Hi,
wud app. someone solving this...
Anthony and Michael sit on the sixmember board of directors for company X. If the board is to be split up into 2 threeperson 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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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.deltaforce wrote:Hi,
wud app. someone solving this...
Anthony and Michael sit on the sixmember board of directors for company X. If the board is to be split up into 2 threeperson 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
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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Let M:Michael be on the comm1deltaforce wrote:Hi,
wud app. someone solving this...
Anthony and Michael sit on the sixmember board of directors for company X. If the board is to be split up into 2 threeperson 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
A: Anthony be on the comm1
x: empty seats
a)M x x  x x x
The remaining positions for comm1 (=2) can be selected in 5C2 ways= 10 ways
b) M A x  x x x
The remaining positions for comm1 (=1) can be selected in 4C1 ways = 4
Hence (4/10) * 100 = [spoiler]40%[/spoiler]
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well, I think u have calculated for both of them staying in one committee say 1. similarly for committee 2 also 20%.deltaforce wrote:Hi,
wud app. someone solving this...
Anthony and Michael sit on the sixmember board of directors for company X. If the board is to be split up into 2 threeperson 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
and total is 40%.
(I assume that both committees are unique)