In how many different ways can 3 men and 4 women be placed into two groups of two people and one group of three people if...

Question

I am in middle school, so simply worded answers would be super helpful!

In how many different ways can 3 men and 4 women be placed into two groups of two people and one group of three people if there must be at least one man and one woman in each group? Note that identically sized groups are indistinguishable.

As I was trying to solve this problem, I realized that the two groups of 2 people had to be made up of 1 man and 1 woman. I then multiplied c(3, 1) by c(4, 1), which equals 12. I am unsure of how to proceed from there, and I don't understand what the last note means.

Answer 1

Strategy: As you realized, there will be two groups with a man and a woman and one group with two women and one man.

1. Choose which two of the women will be in the group with three people.
2. Choose which man is in that group.
3. Choose which of the two remaining men will be in the same group with the remaining woman whose name appears first in an alphabetical list. The remaining group is then determined.

The note about identically sized groups means that the groups are not labeled. Instead, what matters is who is in which group.

Answer 2

Something to keep in mind when dealing with a question like this is that the order in which you select the people does not matter. What matters is the composition, i.e. make sure that each group of 2 has 1 man and 1 woman.

Knowing this, we can safely assume that you will need to use binomial coefficients.

$\binom{n}{k} ~-~\text{The number of ways of picking $k$ unordered outcomes from $n$ possibilities}$

- Wolfram MathWorld

In regards to your attempt, your on the right track in that you first need to find the number of outcomes possible when you pick 1 man from a group of 3 and 1 woman from a group of 4.

What's left to do is determine the number of outcomes possible for the second group of 2 with the remaining men and women. Once, you figure out the number of possible outcomes of the second group of 2, you are simply left to find the final group of 3 which is composed of every man and woman remaining.

One final note to keep in mind is that the order in which you create the groups, i.e. create the groups of 2 first, then the group of 3 last doesn't matter either, so make sure to divide the total number of outcomes by the number of groups factorial (i.e. $3!$) to remove duplicates.

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Now that a couple days have past and assuming the poster figured it out on their own, I'll include the answer here for future reference.

step 1: The first group of two people $\binom{3}{1}\binom{4}{1}$ (3 men to pick from and 4 women to pick from)

step 2: The second group of two people $\binom{2}{1}\binom{3}{1}$
(2 men to pick from and 3 women to pick from because 1 man and 1 woman have been selected to be in the first group of two in step 1)

step 3: The third and final group of three people $\binom{1}{1}\binom{2}{2}$
(1 man and 2 woman remain after step 1 and step 2).

If we multiply the values from each step together and remove the duplicates (mentioned above), we get the following

$$\frac{\binom{3}{1}\binom{4}{1} \cdot \binom{2}{1}\binom{3}{1} \cdot \binom{1}{1}\binom{2}{2}}{3!} = 12 \quad \text{$\square$}$$

Answer 3

Using the same solution as @JetJet 13 we have 72 choices if all the committees were distinguishable. However since there are two committees that are the same we divide by 2 yielding us the answer of 36