Problem Solving

A committee that includes 6 members is about to be divided into 2 subcommittees with 3 members each. On what percent of the possible subcommittees that Michael is a member of is David also a member?

A, 10%
B, 20%
C, 25%
D, 40%
E, 50%

Source: Master GMAT Practice Test (free)
Solution:
Visualize the problem thus: put 3 people in room A and the other 3 in room B. Say Michael is in room A, thus room A has 2 more open places out of 5 (2 in Michael's room plus 3 in the other room) - the chance that David will be in one of the 2 places in Michael's room are 2/5 = 40%

I have gone through the solution provided above but I want a more standard way of solving this problem using Combination formula. Tricks as those suggested here may not be applied to every problem.

kartikshah wrote:A committee that includes 6 members is about to be divided into 2 subcommittees with 3 members each. On what percent of the possible subcommittees that Michael is a member of is David also a member?

A, 10%
B, 20%
C, 25%
D, 40%
E, 50%

Source: Master GMAT Practice Test (free)
Solution:
Visualize the problem thus: put 3 people in room A and the other 3 in room B. Say Michael is in room A, thus room A has 2 more open places out of 5 (2 in Michael's room plus 3 in the other room) - the chance that David will be in one of the 2 places in Michael's room are 2/5 = 40%

I have gone through the solution provided above but I want a more standard way of solving this problem using Combination formula. Tricks as those suggested here may not be applied to every problem.

3-member committees that include Michael:
There are two spaces to fill on this committee and 5 people left.
Number of options = 5C2 = 10.

3-member committees that include Michael and David:
There is one space to fill on this committee and 4 people left.
Number of options = 4.

(committees with both Michael and David)/(committees with Michael only) = 4/10 = 40%.

The correct answer is D.

Hi Mitch,

Thanks for posting the solution. I was wondering what if the question was modified slightly to read as below:
There are 6 members on a committee that is to be subdivided into two committees of 3 members each. Michael and David are two of the six members. What is the possibility that they are always together in the same sub-committee?

Would the possibility be worked out as under:
Total ways of forming the sub-committees = 6C3 = 20 ways
Total ways of forming a sub-committee with M+D+one more member = 4C1 = 4 ways
Possibility = 4/20 = 20%

Thanks for clarifying.

kartikshah wrote:Hi Mitch,

Thanks for posting the solution. I was wondering what if the question was modified slightly to read as below:
There are 6 members on a committee that is to be subdivided into two committees of 3 members each. Michael and David are two of the six members. What is the possibility that they are always together in the same sub-committee?

Would the possibility be worked out as under:
Total ways of forming the sub-committees = 6C3 = 20 ways
Total ways of forming a sub-committee with M+D+one more member = 4C1 = 4 ways
Possibility = 4/20 = 20%

Thanks for clarifying.

Yes. You have done the modified question correctly.

kartikshah wrote:Hi Mitch,

Thanks for posting the solution. I was wondering what if the question was modified slightly to read as below:
There are 6 members on a committee that is to be subdivided into two committees of 3 members each. Michael and David are two of the six members. What is the possibility that they are always together in the same sub-committee?

Would the possibility be worked out as under:
Total ways of forming the sub-committees = 6C3 = 20 ways
Total ways of forming a sub-committee with M+D+one more member = 4C1 = 4 ways
Possibility = 4/20 = 20%

Thanks for clarifying.

If the 6 people are being ASSIGNED to subcommittees A and B, each with 3 members:
The probability that Michael and David are both assigned to subcommittee A is 20%.
The probability that Michael and David are both assigned to subcommittee B is 20%.
The probability that Michael and David are assigned to the SAME subcommittee -- either A OR B -- is 40%.

The key here is that the 6 people are being DIVIDED into TWO groups of 3.
Since there are TWO groups of 3, there are TWICE as many ways to get a favorable outcome.

Here are all of the ways to DIVIDE 6 people M, D, W, X, Y and Z into TWO groups of 3:
MDW - XYZ
MDX - WYZ
MDY - WXZ
MDZ - WXY
MWX - DYZ
MWY - DXZ
MWZ - DXY
MXY - DWZ
MXZ - DWY
MYZ - DWX

In the 4 red groupings, Michael and David are together.
Since there are 10 total ways to DIVIDE the 6 people:
P(Michael and David are in the same group) = 4/10 = 40%.

Using combinatorics:
The number of ways to DIVIDE 6 people into two groups of 3 = 6C3/2! = 10.
The number of ways to form a group of 3 that includes both Michael and David = 4.
P(Michael and David are in the same group) = 4/10 = 40%.

There is a distinction between ASSIGNING people to groups and DIVIDING them into groups.
For more on this distinction, check my posts here:

https://www.beatthegmat.com/combination-t92104.html
https://www.beatthegmat.com/forming-teams-t73034.html