crackgmat007 wrote: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%

OA - B pls explain

The number of way of dividing 6 people into 2 teams of 3 each without any restriction:
(6C3 x 3C3)/2! ( Choose the first 3 from 6 and the rest from 3. Since order does not matter, divide by 2!). This should give 10 ways.

Now fix michael and Anthony and choose 1 person from the remaining 4 to join them. This is 4C1. The remaining 3 people can be chosen from the remaining 3. So 4C1 x 3C3/10 = 4/10 or 40%.

Choose C.

Since order does not matter, divide by 2!).

can you explain the above pls.

Also, answer seems to be B, not sure if OA is wrong...can someone clarify

crackgmat007 wrote:

Since order does not matter, divide by 2!).

can you explain the above pls.

Also, answer seems to be B, not sure if OA is wrong...can someone clarify

AMO PQR is the same as MAO RPQ. Without the 2! you would have both of these in your total list of arrangements and you really need only 1. That is what is meant by order is not important. Regarding the OA,

AM OPQR

Since AM can't move, we have 4 choices of one person to join them. Take O. Then you have one group comprising AMO

You are left with PQR. In how many ways can these three people form a group of 3? Only 1 way. They have no choice even if they hate each other. That means the total ways is 4 x1 =4

Consider another approach. . So we want

AM + 1 of OPQR. imagine you have formed AMO the 4C1 way as above. So you have 1 team already and you want the other team. Fix R. That leaves 2 persons P and Q. You want these 2 to join R . But that can be done in 2C2 ways =1. In order words if AM MUST be together here are all the possibilities.

AMO PQR
AMP OQR
AMQ POR
AMR POQ

The OA says 30%. To have 30% we must have 3/10. I don't see how you get a 3 in the numerator.How can you only have 3 in the numberator when you have 4 possibilities?

Yup.. I also think that the answer should be 40%..

Once you select that Michael is on one team, number of ways to fill other two people is 5C2 = 10

If Michael and Antony are together.. then there are 4 different ways you can fill the last position..

so possible committees = 4/10 = 40%

dumb.doofus wrote:Yup.. I also think that the answer should be 40%..

Once you select that Michael is on one team, number of ways to fill other two people is 5C2 = 10

If Michael and Antony are together.. then there are 4 different ways you can fill the last position..

so possible committees = 4/10 = 40%

Without using the combinations formula:

Once we fix M, we have 2 spots to fill.

The chance of picking A 2nd is:

1/5 * 4/4 = 1/5

The chance of picking A 3rd is:

4/5 * 1/4 = 1/5

Since we can pick him either 2nd or 3rd, the chance of picking A at all is 1/5 + 1/5 = 2/5 = 40%