Permutation percentage combination

[deltaforce]

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

[ladhanivishal]

I am gettin 20% as well

[gabriel]

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.

[kanha81]

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]

[real2008]

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)