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

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?

A. 20%
B. 30%
C. 40%
D. 50%
E. 60%

First approach:

Let's take the group with Michael: there is a place for two other members and one of them should be taken by Anthony, as there are total of 5 people left, hence there is probability of \(\frac{2}{5}=40\%\).

Second approach:

Again in Michael's group 2 places are left, # of selections of 2 out of 5 is \(C^2_5=10\) = total # of outcomes.

Select Anthony - \(C^1_1=1\), select any third member out of 4 - \(C^1_4=4\), total # \(=C^1_1*C^1_4=4\) - total # of winning outcomes.

\(P=\frac{# \ of \ winning \ outcomes}{total \ # \ of \ outcomes}=\frac{4}{10}=40\%\)

Third approach:

Michael's group:
Select Anthony as a second member out of 5 - 1/5 and any other as a third one out of 4 left 4/4, total \(=\frac{1}{5}*\frac{4}{4}=\frac{1}{5}\);

Select any member but Anthony as second member out of 5 - 4/5 and Anthony as a third out of 4 left 1/4, \(total=\frac{4}{5}*\frac{1}{4}=\frac{1}{5}\);

\(Sum=\frac{1}{5}+\frac{1}{5}=\frac{2}{5}=40\%\)

Fourth approach:

Total # of splitting group of 6 into two groups of 3: \(\frac{C^3_6*C^_3}{2!}=10\);

# of groups with Michael and Anthony: \(C^1_1*C^1_1*C^1_4=4\).

\(P=\frac{4}{10}=40\%\)

Answer: C.

Hope it helps.