A committee of three people is to be chosen from four married couples.

Solution:

4 couples=>8 people

1st person-can be selected from the 8 people => 8 ways

2nd person- should not be a spouse of the first and hence we have=>6 ways to choose him/her (7-1=6)

3rd person- should not be a spouse of either of the 2, so we can choose him in =>4 ways.

Total no. of ways we can choose the people will be 8*6*4 ways

Since order is not important, total number of groups = 8*6*4/3!

=32 (option e)

Say, for example, that each person is assigned the same letter as that particular person's spouse. (AA, BB, CC, DD, EE)

First, we calculate all the possibilities to choose three people out of eight (4 couples x 2 people).

\(\frac{{8!}}{{3!(8-3)!}}\) = 56 possibilities.

Next, all we need to do is to calculate the number of possibilities that two people who are married to each other both serve on the committee.
\(8*1*6= 48\) (8 because we can initially choose all people, 1 represents the previously chosen person's spouse, and 6 represent the remaining six people)

Next, we have to divide 48 by 2! = 24, because we definitely do not want to double count the same letter. For instance, the chosen committee comprises of A, A, and D.

As a result, the answer will be \(56 - 24 = 32\) (E)

Please correct me if I am wrong. Thanks!

Thank you

I think you listed five couples initially (you included "EE") instead of four, but otherwise your solution is good. There's a slightly easier way to count the number of committees that include a married couple (i.e. the number of committees we do not want to count here). We have four couples, so we have 4 choices for the two married people who will serve on the committee, and then we'll have 6 people left to fill the third slot on the committee, so there are 4*6 = 24 committees we can make that include one married couple. Then, as you say, there must be 8C3 - 24 = 56 - 24 = 32 committees which do not include a married couple.

Could somebody please explain how "since the order is not important" translates into "we divide the total by 3! ways"?

If you take three things, say the letters A, B and C, there are 3! = 6 different orders you can put them in:

ABC
ACB
BAC
BCA
CAB
CBA

Now say A, B and C are three of the ten employees of a company (short for Ali, Beka and Carlos, say). You might be asked to count how many ways you can pick three of these ten employees where order is important, or where order is not important. So you might be asked:

• how many selections are possible of a President, Vice-President and Treasurer (if different people occupy each role)? Then order is important, and we have 10 choices for President, 9 for Vice-President, and 8 for Treasurer, for 10*9*8 possibilities in total. Notice when we happen to pick Ali, Beka and Carlos for the three positions, we have 6 different possibilities -- the six I listed above. For example, if we pick them in the order ABC, then Ali is President, Beka is VP, and Carlos is Treasurer, while if we pick them in the order CAB, Carlos is President, Ali is VP, and Beka is Treasurer.

• how many selections are possible for a board of 3 directors? Then order is not important -- we have the same Board no matter what order we list our names in. Now notice that when we happen to pick Ali, Beka and Carlos, there is only one Board we can make. The six possibilities I listed above (ABC, ACB, BAC, BCA, CAB, CBA) are all the same Board. So when order matters, when we pick 3 specific people, we get 3! = 6 different possibilities, but when order does not matter, we only get 1 possibility. When order matters, the number of possibilities is 3! = 6 times larger than when order doesn't matter. So if we counted pretending order matters, we can just divide by 3! to find the number of possibilities when order does not matter.

In general, if you're picking k things, and their order doesn't matter, you can count all your possibilities by pretending first that order does matter, and then dividing by k! to account for the fact that order does not matter.

You can approach this problem using the box-and-fill-method (courtesy of Target Test Prep).

There are three spots available.

For the first spot, we can select from a pool of eight people.
For the second spot, we can select from a remaining pool of seven people minus one, which is the husband/wife of the person we first selected, so six.
For the third spot, we can select from a pool of six people minus the husbands/wifes of the first two people we selected, so four.

Multiply and, since this is a combination problem, divide by 3!.

8 * 6 * 4/3! = 8 * 4 = 32

Just because I think it's important to credit these things correctly: that method dates back more than 1000 years in Indian mathematics, and all of the combinatorial methods people use on the GMAT were known by the time Pascal and Leibniz wrote De Arte Combinatoria in 1666, many of them long earlier. The method you cite appears, under different names, in every decent GMAT book, including many written before that prep company existed. I'd be surprised if there was a single mathematical method in any GMAT prep material that a prep company deserves credit for. The only 'methods' I know of that are original to prep companies are 'methods' specific to the GMAT (e.g. "guess D on Data Sufficiency inequality questions") and most of those don't work.

Bunuel, I have a doubt here. I solved it like this, 4 couples - 4 women and 4 men.
1) 4C2 * 2C1 - 12 (2 women, 1 man)
2) 4C1 * 3C2 - 12 (1 woman, 2 men)
3) 4C0 * 4C4 - 1 (0 Women, 4 men)
4) 4C4 * 4C0 - 1 (4 women, 0 men)

I am getting 26, what is wrong with this approach?

Hi anuhyadixit,

Your approach can absolutely get you to the correct answer - but you have made a slight error in your calculations. The group is supposed to be 3 people (and in your 3rd and 4th calculations, you assume that the group is 4 people, which is why you're only getting 1 outcome for each of those two options). If you calculate 4c3, then you will have 4 groups of all women and 4 groups of all men, which will bring the total to 32.

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: Rich.C@empowergmat.com

ScottTargetTestPrep, I tried to solve this question with the approach discussed in TTP. Total = Ways Together + Ways Not Together.

I didn´t know how to calculate the number of ways together. Could you walk me through that approach?

Thank you in advance!

I see your question was answered below! I would like to add, that I prefer my method I used since I find it to be the most straightforward way to solve this problem.

I tried to solve this sum by using combinations.
My solution: 8c1 * 6c1 * 4c1= 56
In the solutions we have divided this by 3!. I dont understand the logic behind this. Since combinations already ignores order, why are we dividing again by 3!?

Combinations does ignore order, so if you chose 3 out of 8, it will be 8C3. But here in 8*6*4, you are doing arrangements.
So, if you had three people A, B and C, you would get one of the order was A, B,C and other will be BAC and so on.

Now, the members are same A, B and C, but they are taken as different because we take order to matter.
A could be chosen in 8C1, B in 6C1 and C in 4C1. But it is possible that B is chosen in 8C1, A in 6C1 and C in 4C1 and so on.

A, B and C can be arranged in 3*2*1 or 3! ways. So if order does not matter, we divide the ways(where order matters) by 3!­

How would the solution change if the question did not say A (1) team of 3 is to be selected?
Instead just how many committes..... would that be 3C8 - (1C4*6) + (1C3 *4)? or 3C8 - [(2C2)^4 * 1C6]?
2C2^4: there are 4 couple and we choose both members of the couple
1C6 ways to choose 1 member out of 6 persons

We could make 2 committes with 2 siblings + another person. Then there would be one spare couple of siblings unchosen.
Right? I am confused....
Any clarification would help a lot! Thanks!­