TITLE: How to find all possible combinations of a set of options?
QUESTION [17 upvotes]: If you are given these options for dinner:

Pie
Cake
Muffin
Ice Cream

And you can have any of these options as you like and you can't have more than one of the same item. Then what are all the possible combinations?
For example you can have:
Pie
Pie, Cake
Pie, Cake, Muffin
Pie, Cake, Muffin, Ice Cream
Pie, Ice Cream
Cake, Muffin

etc...

Is there a way to find all possible combinations?

REPLY [3 votes]: 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...
$$\style{font-family:inherit}{\text{# Combinations}} = 4C1 + 4C2 + 4C3 + 4C4 = \left(\begin{smallmatrix} 4 \\ 1 \end{smallmatrix}\right) + \left(\begin{smallmatrix} 4 \\ 2 \end{smallmatrix}\right) + \left(\begin{smallmatrix} 4 \\ 3 \end{smallmatrix}\right) + \left(\begin{smallmatrix} 4 \\ 4 \end{smallmatrix}\right) = 15$$