BTGModeratorVI wrote: ↑

Fri Apr 10, 2020 8:07 am

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%

Answer: C
Source: Manhattan prep

We can think of this as a probability question: What is the probability that Anthony and Michael are on the same subcommittee?

Now, assume that we're creating subcommittees.
We want to place 6 people in the following spaces:
_ _ _ | _ _ _

First, we place Michael in one subcommittee; it makes no difference which one:
M _ _ | _ _ _

When we go to place Anthony, we see that there are 5 spaces remaining.
2 spaces are on the same subcommittee as Michael.
So the probability that they are on the same subcommittee is 2/5 = 40%

Answer: C

Cheers,
Brent

BTGModeratorVI wrote: ↑

Fri Apr 10, 2020 8:07 am

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%

Answer: C
Source: Manhattan prep

Solution:

Let’s first determine the number of ways 2 three-person subcommittees can be formed from 6 people. The number of ways 3 people can be selected from 6 people for the first committee is 6C3 = (6 x 5 x 4)/(3 x 2) = 20. The number of ways 3 people can be selected from the remaining 3 people for the second committee is 3C3 = 1. Thus, the number of ways 2 three-person subcommittees can be formed from 6 people is 20 x 1 = 20, if the order of selecting the committees matters. However, since the order of selecting the committees doesn’t matter, we have to divide by 2! = 2. Thus, the number of ways 2 three-person subcommittees can be formed from 6 people is 20/2 = 10.

Since only a total of 10 committees can be formed, we can list all of these committees and see how many of them have Anthony and Michael on the same committee. We can let A be Anthony, M be Michael, and B, C, D, and E be the other 4 people.

1) A-B-C, D-E-M
2) A-B-D, C-E-M
3) A-B-E, C-D-M
4) A-B-M, C-D-E
5) A-C-D, B-E-M
6) A-C-E, B-D-M
7) A-C-M, B-D-E
8) A-D-E, B-C-M
9) A-D-M, B-C-E
10) A-E-M, B-C-D

We can see that from the 10 committees that can be formed, 4 of them (in bold) include both Anthony and Michael. Thus, the probability is 4/10 = 40%

Alternate Solution:

To determine the number of subcommittees that include Anthony, assume Anthony is already selected. The remaining two spots on the committee can be filled from the remaining five people in 5C2 = 5!/(2!*3!) = (5 x 4)/2 = 10 ways.

To determine the number of subcommittees that include Michael as well as Anthony, assume both Anthony and Michael are selected. The remaining one spot can be filled from the remaining four people in 4C1 = 4 ways.

Thus, 4/10 = 40% of the committees that include Anthony also include Michael.

Answer: C