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?




This is how I approached the first part of the problem.

There are 3 seats available. 1 seat is occupied by Michael.
So we are left with 2 seats and 5 members.
i.e M X 5 X 4
or 1 X 5 X 4
This is equal to 20.
What is wrong with my approach.

Permutation and combination is my weakest topic. I am really not sure how to distinguish between perm. and comb.

Pls help......

Hello,

My solution is that ... because 6 people are split into 2 groups so I just try to find the probability to pick 1 group neither Michael nor Anthony. That's also the rate of M & A appear in other group.

x x x x M A

(3 picks not permutation)
1st : x --- 4/6
2nd: xx --- (4/6) * (3/5)
3rd : xxx --- (4/6) * (3/5) * (2/4) = 1/5 = 20%

The good news is that Permutations and Combinations doesn't show up all that much on the GMAT. You might get one or two on the whole exam. So this one question type won't be what makes or breaks your score. So I wouldn't worry about this too much; focus your attention to other areas of weakness.

A quick answer to your question about Permutations and Combinations. Permutations is used when order matters. Combinations is used when order does not matter.

In this case, if we want to form a subcommittee with members A, B, and C, it doesn't really matter if I select A, B, then C; or A, C, then B; or B, C, then A; or C, B, then A. All three options give me the same subcommittee. This is a combinations problem.

However, if in this committee, we select a President, a VP, and a Secretary, the order does matter. Because A, B, then C is different from B, C, then A (in order of Pres, VP, and Sec). This would be a permutations problem.

If you follow MGMAT's Strategy Guides, we recommend the Anagram Grid. The beauty of this approach is that we don't need to know either permutations or combinations to do the problem; it automatically works itself out.

Let's suppose the six member board is Anthony (A), Michael (M) and the other four (B, C, D, and E). When we solve this problem, we are looking for the percent of all possibilities. So basically we need to calculate
[(Number of subcommittees that include A &M) / (Number of subcommittees that include M)] * 100%

We want to first find the number of subcommittees that have A and M in it. Well, if we already know A and M are in it, then there are only four others (B, C, D, or E) that can join them for the third spot in that subcommittees. There are only 4 subcommittes with both A and M.

Now if we want to find the number of subcommittees that include M, we have two spots remaining that can be filled with five Board members (A, B, C, D, or E). Using the Anagram approach, we can label "Y" if they are selected on the subcommittee, or "N" if they are not.

A, B, C, D, E
Y, Y, N, N, N

So the question is how many anagrams we can form from YYNNN? This will be 5!/(2!3!) = 10.

So the percent of subcommittees with M that also include A is (4 / 10)*100% = 40%.

The combinations approach would be:
(4C1)/(5C2) = [(4!)/(3!1!)]/[(5!)/(3!2!)] = 4 / 10 = 40%

_________________
Ben Ku
Instructor
ManhattanGMAT

Thanks for the expln.. Ben.

You're welcome!

_________________
Ben Ku
Instructor
ManhattanGMAT

This is a v helpful explanation by Ben.
However, I had a question. The Q states that the 6 member board is going to be broken up into TWO 3-person subcommittees. therefore, shouldn't be doubling the 40% to also account for the fact that Anthony and Michael may both be present in the 2nd of the two subcommittees?

Thanks for the explanation.

But I still havent understood the way to calculate no. of ways Michael will be selected.

If I dont use Anagram technique :

If 1 position is taken by Michael , then there are 2 positions left in that committee which can be taken by 5 ways and 4 ways.

SO total no . of ways = 5x 4 =20.
M 5 4
-- -- -- -- -- --

PLease explain the flaw in above technique.

Ben,

Please appraise my explanation .

Michael will be on one of the two committees for sure. Which will leave two seats on the 3 member committee. thus Anthony has to be one of the 2 members of the 5 remaining members in order to be on the same committee as Michael.
i.e. 2/5 or 40% .

Okay folks, here's the deal:
There are 3 chairs on one side of the room and 3 chairs on the other side. These are our two committees. Michael is already seated in one of the chairs. We want to know what the probability is that Anthony will be on the same committee. Well, how many empty chairs are there for Anthony to sit in? And how many are on the same side as Michael? Out of 5 total chairs, there are 2 next to Michael. Thus a 2/5 or 40% chance that Anthony will be in the same committee as Michael. Let me know if this still doesn't clear things up..

_________________
Tim Sanders
Manhattan GMAT Instructor

Hi to all,

I would like to learn if my approach to problem is correct. Below is the how I attacked this question.

1-) I assumed Ant. and Mich. as one person, so there might be 5!=120 possible group options.

2-) The problem says 2 subgroups;
A, M, 4
3, 2, 1 4!= 24 subgroups Ant and Mich together

3-) another 4! since A and M might be in second subgroup too
4, 3, 2
A, M, 1

4-) 48/120 = 40 %

It seems long solution but it took less than 1 minute to apply.
However I am not sure if this approach is right or I found the correct answer by coincidence. I appreciate and thank your comments in advance...
Btw I am new to the MGMAT.

Sorry, no.

This method will only work when (unclaimed seats in Michael's group:unclaimed seats elsewhere) = (number of groups: one less than the number of members). It just works by coincidence.

Also, 4! does not equal the number of subgroups that include both Michael and Anthony. You are making two mistakes here, one of which is to conflate permutations and combinations. There are four groups that include both M and A, though each of these could be arranged in 6 different ways, if order mattered. It doesn't.

Tim - That's an awesome explanation! I feel other methods might take more than 2 mins :)

Very nice work all!

Thank you,

_________________
Jamie Nelson
ManhattanGMAT Instructor

im very bad with combinations and permutations

but i used the following approach and would like to know if my reasoning is correct:

let the seats be
A1 A2 A3 and B1 B2 B3
and the following be where A n M can be seated
M A -- -- -- --
-- M A -- -- --
-- -- M A -- --
-- M -- -- A --
M -- -- -- A --

so basically, what i have done is listed out all the seats where M sits and where A and M are not in the same sub committee I have made A sit on the other sub committee in any manner (because thats not what we are concerned with-in this question)
so total ways in which M can sit=5
and committees where A sits with M=2
therefore, 2/5 *100=percentage=40%

am i correct?

While your answer is correct, I'm not understanding how you are getting 5 total ways that M can sit, considering that we don't care about order or arrangement in this problem. Can you please explain?

_________________
Jamie Nelson
ManhattanGMAT Instructor