Anthony and Michael

[maihuna]

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%

[sonu_thekool]

I am getting 30% (B).

2 sets of 3 member parties out of 6 is equivalent to AAABBB since who goes into which committee is irrelevant.

So, we can form 6! / (3! * 3!) = 20 total ways of coming up with 3 member committees among 6 people.

Now, all the ways of getting Anthony and Michael on the same team among both the sets is (3! / 2!) + (3! + 2!) = 3 + 3 = 6

6 out of 20 would be 30% of all sub committees will have Anthony and Michael on the same team.

What is the OA please ? I am not sure if this is the right approach. Any better way folks ?

[maihuna]

wrong

[Brent@GMATPrepNow]

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%

It seems that we can look at 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 _ _ | _ _ _
Now we 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%

[x2suresh]

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%

It seems that we can look at 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 _ _ | _ _ _
Now we 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%

I like your method.

Here is what I did.

Total No. of ways = 6C3 *3C3 = 20

AM XXXX

No. of ways AM together =
= 2*2C2*4C1*3C3=8

% = 8/20 =40%

[sureshbala]

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%

Total number of Chances
Since the subcommittee must contain Michael, the remaining 2 persons could be any two persons from the remaining 5. So total number of subcommittees containing Michael = 5C2 = 10.

Favorable Chances

Since the subcommittee must contain both Michael and Anthony, the remaining one person can be anyone from the reamaining 4. So favorable chances = 4.

So probability = 4/10 = 40%[/u]

[LalaB]

1) 2 subcommities with 3 ppl is below-
_ _ _ //_ _ _

let M be the first one. then the chances of A to be the second are 1/5, the 3d place 4C4=1

2) 1C1*1C1*4C1/6C3=1/5

[Brent@GMATPrepNow]

1) 2 subcommities with 3 ppl is below-
_ _ _ //_ _ _

let M be the first one. then the chances of A to be the second are 1/5, the 3d place 4C4=1

2) 1C1*1C1*4C1/6C3=1/5

If I understand your calculations correctly, then you're half way there.
This is the probability that A and M are both on committee #1.
The probability that A and M are both on committee #2 equals 1/5 as well.

So, the probability that A and M are on the same committee = 1/5 + 1/5 = 2/5

Cheers,
Brent

[Scott@TargetTestPrep]

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%

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