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Anthony and Michael

by logitech » Tue Nov 25, 2008 10:48 pm
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?

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by jimmiejaz » Wed Nov 26, 2008 1:37 am
Let, Michael and Anthony be considered as a single unit. Then we have to select 1 person from remaining 4 to form a 3 member subcommitee.
1 person can be selected in 4C1 =4 ways.
Now, according to question, 2 subcommitees are to be formed, so total ways = 4*2 = 8

Total no of ways of selecting 3 people from 6 = 6C3=20
Percentage = 8/20 = 40%
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by logitech » Wed Nov 26, 2008 8:24 am
jimmiejaz wrote:Let, Michael and Anthony be considered as a single unit. Then we have to select 1 person from remaining 4 to form a 3 member subcommitee.
1 person can be selected in 4C1 =4 ways.
Now, according to question, 2 subcommitees are to be formed, so total ways = 4*2 = 8

Total no of ways of selecting 3 people from 6 = 6C3=20
Percentage = 8/20 = 40%
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by mental » Sun Nov 30, 2008 5:40 am
what percent of all the possible subcommittees that include Michael also include Anthony?
Total number of 3 member committees that include Michael:
5C2 = 10 (with Michael as the third member)

Total number of 3 member committees that include Michael and Anthony:
4C1 = 4 (with Michael and Anthony as the two members)

now 4 is what percent of 10? = 40%
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by sumitkhurana » Mon Jan 26, 2009 1:02 am
jimmiejaz, logitech

Could you please explain this step:-

Now, according to question, 2 subcommitees are to be formed, so total ways = 4*2 = 8

Question - Why did you do this ? Once you have Michael and Anthony together, there are just 4 different ways in which 2 subcommitees could be formed. Why multiply by 2 ? The second subcommittee is formed automatically once you have chose the third person who would be with Anthony and Michael.
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