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Anthony and Michael sit on the six-member board of directors

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Joined: 15 Jul 2008
Posts: 208
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Kudos [?]: 31 [0], given: 0

Anthony and Michael sit on the six-member board of directors [#permalink]  23 Jul 2008, 04:50
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?

20%
30%
40%
50%
60%

My convoluted thought process was...
Number of ways of selecting a group 3 out of 6 is 6C3 = 20 ways. But since we are to select 2 groups of 3, each selection from 6C3 ways would by default give another one. So the number of ways of getting 2 group of 3 each = 6C3/2=10.

Keeping Anthony and Michael in one group, one of the remaining 4 can be chosen in 4 ways to form the desired group. So #of favorable events = 4

4/10 = 40% = OA

I believe there has to be a better line of thinking. i wasn't able to beat it under 3 min.

Also, in this case it worked.
If it had been something like 3 groups of 2 members each..(any thoughts on this ? i get 1/5)
then it would take me longer to tackle this. Is there a better way to deal with this kind of problem ?

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Joined: 04 Jan 2005
Posts: 283
Location: Milan
Schools: Wharton, LBS, UChicago, Kellogg MMM (Donald Jacobs Scholarship), Stanford, HBS
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Re: Combinations [#permalink]  23 Jul 2008, 05:00
bhushangirl,

my reasoning for this was:

1. Michael is chosen for one of the two subcommittees (which happens in any case)
2. There are two seats left in that subcommitee, and five people remaining to be chosen out of with equal possibilities.

2/5 = 40%
Re: Combinations   [#permalink] 23 Jul 2008, 05:00
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