Preliminary
- The subsets of a set $A=A$'s power set $=\mathcal{P}\left(A\right)$.
- The size of a set $A=A$'s cardinality $=card\left(A\right)$.
- Let your options be represented by the set $X = \left\{P,C,M,I\right\}$.
This table describes what to do for any basic combinatorics/permutations q...
$$
\begin{array}{l}
\begin{array}{c|c}
\hskip36.5pt & \hskip42.5pt\style{font-family:inherit}{\text{Ordering}}
\end{array} \\[-7pt]\hline\hskip-5.5pt
\begin{array}{c|c|c}
\style{font-family:inherit}{\text{Repetition}} & \style{font-family:inherit}{\text{w/}}
& \style{font-family:inherit}{\text{w/o}} \\\hline
\style{font-family:inherit}{\text{w/}} & P_r^n=n^r
& C_r^n=\left(\!\left(\begin{smallmatrix} n \\ r \end{smallmatrix}\right)\!\right)=\left(\begin{smallmatrix} n+r-1 \\ r \end{smallmatrix}\right) \\[0pt]\hline
\style{font-family:inherit}{\text{w/o}} & nPr=\frac{n!}{(n-r)!}
& nCr=\left(\begin{smallmatrix} n \\ r \end{smallmatrix}\right)=\frac{n!}{r!(n-r)!}
\end{array}\hskip-5.5pt
\end{array}
$$
Remark
I will provide my own answer; but first, I will assign some notation to the already accepted answer (which correctly stated that the q could be thought of as the # sequences "(a,b,c,d)" s.t. a,b,c,d $=$ Y/N, i.e., w/ repetition & w/ ordering, subtracted by 1)...
- $\mathcal{P}\left(X\right) = \left\{\left\{\right\},\left\{P\right\},\left\{C\right\},\left\{M\right\},\left\{I\right\},\left\{P,C\right\},\left\{P,M\right\},\left\{P,I\right\},\left\{C,M\right\},\left\{C,I\right\},\left\{I,M\right\}\left\{P,C,M\right\},\left\{P,C,I\right\},\left\{P,M,I\right\},\left\{C,M,I\right\},\left\{P,C,M,I\right\}\right\}\hskip-2pt\style{font-family:inherit}{\text{.}}$
- # Combinations w/ empty set $= card\left(\mathcal{P}\left(X\right)\right) = 16$.
- # Combinations w/o empty set = $card\left(\mathcal{P}\left(X\right)\right) - 1 = 15$.
Answer
Let's consider the q in its intuitive (some would say "raw") form, i.e., 1 about finding the # combinations of options from $X$ w/o repetition or ordering. The table says to use the Binomial Coefficient ($nCr$) but we will have to use it 4 times & + up the results to account for your 4 differently sized subsets...
$$\style{font-family:inherit}{\text{# Combinations}} = 4C1 + 4C2 + 4C3 + 4C4 = \left(\begin{smallmatrix} 4 \\ 1 \end{smallmatrix}\right) + \left(\begin{smallmatrix} 4 \\ 2 \end{smallmatrix}\right) + \left(\begin{smallmatrix} 4 \\ 3 \end{smallmatrix}\right) + \left(\begin{smallmatrix} 4 \\ 4 \end{smallmatrix}\right) = 15$$