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

The way you are doing is wrong because 8*6*4=192 will contain duplication and to get rid of them you should divide this number by the factorial of the # of people - 3! --> 192/3!=32.

Consider this: there are two couples and we want to choose 2 people not married to each other.
Couples: \(A_1\), \(A_2\) and \(B_1\), \(B_2\). Committees possible:

\(A_1,B_1\);
\(A_1,B_2\);
\(A_2,B_1\);
\(A_2,B_2\).

Only 4 such committees are possible.

If we do the way you are doing we'll get: 4*2=8. And to get the right answer we should divide 8 by 2! --> 8/2!=4.

You can see the similar problem at:
committee-of-88772.html#p669797

Hope it helps.

So there are 3 people we need to choose from 8.

Case 1 : all 3 are men ... C(4,3)=4 ways
Case 2 : all 3 are women ... C(4,3)=4 ways
Case 3 : 2 men 1 woman ... Choose men in C(4,2) ways, then we can only choose the woman in 2 ways, since their wives can't be chosen .., hence 12x2=24 ways
Case 4 : 2 women 1 man ... Exactly same logic as case 3, 12 ways

Total ways = 4+4+12+12 = 32

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It's not clear what is the logic behind the formula.

Reversed approach would be:
There are 8C3=56 ways to select 3 people out of 8 without any restriction;
There are 4C1*6=24 ways there to be a couple among 3 members: 4C1 ways to select a couple out of 4, which will be in the committee and 6 ways to select the third remaining member (since there will be 6 members left after we select a couple out of 8 people).

56-24=32.

Hope it's clear.

Hi everybody,
I want to solve this problem with some other method. But i am definately wrong somewhere in this method... i dont understand where.

i can choose first member of the comitee in 8 ways..
removing the spouse of the selected person second member can be chosen in 6 ways....
third member in 4 ways.....
so 8*6*4 which is not answer can someone explain why?

Person (p1 p2 p3 p4 p5 p6 p7 p8)
No of ways to choose 1st Person: Any 8
No of ways to choose 2nd Person: 6 (Pair of 1st person can not be considered so we need to exclude 1 pair)
No of ways to choose 3rd Person: 4 (Pair of 1st & 2nd Person can not be considered so we need to exclude 2 pair)
No of ways : 8X6X4 (Now we have done a permutation)
But here order of the team member is not important and 3 person can arrange themselves in 3! ways. So need to divide the permutation by 3!.
Ans: 8*6*4/3! = 32

is this a correct way to get the answer? or was it just coincidence:

8C3 - 4(4C2) = 56 - 4(6) = 32

Hi Bunuel,

I'm confused by this step, which is also outlined above. "There are 4C1*6=24 ways there to be a couple among 3 members:". Can you please elaborate on this?

First couple: \(A_1,A_2\);
Second couple: \(B_1,B_2\);
Third couple: \(C_1,C_2\);
Fourth couple: \(D_1,D_2\).

We want to select 3 people: a couple and one more.

We can select any from 4 couples (4 options) and for the third member we can select any from the remaining 6 people. For example if we select \(A_1,A_2\), then we can select third member from \(B_1,B_2,C_1,C_2,D_1,D_2\).

Hope it's clear.

We can solve this in two general steps.

1) We can pretend order matters (which can be much easier to visualize)
2) Then we can think about what happens if order doesn't matter.

1) If order matters, we can visualize three chairs or slots or whatever and consider them one at a time: Anybody can sit in the first chair, so there are 8 options for chair one. But now that somebody is in chair one, there are only 6 options for chair two (it can't be the person in chair one and it can't be that person's spouce). And for every set of people occupying chairs one and two, there are only 4 remaining people allowed to sit in chair three.

That means that there are 8*6*4 ways to fill the three chairs.

2) But we're looking for 'committees.' and for committees order doesn't matter (whereas it does when we're visualizing chairs). For example, Andy, Beth, Charlie would be the same committee as Beth, Andy Charlie.

In fact there are six ways that Andy Beth and Charlie could fill the chairs:

ABC
ACB
BAC
BCA
CAB
CBA

So for each committee, we counted 6 different ways for them to fill the chairs.

In other words, our 8*6*4 number is too big by a factor of 6.

