Combinatorial argument why is the following below true?

Question

$\frac{12!}{2^6 \cdot 3! \cdot 3!} = {12 \choose 6} \cdot 5^2 \cdot 3^2 \cdot 1^2 $

I'm trying to formulate a combinatorial argument and have singled out a case of the two formulas above hoping to see if anyone can explain why the above is true. If so it may help me solve the argument for a general case. Any insight or tips would be much appreciated.

Here is what I know so far from the left side of the equation: $\frac{12!}{2^6 \cdot 3! \cdot 3!}$ it seems like we are dispersing $12$ distinct objects to two bins of size $3$ but also dividing all the possible subsets of size $6$ (denoted by $2^6$)? Don't know if that is true but this is what I'm inferring right now.

And the right side seems to first pick out a subset of size $6$, and a bit lost from there on what the $5^2 \cdot 3^2 \cdot 1^2$ denotes and how it can be interpreted with the subset of size $6$ using the product rule.

Answer 1

I believe this equality cannot be easy explained in combinatorial terms. More easy way is to show this directly: $$\frac{12!}{2^6\cdot 3!\cdot 3!}=\frac{12!}{6!\cdot6!}\cdot\left(\frac{6!}{2^3\cdot 3!}\right)^2= {12 \choose 6}\cdot\left(\frac{6!!\cdot 5!!}{2^3\cdot 3!}\right)^2={12 \choose 6}\cdot(5!!)^2={12 \choose 6}\cdot 5^2\cdot 3^2\cdot 1^2$$

Answer 2

You can factor the $2^6$ out of the $12!$ to give

$$\binom{6}{3}11!!=\binom{12}{6}5^23^21^2$$

which is pair $12$ people into $6$ groups, and split them into two rooms, as per @NFTaussig's answer.

This can be generalized:

$$\binom{2k}{k}(4k-1)!!=\binom{4k}{2k}(2k-1)!!^2$$

The double factorials cancel a bit to give, for example

$$\binom{6}{3}\frac{11\cdot9\cdot7}{5\cdot3\cdot1}=\binom{12}{6}$$

Answer 3

Suppose we have $12$ people. We will place six people each in two labeled rooms, $A$ and $B$, and then place the people in each room in three pairs.

There are $\binom{12}{6}$ ways of selecting which six of the twelve people will be placed in room $A$. The rest must be placed in room $B$. Line up the six people in room $A$ in some order, say alphabetically. There are five ways to pair one of the other people in room $A$ with the first person in the line. Remove those people from the line. That leaves four people in that line. There are three ways to pair one of the other three people in the line with the first person remaining in the line. Remove those people from the line. The remaining two people can be paired in one way. Hence, the six people in room $A$ can be placed in three pairs in $5 \cdot 3 \cdot 1$ ways. The six people in room $B$ can be placed in pairs in $5 \cdot 3 \cdot 1$ ways. Hence, the number of ways of placing six people in room $A$ and six people in room $B$, then placing the people in each room in three pairs is $$\binom{12}{6}(5 \cdot 3 \cdot 1)^2 = \binom{12}{6}5^2 \cdot 3^2 \cdot 1^2$$

There are $$\binom{12}{2}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{12!}{2!10!} \cdot \frac{10!}{2!8!} \cdot \frac{8!}{2!6!} \cdot \frac{6!}{2!4!} \cdot \frac{4!}{2!2!} \cdot \frac{2!}{2!0!} = \frac{12!}{2^6}$$ ways to make an ordered selection of six pairs of people from the $12$ available people. We will place the first three selected pairs in room $A$ and the remaining three selected pairs in room $B$. However, doing so counts the same three pairs placed in room $A$ in $3!$ ways since the order in which we select the three pairs we place in room $A$ does not matter. Similarly, we have counted the same three pairs of people placed in room $B$ in $3!$ ways. Hence, by making an ordered selection of pairs, we have counted each valid arrangement $3!3!$ ways. Hence, the number of ways of placing six people in room $A$ and six people in room $B$, then placing the people in each room in three pairs is $$\frac{1}{3!3!}\binom{12}{2}\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{12!}{2^6 \cdot 3! \cdot 3!}$$

We have counted the same set of arrangements in two different ways. Equating the two expressions gives the result.