Oops. Based on what you wrote first, I think you intended to say "combination over permutation".
With combinations, order does not matter.
Red,Blue,Green,Yellow,Orange is the same combination as
Green,Blue,Red,Yellow,Orange. We would not count both of them, if counting combinations. (However, they are two different "permutations".)
Is this exercise from a math class?
Are you familiar with any of the notations below?
nCr
n Choose r
C(n,r)
They're all different ways of representing the same thing: number of combinations (subsets) that can be formed by picking r objects from a set of n objects.
Are you familiar with factorials? There's a formula for calculating nCr (combinations); it uses factorials.
\(\displaystyle C(n,r) = \dfrac{n!}{r! (n-r)!}\)
You could do this calculation for every possible value of r (from zero through 33), and add the results. In other words:
How many ways to choose 0 boxes from 33 boxes? C(33,0) = 1
How many ways to choose 1 boxes from 33 boxes? C(33,1) = 33
How many ways to choose 2 boxes from 33 boxes? C(33,2) = 528
How many ways to choose 3 boxes from 33 boxes? C(33,3) = 5456
… et cetera …
How many ways to choose 30 boxes from 33 boxes? C(33,30) = 5456
How many ways to choose 31 boxes from 33 boxes? C(33,31) = 528
How many ways to choose 32 boxes from 33 boxes? C(33,32) = 33
How many ways to choose 33 boxes from 33 boxes? C(33,33) = 1
Oh look! A pattern. Perhaps there's a way to cut your work in half? Think about that!
There's also an overall shortcut, to get the grand total without using C(n,r), but I hesitate to discuss it without knowing whether you're doing homework. (The shortcut has to do with Pascal's Triangle and powers of two.)
PS: The total you're looking for is more than 8.5 billion.