Is my analysis of the following probability problem correct? I am trying to learn probability, and was reading some lecture notes, while I came across the following example-

So I tried to understand the problem and the solution,
For sampling without replacement, 


*

*The number of outcomes is $6!$ because with each draw one element is removed from the sample space and thus the choice reduces. For first draw there are $6$ choices, for second draw there are $5$ choices etc. 

*The number of good outcomes means in a manner so that the outcomes spell the word PEPPER. Right? So out of $3$ P's, for the first position we can choose it in $3$ ways, then for the $3rd$ position we left with $2$ choices for $P$ etc. Similarly for $E$ also.
Am I right with this kind of analysis? I am not quite satisfied. Can anyone provide me a better analysis? Thanks.
 A: That's all correct but I suppose you could spell out the enumerations a little more exhaustively if you wanted it to be clearer.
Re Q1, there are $6!$ total outcomes. The number successful is:


*
*3 ways to choose P

*2 ways to choose E

*2 ways to choose P

*1 way to choose next P

*1 way to choose E

*1 way to choose R


Therefore it's $3\times2\times2=12$.
I'm not sure writing this as $3!2!$ in the question is helpful.
The probability of spelling PEPPER is $\frac{12}{6!}=\frac{1}{60}$
For sampling with replacement, you have 6 total ways of choosing each piece of paper so the total outcomes number $6^6$
The successful ones out of those are:


*
*3 ways to choose P

*2 ways to choose E

*3 ways to choose P

*3 way to choose next P

*2 way to choose E

*1 way to choose R


Gives $3\times2\times3\times3\times2\times1=3^32^2=108$ successful outcomes out of $6^6$, which equals $$\frac{1}{2\times6^3}$$
A: For sampling without replacement, one can argue as follows: The total number of cases is same as number of ways of arranging $P, P, P, E, E, R$ in a row. This is standard "MISSISSIPPI" problem. The number of ways is $\frac{6!}{3!2!} = 60$. Since only one outcome $PEPPER$ is favorable, the probability is $\frac{1}{60}$.
For sampling with replacement, in each draw, the letter can be any one of $P, E, R$. The probability of drawing $P, E, P, P, E , R$ in that order is 
\begin{equation*}
\frac{3}{6}\times \frac{2}{6}\times \frac{3}{6} \times\frac{3}{6} \times\frac{2}{6} \times \frac{1}{6} = \frac{1}{2^4 \times 3^3} = \frac{1}{2\times 6^3}
\end{equation*}
