TITLE: Permutations with Repetition Formula
QUESTION [6 upvotes]: Let us say we are trying to compute the unique permutations of the word:
$$PEPPER$$
We cannot compute this by relabeling:
$$P_1E_1P_2P_3E_2R$$
Because this would simply be $6!$ and does not take into account repetition. However, I have seen the formula:
$$
\frac{6!}{3! 2!}
$$
I.e take the total number of positions, and then divide by the product of factorials of repeated elements. My question is, why does this work? Could anyone supply a combinatorial argument?

REPLY [3 votes]: Right, with repetition there are $6!$ ways. To convert from "with repetition" to "without repetition", you need to remove the equivalents. For example, consider $P_1P_2P_3E_1E_2R$. Now you need to ignore other choices with the three letters $P$ in the first three positions. Overall, there are $3!$ choices with the first three letters $P$:
$$\color{red}{P_1}\color{green}{P_2}\color{blue}{P_3}E_1E_2R \equiv \\
\color{red}{P_1}\color{blue}{P_3}\color{green}{P_2}E_1E_2R \equiv \\
\color{green}{P_2}\color{red}{P_1}\color{blue}{P_3}E_1E_2R \equiv \\
\color{green}{P_2}\color{blue}{P_3}\color{red}{P_1}E_1E_2R \equiv \\
\color{blue}{P_3}\color{red}{P_1}\color{green}{P_2}E_1E_2R \equiv \\
\color{blue}{P_3}\color{green}{P_2}\color{red}{P_1}E_1E_2R.\ \  $$
Hence, you must divide by $3!$. Similarly, for other letters.