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@moth
I think i've got it now?
If you are allowed up to k toppings, then you can pick a combination of m (m <= k) toppings from the n available, then distribute up to k-m 'extras' among the m toppings (e.g convert bacon & mushroom to bacon and tripple mushroom (4 toppings)).
My textbook tells me* there are (e+m-1)!/e!(m-1)! ways of sharing out e extras among the m toppings chosen.
So with n possible toppings and k max toppings per pizza:
sum from m = 1 to k { sum from e = 0 to k-m { n!(e+m-1)! / (n-m)!m!e!(m-1)!}}
1 for the null pizza
The permutations formula n! only works if all n items are distinguishable. If m of them are identical then you need to divide by the m! ways they can be arranged. Putting e balls into m boxes is equivalent to picking a permutation of e identical balls and m-1 identical box-dividers.