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Re: Anthony and Michael sit on the six member board of directors
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19 Jul 2012, 12:16

1

I did it this way:

Total ways to choose 1st subcommittee = 6C3 Total ways to choose 2nd subcommittee = 1 (since only 3 people left) Possible subcommittees = 1*6C3

Total ways to choose which out of the two subcommittee will Anthony and Michael be in= 2C1 Total ways to choose the remaining 1 person on that subcommittee = 4C1

Re: Anthony and Michael sit on the six member board of directors
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19 Jul 2012, 20:32

5

Totak ways to select 3 members from 6 members = 6C3 = 6!/3!*3! = 20 Now if the sis members are A,M,X1,X2,X3,X4 , take A & M together as a sing le unit Y. Y = A,M, they can be selected as 2! ways = 2 Grups can be made as = YX1, YX2,YX3 & YX4 = 4 Total grups = 4*2=8

Re: Anthony and Michael sit on the six member board of directors
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20 Jul 2012, 05:57

manwiththeharmonica wrote:

I did it this way:

Total ways to choose 1st subcommittee = 6C3 Total ways to choose 2nd subcommittee = 1 (since only 3 people left) Possible subcommittees = 1*6C3

Total ways to choose which out of the two subcommittee will Anthony and Michael be in= 2C1 Total ways to choose the remaining 1 person on that subcommittee = 4C1

P = (2C1*4C1)/(1*6C3) = 2/5 (40%)

this is wrong, please do not follow this, i did not read the question correctly.

Re: Anthony and Michael sit on the six member board of directors
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01 Aug 2018, 06:42

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