As stated in the comments, both of your explanations for the answer to part (a) are correct.
How many arrangements are possible using five letters from MISSISSIPPI.
The number $5$ can be partitioned in $7$ ways.
\begin{align*}
5 & = \color{red}{5}\\
& = 4 + 1\\
& = 3 + 2\\
& = 3 + 1 + 1\\
& = 2 + 2 + 1\\
& = 2 + 1 + 1 + 1\\
& = \color{red}{1 + 1 + 1 + 1 + 1}
\end{align*}
However, the word MISSISSIPPI does not contain five copies of the same letter, nor does it contain five different letters, so we do not have to worry about the first and last cases.
One letter appears four times and a different letter appears once: There are two ways to pick which letter will appear four times, I or S. There are $\binom{5}{4}$ ways to choose the positions of that letter. There are three ways to select the letter that will fill the remaining position in the word. Hence, there are
$$\binom{2}{1}\binom{5}{4}\binom{3}{1}$$
such arrangements.
One letter appears three times and a different letter appears twice: There are two ways to pick which letter will appear three times, I or S. There are $\binom{5}{3}$ ways to select where in the word it will appear. There are two ways to choose which letter will appear in the remaining two positions, P and the letter not selected earlier from the set $\{I, S\}$. Hence, there are
$$\binom{2}{1}\binom{5}{3}\binom{2}{1}$$
such arrangements.
One letter appears three times and two other letters each appear once: There are two ways to choose the letter that appears three times. There are $\binom{5}{3}$ ways to choose the positions for that letter. There are three ways to pick the letter that will fill the leftmost open position. There are two ways to pick the letter that will fill the remaining open position. Hence, there are
$$\binom{2}{1}\binom{5}{3}\binom{3}{1}\binom{2}{1}$$
such arrangements.
Two letters each appear twice and one letter appears once: There are $\binom{3}{2}$ ways to pick which two letters appear twice from the set $\{I, P, S\}$. There are $\binom{5}{2}$ ways to choose the positions of the selected letter that appears first in an alphabetical list and $\binom{3}{2}$ ways to choose the positions of the other selected letter. There are $2$ ways to pick the letter that fills the remaining open position. Hence, there are
$$\binom{3}{2}\binom{5}{2}\binom{3}{2}\binom{2}{1}$$
such arrangements.
One letter appears twice and the other three letters each appear once: There are three ways to choose the letter that appears twice from the set $\{I, P, S\}$. There are $\binom{5}{2}$ ways to choose its positions. There are $3!$ ways to arrange the remaining letters in the remaining positions. Hence, there are
$$\binom{3}{1}\binom{5}{2}3!$$
such arrangements.
Since these cases are mutually exclusive and exhaustive, the desired answer can be found by adding the above results.
