# Number of groups

#### goosey00

##### Member
I am not quite sure if I am using the correct formula. The problem is -A class of 30 students(12 male and 18 female) are put into groups of 3. How many combinations can be formed if the requirement is that no group can be entirely male or female? I get 4060 since it doesn't matter the order except it can't be the entire amount. Is that correct? That is the part I get stumped on.

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#### Jameson

Staff member
I am not quite sure if I am using the correct formula. The problem is -A class of 30 students(12 male and 18 female) are put into groups of 3. How many combinations can be formed if the requirement is that no group can be entirely male or female? I get 4060 since it doesn't matter the order except it can't be the entire amount. Is that correct? That is the part I get stumped on.
Hi goosey00,

I think I have the way to do this but it involves using combinations and I'm not sure if you are able to use them. If there can't be all male or all female groups that means that the group can either have 2F+1M or 2M+1F.

# of ways for 2F+1M: $$\displaystyle \binom{18}{2} \binom{12}{1}$$
# of ways for 2M+1F: $$\displaystyle \binom{12}{2} \binom{18}{1}$$

Now you just add those up. There should be another way to do this where you divide your answer, $$\displaystyle \binom{30}{3}$$ by something but I don't see it yet.

#### goosey00

##### Member
Hi goosey00,

I think I have the way to do this but it involves using combinations and I'm not sure if you are able to use them. If there can't be all male or all female groups that means that the group can either have 2F+1M or 2M+1F.

# of ways for 2F+1M: $$\displaystyle \binom{18}{2} \binom{12}{1}$$
# of ways for 2M+1F: $$\displaystyle \binom{12}{2} \binom{18}{1}$$

Now you just add those up. There should be another way to do this where you divide your answer, $$\displaystyle \binom{30}{3}$$ by something but I don't see it yet.
Other ways we were doing similar problems is using either either nPr or nCr formula. Thats where I got this answer. Is that wrong?

#### Jameson

Staff member
Other ways we were doing similar problems is using either either nPr or nCr formula. Thats where I got this answer. Is that wrong?
$$\displaystyle \binom{18}{2}$$ means 18C2 so that's the formula I'm referring to when I write that.

#### Plato

##### Well-known member
MHB Math Helper
I am not quite sure if I am using the correct formula. The problem is -A class of 30 students(12 male and 18 female) are put into groups of 3. How many combinations can be formed if the requirement is that no group can be entirely male or female? I get 4060 since it doesn't matter the order except it can't be the entire amount. Is that correct? That is the part I get stumped on.
I did not answer yesterday, because I wanted to see what you may find.

It is must more complicated that it appears. There are called unordered partitions.

A set of thirty people can be divided into three subset of ten each in $\dfrac{30!}{6(10!)^3}=925166131890$ ways.
Now that counts many many single-sex groupings.

There $\binom{12}{10}\dfrac{20!}{4(10!)^2}$ to have at least one all male group.

There $\binom{18}{10}\dfrac{20!}{4(10!)^2}$ to have at least one all female group.

There $\binom{12}{10} \binom{18}{10}$ to have two single-sex groups.

So to answer this whole question:
$\dfrac{30!}{6(10!)^3}-\binom{12}{10}\dfrac{20!}{4(10!)^2}-\binom{18}{10}\dfrac{20!}{4(10!)^2}+\binom{12}{10} \binom{18}{10}$

#### Jameson

Staff member
Thank you for posting, Plato. I see now how I misinterpreted the problem. It's not the number of ways to choose 1 group of 3, but 10 groups of 3.

#### Plato

##### Well-known member
MHB Math Helper
Thank you for posting, Plato. I see now how I misinterpreted the problem. It's not the number of ways to choose 1 group of 3, but 10 groups of 3.

I don't know which it is. I was working off your reply. If it is ten groups of thee then my reply is wrong. The total there is $\dfrac{30!}{(3!)^{10}(10!)}=1208883745669600000$.
That really complicates the question, because there could be many more same-sex groups. Using inclusion/exclusion there is a nightmare.

#### Jameson

Staff member
I don't know which it is. I was working off your reply. If it is ten groups of thee then my reply is wrong. The total there is $\dfrac{30!}{(3!)^{10}(10!)}=1208883745669600000$.
That really complicates the question, because there could be many more same-sex groups. Using inclusion/exclusion there is a nightmare.
I don't follow your other reply at all then. If you the situation where you have 30 objects and want to find the number of ways to arrange 3 objects you can make from the 30, that is simply $$\displaystyle \binom{30}{3}$$. I broke my previous reply into the 2F+1M and 2M+1F cases to account for the restriction that you can't have 3M or 3F groups. Due to this restriction the answer should be less than $$\displaystyle \binom{30}{3}$$ it seems to me.

A set of thirty people can be divided into three subset of ten each in $\dfrac{30!}{6(10!)^3}=925166131890$ ways.
The part in bold makes me think that your reply was yet another interpretation. It appears to be breaking into 3 groups of 10.

So it could be 1 group 3, 3 groups of 10 or 10 groups of 3. I still kind of lean towards the first interpretation since the OP is just beginning to study combinatorics.

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#### Plato

##### Well-known member
MHB Math Helper
The part in bold makes me think that your reply was yet another interpretation. It appears to be breaking into 3 groups of 10.
So it could be 1 group 3, 3 groups of 10 or 10 groups of 3. I still kind of lean towards the first interpretation since the OP is just beginning to study combinatorics.
Three groups of ten each is doable. That is what I thought it meant, particularly at this level. Even tough unordered partitions are advanced.

BUT ten groups of three each is almost impossible at this level with the restrictions.
Here are some combinations. Because we cannot two females in each group.
We cannot have one male in each group. Thus working out all possible combinations is a nightmare. Then the possible over-counts can be many.

Experience tells me that the three group question was intended.