Why do we divide by $2^2$ for $(6, 6, 5, 5)$ to find the permutations?

Question

Why do we divide by ${2^2}$ and are we using the permutations formula?

$(6, 6, 5, 5)$

$\frac{4!}{2^2}= {6}$

Answer 1

Rather than a "division by symmetry" argument, I find it far more convincing to avoid division entirely.

To count the number of arrangements of the digits $5,5,6,6$ you can count this directly by deciding in which two of the four available positions the $5$'s are occupying. This can be done in $\binom{4}{2}=6$ ways. The $6$'s will then occupy whatever spaces are left.

Even more generally, the number of ways of arranging $k_1$ copies of $1$, $k_2$ copies of $2$, on up to $k_n$ copies of $n$ where $k_1+k_2+\dots+k_n = N$ can be counted direction using the multinomial coefficient $\binom{N}{k_1,k_2,\dots,k_n}=\frac{N!}{k_1!k_2!\cdots k_n!}$ and this is seen to be equal to having chosen where the $1$'s are placed, multiplying by the number of ways of where the $2$'s are placed from those still available spots, etc... to give:

$$\binom{N}{k_1,k_2,\dots,k_n} = \binom{N}{k_1}\binom{N-k_1}{k_2}\binom{N-k_1-k_2}{k_3}\cdots\binom{N-k_1-\dots-k_{n-1}}{k_n}$$

Of course, the algebra is made much simpler to just take note of the final result and use $\binom{N}{k_1,k_2,\dots,k_n} = \frac{N!}{k_1!k_2!\dots k_n!}$ directly.

Answer 2

The correct way to write it is $$ \frac{4!}{2!2!} $$

The terms in denominator reflect the fact that permutation of identical elements does not matter. Since the number in numerator counts all permutations as if all the elements were distinct, it should be divided by numbers of all possible permutations in every group of identical elements (counted as if the group elements were distinct).

For $(1,2,2,3,3,3)$ it would be $$ \frac {6!}{1!2!3!}$$ and so on.

Answer 3

Because the four permutations $$ (\color{red}6, \color{blue}6, \color{red}5, \color{blue}5)\\ (\color{red}6, \color{blue}6, \color{blue}5, \color{red}5)\\ (\color{blue}6, \color{red}6, \color{red}5, \color{blue}5)\\ (\color{blue}6, \color{red}6, \color{blue}5, \color{red}5) $$ are considered to be the same.