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?
Best Answer
Since the word $PEPPER$ has six letters, we have six positions to fill. We can select which three are occupied by the three $P$s in $\binom{6}{3}$ ways. That leaves three positions to fill. We can choose which two of them are occupied by the two $E$s in $\binom{3}{2}$ ways. The remaining position must be occupied by the $R$. Hence, the number of distinguishable ways the letters of the word $PEPPER$ can be arranged is $$\binom{6}{3}\binom{3}{2}\binom{1}{1} = \frac{6!}{3!3!} \cdot \frac{3!}{2!1!} \cdot \frac{1!}{1!0!} = \frac{6!}{3!2!1!0!} = \frac{6!}{3!2!1!}$$ The factor of $3!$ in the denominator represents the number of ways the three $P$s could be permuted among themselves within a given arrangement of the letters of the word $PEPPER$ without producing an arrangement that is distinguishable from that arrangement. The factors of $2!$ and $1!$ play an analogous role for the two $E$s and one $R$ in an arrangement of the letters of the word $PEPPER$.