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Anthony and Michael sit on the six-member board of directors [#permalink]
18 Dec 2005, 01:49

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 1 sessions

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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?

If we put M in the team we need 2 more people which can be selected in 5C2 or 10 ways.When M and A are together we need one more person who0 can be selected in 4 ways. Seems like 40%

2 out of 5 will be with Miachel. Everyone has the same chance. So each one's chance to be with Miachel is 2/5. Thus Anthony's chance to be with Miachel is 2/5.

There is a simple and there is a really complicated way to solving this problem.

Simple way:

subdivide the committee into two three-person teams.

_ _ _ and _ _ _

Fix Michael on any slot of either team.

That leaves 2 spots remaining for Anthony to be on the same team and five total possible spots.

2/5=40%

OA is C.

The complicated way and OE:

It is important to first note that our point of reference in this question is all the possible subcommittees that include Michael. We are asked to find what percent of these subcommittees also include Anthony.

Let's first find out how many possible subcommittees there are that must include Michael. If Michael must be on each of the three person committees that we are considering, we are essentially choosing people to fill the two remaining spots of the committee. Therefore, the number of possible committees can be found by considering the number of different two people groups that can be formed from a pool of 5 candidates (not 6 since Michael was already chosen).

Using the anagram method to solve this combinations question, we assign 5 letters to the various board members in the first row. In the second row, two of the board members get assigned a Y to signify that they were chosen and the remaining 3 get an N, to signify that they were not chosen:

A
B
C
D
E

Y
Y
N
N
N

The number of different combinations of two person committees from a group of 5 board members would be the number of possible anagrams that could be formed from the word YYNNN = 5! / (3!2!) = 10. Therefore there are 10 possible committees that include Michael.

Out of these 10 possible committees, of how many will Anthony also be a member? If we assume that Anthony and Michael must be a member of the three person committee, there is only one remaining place to fill. Since there are four other board members, there are four possible three person committees with both Anthony and Michael. Of the 10 committees that include Michael, 4/10 or 40% also include Anthony.

# of possible to form 3 person team from 6 people = 6C3 = 20. That leaves another 20 combinations for the other team. Total = 40 possibilites (20 team A combination, 20 team B combination).

If Micheal is in team A, then Anthony has to be in team A, and team A will have only 4 ways to form (due to 4 people remaining).

But Micheal and Anthony can be in team A or team B. So % = (4/20)*2*100% = 40%

I think that's my problem. I usually see terms as "Michael also Anthony" as dependant, namely "Anthony under the condition that Michael". That's simply wrong in this cases.