Permutations with Repeating Letters 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.
 A: Let's do it smaller with $PPE$. If the letters $P$ are given an index then there are $3!=6$ possibilities:


*

*$P_1P_2E$

*$P_2P_1E$

*$P_1EP_2$

*$P_2EP_1$

*$EP_1P_2$

*$EP_2P_1$


If the indices are taken away then $P_1P_2E$ and $P_2P_2E$ both become $PPE$. It appears that possibility $PPE$ has been counted $2!=2$ times. To repair this we must divide by $2!$ and get $3$ as answer. This agrees with the fact that there are $3$ possibilities:


*

*$PPE$

*$PEP$

*$EPP$


Likewise in $PEPPER$ every permutation is originally counted $3!2!$ times, so we must divide $6!$ by $3!2!$ to repair.
A: 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.
