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

@bunuel, how very generous with the explanations and the samples - terrific, thanks!

Hi Bunnuel,

On approach 3, are you implying that the order does matter? By calculating the odds of picking Anthony in the second position and anyone else in the third position AND the odds of picking anyone except Anthony in the second position and Anthony in the 3rd position you are basically establishing that the order does matter (whether Anthony is in the second or 3rd position).

Can you clarify? I am sure there is something wronog in my thought process. Thanks.

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 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 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 Michae,l 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%

Answer: C

Backward logic approach:
Probability of selecting Antony = 1-Probability of not selecting Antony
So, 1-(4/5*3/4)=2/5

Total ways of selecting 3 people from 6 people = 6C3 = 20
M and A constitute a set as they will always be together.

Total committees to be formed = 2

Selecting any one committee from 2 committees = 2C1 = 2
Select 1 person from the remaining 4 people = 4C1 = 4

Required ways = 4C1 * 2C1 = 8
Total way = 6C3 = 20

8/20 = 2/5

Can you explain the line "# of groups with Michael and Anthony: \(C^1_1*C^1_1*C^1_4=4\)."

how did you arrive at C^1 and C^4

1 way to choose Michael, 1 way to choose Anthony, and 4 ways to choose the third member of the subcommittee from the remaining 4 members.

There are 20 possibilities in creating two sub-committees. When you fix Michael and Anthony in the first committee, there are 4 ways to pick the 3rd person, hence 4 ways the first committee can be formed. Same goes for the 2nd committee: if Michael and Anthony are both in the 2nd committee, there are 4 ways to form it. So 4 ways to form the first committee, 4 ways to form the second committee makes it 8 ways to form two committees in which both Michael and Anthony are in one. 8/20 = 40%.

Another alternative approach:

Assume that both the persons are in the same group.
We need to select any 1 group from the 2 available groups = 2C1 = 2

For the remaining person, we need to select just 1 person from 4 persons = 4C1 = 4

Ways to select 3 persons from 6 persons = 6C3 = 20

Answer = (4*2)/20 = 8/20 = 40%

kamilaak

Given: Anthony and Michael sit on the six member board of directors for company X.
Asked: 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?

Total possible sub-committees = 6C3 = 20
All the possible subcommittees that include Michael also include Anthony = 4C1 *2= 4*2

The percent of all the possible subcommittees that include Michael also include Anthony = 8/20 = 2/5 = 40%

IMO C
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