From a group of 4 married couples, a team of four

[Saurabh Malpani]

Hi All,

Source of teh question is: GMAT Prep

Question: From a group of 4 married couples, a team of four is to be selected such that only one person from a couple is selected.

A) 16
b) 24
c) 26
d) 30
e) 32

Please explain
Thanks
Saurabh Malpani

[esledge]

Hi Saurabh,

In order to select a team of 4 such that no married couple is on the team together, we would have to pick one spouse from each of the 4 couples. Of course, from each couple, we could chose either spouse to serve on the team. So there are two possible choices from four couples, or 2^4 = 16 possible teams.

To illustrate, if H stands for Husband and W for Wife, and the numbers 1, 2, 3, and 4 label which couple the person is from, then the 16 possible teams are:

1) H1 H2 H3 H4
2) H1 H2 H3 W4
3) H1 H2 W3 H4
4) H1 H2 W3 W4
5) H1 W2 H3 H4
6) H1 W2 H3 W4
7) H1 W2 W3 H4
8) H1 W2 W3 W4
9) W1 H2 H3 H4
10) W1 H2 H3 W4
11) W1 H2 W3 H4
12) W1 H2 W3 W4
13) W1 W2 H3 H4
14) W1 W2 H3 W4
15) W1 W2 W3 H4
16) W1 W2 W3 W4

[Saurabh Malpani]

Hi,

Thanks for the reponse...but say if we have to select 3 people out of 8 such that a couple cannot serve together.

Then will it be 2^3?

Here is the exact question:

A committee of three people is to be chose from four married couples. What is the number of different committees if two people who are married to each other cannot serve on the committe?

Hi Saurabh,

In order to select a team of 4 such that no married couple is on the team together, we would have to pick one spouse from each of the 4 couples. Of course, from each couple, we could chose either spouse to serve on the team. So there are two possible choices from four couples, or 2^4 = 16 possible teams.

To illustrate, if H stands for Husband and W for Wife, and the numbers 1, 2, 3, and 4 label which couple the person is from, then the 16 possible teams are:

1) H1 H2 H3 H4
2) H1 H2 H3 W4
3) H1 H2 W3 H4
4) H1 H2 W3 W4
5) H1 W2 H3 H4
6) H1 W2 H3 W4
7) H1 W2 W3 H4
8) H1 W2 W3 W4
9) W1 H2 H3 H4
10) W1 H2 H3 W4
11) W1 H2 W3 H4
12) W1 H2 W3 W4
13) W1 W2 H3 H4
14) W1 W2 H3 W4
15) W1 W2 W3 H4
16) W1 W2 W3 W4

[StaceyKoprince]

Hi, Saurabh

Is the second question still from GMATPrep? Also, please post the entire question, including answer choices - we need to record full questions so others can learn from them too (and sometimes the answer choices can help suggest the easiest approach to a problem).

[Saurabh Malpani]

Yes Stacey the problem is from GMAT Prep ...I typed the wrong problem first time around (4 instead of 3)...because I was at work and didn't have access to GMAT prep.

The options are:

8
16
26
30
32

Hi, Saurabh

Is the second question still from GMATPrep? Also, please post the entire question, including answer choices - we need to record full questions so others can learn from them too (and sometimes the answer choices can help suggest the easiest approach to a problem).

[Guest]

Hi Saurabh -

This problem isn't too bad from a time point of view. I took a top-down approach to it - that is I counted the total number of ways to select a committee of three people from eight, then subtracted the "bad" cases where a couple would be on the committee together:

Total: 8C3 = 56

Now we need to look at the bad cases. Let's consider the case where couple #1 is on the committee. The third person can be chosen from any of the remaining six. So there are 6C1 = 6 ways of choosing the committee where both members of couple 1 are on the committee. The same reasoning holds for couples 2,3 and 4. So there are 6*4 = 24 "bad" ways of choosing the committee.

If there are 8C3 = 56 total ways of choosing the committee, and 24 "bad" ways of choosing the committee where a couple is on the committee, then that leaves us with 32 ways of choosing the committee where both members of the committee are NOT on the board.

-------------------------------------------------------------------

After spending about 2.5 minutes solving the problem from a "top-down" approach, I realized it's probably faster to go with a "bottom-up" approach: The fist committee member can be selected from any of the 8 candidates. The second can be selected from 6 candidates (not 7, since the spouse of the first candidate selected isn't allowed.). The third member can be selected from 4 candidates (excluding the spouses of the first two selected). So this gives 8*6*4 = 192. Order does not matter, so you need to divide by 3! = 6. 192/6 = 32.

All of these problems can be solved either way - I find I have an uncanny knack for choosing the one that is the more computationally intensive!

Cheers,

Jeff

[christiancryan]

I like Jeff's second way of doing this. Often, to apply constraints, use a "successive choice" method to write down the number of choices you have at each step, AS IF you were filling distinguishable seats. So, you get 8*6*4. THEN worry about whether order matters or not. Since it does not in this question (we're just choosing a committee or a group), you divide by the factorial of the group size (3! in this case). This method allows you to handle many "GMAT-like" constraints more easily. Even the most difficult combinatorics problems on the GMAT don't ask you to handle lots & lots of subcases and then add them up (or subtract them to get rid of illegal arrangements), so if you can apply the constraint correctly using "successive choice" and then, if necessary, divide by the factorial of the group (to make order not matter), then you'll often be better off.