TITLE: Permutations with Repeating Letters
QUESTION [5 upvotes]: This question is taken from A First Course in Probability (8e) by Ross.
How many different arrangements can be formed form the letter PEPPER?
I understand that there are 6! permutations of the letters when the repeated letters are distinguishable from each other. And that for each of these permutations, there are $(3!)(2!)$ permutations within the Ps and Es. This means that the 6! total permutations accounts for the $(3!)(2!)$ internal permutations. Then, the explanation in the text states that there are $\frac{6!}{(3!)(2!)} = 60$ possible letter arrangements of the letters PEPPER. 
I don't understand this last part. I thought that since the internal permutations were accounted for the total possible letter arrangements would be the $1 - \frac{(3!)(2!)}{6!}$. Can someone please explain the logic behind the last part? Thank you.

REPLY [5 votes]: Suppose we want to form a permutation of $P_1E_1P_2P_3E_2R$.  
We know there are $6!$ ways to do this, since the letters are all distinct, but we could also do this by 
1) forming a permutation of $PEPPER$, which can be done in, say, x ways;
2) assigning subscripts to the P's, which can be done in $3!$ ways; and then
3) assigning subscripts to the E's, which can be done in $2!$ ways.
Therefore $6!=x(3!)(2!)\;,$ so $\displaystyle x=\frac{6!}{3!2!}$

Here is an alternate way to do this:
1) Choose the places for the E's, which can be done in $\dbinom{6}{2}$ ways.
2) Next choose the place for the R, which can be done in $\dbinom{4}{1}=4$ ways.
The P's must go in the remaining places, so we get $\dbinom{6}{2}\cdot4$ permutations of PEPPER.