Therefore, the answer is (8*6*4)/6 = 32

We are given that there are 4 married couples (or 8 people), and we need to determine the number of ways of choosing 3 people in which no 2 people are a married couple. So this is a special combination problem. Before we tackle this problem, let’s review a combination problem with no restrictions.

With no restrictions, the number of ways of choosing 3 people from 8 is 8C3, which is calculated as follows:

8C3 = 8!/[3!(8-3)!] = (8 x 7 x 6)/3! = 56

8, 7, and 6 in the numerator represent the number of ways in which the first, second, and third person can be chosen, respectively. We divide the numerator by 3! because, in a combination problem, we do not care about the order in which the 3 people are chosen.

However, in this (special combination) problem, 3 people are chosen in which no married couple can serve together on the committee. The first person could be any one of the 8 people. However, once a person is selected, that person’s spouse cannot also be selected for the committee. This reduces the choice of the second person to 6 possible people (one person has already been selected and that person’s spouse now cannot be selected). Once the second person is chosen for the committee, that person’s spouse cannot be chosen. This reduces the number of people who could be chosen as the third person to 4. Therefore, the number of ways of choosing these 3 people is:

(8 x 6 x 4)/3! = 32

Thus, there are 32 ways to choose such a committee.

Answer: E

Hi All,

This question is fairly high-concept, but the math behind it isn't too complex. You have to be organized and you have to remember that we're forming GROUPS of people, so 'duplicate entries' are not allowed.

From the prompt, we have 4 married couples, but we have to form a group of 3 without putting a married couple in the group. Let's work through the logic in pieces...

The 1st person in the group can be ANY of the 8 people. Once we pick one of those 8, then we CANNOT pick the married partner to that person...

For the 2nd person in the group, we now have 6 people to choose from. Once we pick one of those 6, then we CANNOT pick the married partner to that person...

For the 3rd person in the group, we now have 4 people to choose from.

We now multiply those three numbers together: (8)(6)(4) = 192

We're NOT done though. Remember that we're forming GROUPS of people, and the 192 options we've figured out so far contain LOTS of duplicates... If you have 3 people: A, B and C, then that is just ONE group. However, in the above approach, you can end up with that group in 6 different ways:

ABC
ACB
BAC
BCA
CAB
CBA

Thus, every actual option has been counted 6 times, so we have to divide that total by 6...

192/6 = 32

Final Answer:
GMAT assassins aren't born, they're made,
Rich

Take the task of selecting the 3 committee members and break it into stages.

Stage 1: Select the 3 couples from which we will select 1 spouse each.
There are 4 couples, and we must select 3 of them. Since the order in which we select the 3 couples does not matter, we can use COMBINATIONS
We can select 3 couples from 4 couples in 4C3 ways (4 ways)

Stage 2: Take one of the 3 selected couples and choose 1 person to be on the committee.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.

Stage 3: Take one of the 3 selected couples and choose 1 person to be on the committee.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.

Stage 4: Take one of the 3 selected couples and choose 1 person to be on the committee.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.

By the Fundamental Counting Principle (FCP) we can complete all 4 stages (and thus create a 3-person committee) in (4)(2)(2)(2) ways (= 32 ways)

Answer = E

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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We have 4 couples. Let the couples be-

AB CD EF GH

Number of ways to select ANY 3 out of 8 is 8C3 = 56

Number of ways we select the 3 such that there is at least one couple =
Number of ways to select a couple = 4 ways
Number of ways to select the third member = 6 ways (as after picking one couple, 6 will remain)
Thus Number of ways to select the group such that there is at least one couple = 4*6 = 24

Hence, the number of ways to select the group such that there are no couples = 56 - 24 = 32

Answer E

Choosing 3 out of 8 can be done in 8C3= 56 ways. From this we need to remove the ones which have couples who are married to each other. Let's say the couples are (M1F1) (M2 F2) (M3F3) and (M4F4).
If we choose M1F1 then there are 6 choices for the third member. Similarly there will be 6 cases in which M2F2 will be there and 6 in which M3F3 will be present and 6 in which M4F4 will be present. That makes it 24 sets in which couples who are married to each other would be present. So subtracting this from 56 we have 32 sets or groups in which no couple would be present. Hence E

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