10:59 pm
Hi,
wud app. someone solving this...

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?

Ans - [spoiler]40%[/spoiler]

the way i approached this problem is

nos of ways to total comm. is 6c3 * 3c3 = 20
nos of ways where bot mike and anthony are on the same team = 4c1 *2c2 = 4

=4/20 * 100 = 20% of all teams

But the answer is something else. what's wrong with my approach?

Thanks
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by ladhanivishal » Mon Jul 06, 2009 12:11 am
I am gettin 20% as well
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Re: Permutation percentage combination

by gabriel » Mon Jul 06, 2009 12:58 am
deltaforce wrote:Hi,
wud app. someone solving this...

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?

Ans - [spoiler]40%[/spoiler]

the way i approached this problem is

nos of ways to total comm. is 6c3 * 3c3 = 20
nos of ways where bot mike and anthony are on the same team = 4c1 *2c2 = 4

=4/20 * 100 = 20% of all teams

But the answer is something else. what's wrong with my approach?

Thanks
Well two things, first the question is not asking for the total number of committees but the number of commiittees that includes Micheal. So the denominator of your equation should be 1*5c2 = 10, the numerator remains the same.

Second this question has been discussed several times before on the forum. So please do use the search function before you post a question.
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Re: Permutation percentage combination

by kanha81 » Mon Jul 06, 2009 9:56 am
deltaforce wrote:Hi,
wud app. someone solving this...

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?

Ans - [spoiler]40%[/spoiler]

the way i approached this problem is

nos of ways to total comm. is 6c3 * 3c3 = 20
nos of ways where bot mike and anthony are on the same team = 4c1 *2c2 = 4

=4/20 * 100 = 20% of all teams

But the answer is something else. what's wrong with my approach?

Thanks
Let M:Michael be on the comm-1
A: Anthony be on the comm-1
x: empty seats

a)M x x | x x x

The remaining positions for comm-1 (=2) can be selected in 5C2 ways= 10 ways

b) M A x | x x x

The remaining positions for comm-1 (=1) can be selected in 4C1 ways = 4

Hence (4/10) * 100 = [spoiler]40%[/spoiler]
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Re: Permutation percentage combination

by real2008 » Tue Jul 07, 2009 12:50 am
deltaforce wrote:Hi,
wud app. someone solving this...

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?

Ans - [spoiler]40%[/spoiler]

the way i approached this problem is

nos of ways to total comm. is 6c3 * 3c3 = 20
nos of ways where bot mike and anthony are on the same team = 4c1 *2c2 = 4

=4/20 * 100 = 20% of all teams

But the answer is something else. what's wrong with my approach?

Thanks
well, I think u have calculated for both of them staying in one committee say 1. similarly for committee 2 also 20%.

and total is 40%.

(I assume that both committees are unique)
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