Thread: Splitting 18 children into 3 equally sized groups. Is my answer correct?

1. Splitting 18 children into 3 equally sized groups. Is my answer correct?

In how many can a group of 18 children be split into 3 groups of equal size? (The names of the groups isn't important, only who is with who).

We'll line the kids up in a line. There are 18! ways to do this.
The first 6 kids are group 1, the 2nd 6 are group 2 and the 3rd 6 are group 3.
We'll divide 18! by 3! because all that matters is whom is whom and not the name of the group.
We'll divide this number by 6! because the group ABCDEF is the same as BACDEF.

So in total we have 18!/(3!*6!) ways of splitting up the kids. Is this correct?

Thank you

2. Originally Posted by durrrrrrrr
In how many can a group of 18 children be split into 3 groups of equal size? (The names of the groups isn't important, only who is with who).
Your question is commonly known as involving unordered partitions.
Suppose that we have $\displaystyle N$ distinct objects to be grouped into $\displaystyle j$ cells of $\displaystyle k$ each.
Clearly it must be true that $\displaystyle N=k\cdot j~$.
The number of ways that can be done is $\displaystyle \dfrac{N!}{(k!)^j\,(j!)}$.

3. So the answer is 18!/((6!^3)*3!). I think I understand my mistake.

4. QQ

Hello, durrrrrrrr!

Your reasoning is correct . . . but incomplete.

$\displaystyle \text{In how many ways can a group of 18 children be split into 3 groups of 6?}$
$\displaystyle \text{(The na{m}es of the groups isn't important, only who is with who).}$

$\displaystyle \text{Is this the correct answer?}$

$\displaystyle \text}We line the kids up in a row.\: There are 18! ways to do this.}$

For example: .$\displaystyle ABCD{E}FGHIJKLMNOPQR$

$\displaystyle \text{The first 6 kids are one group, the second 6 are in another group,}$
$\displaystyle \text{and the last 6 are in yet another group.}$

So we have: .$\displaystyle |\,ABCD{E}F\,|\,GHIJKL\,|\,MNOPQR\,|$

$\displaystyle \text{We divide }18!\text{ by }3!\text{, because the groups are indisguishable.}$ . Right!

$\displaystyle \text{We divide this number by }6!\text{ because }ABCD{E}F\text{ is the same as }BACD{E}F.$
. . Keep going!

We divide by $\displaystyle 6!$ again, because $\displaystyle GHIJKL$ is the same as $\displaystyle HGIJKL.$

We divide by $\displaystyle 6!$ again, because $\displaystyle MNOPQR$ is the same as $\displaystyle NMOPQR.$

And that's why the answer is: .$\displaystyle \dfrac{18!}{3!(6!)^3}$