I have been trying this for a while now. Using the formula for permutations, I am getting P(31, 2) = 10,230, but this seems way too high...
An ice cream shop has 31 different flavors of ice cream and two kinds of cones. The rules are, cones can be one, two or three-scoops (but no larger), and if two scoops are the same kind of ice cream they are indistinguishable. Otherwise the order of the scoops matters. How many different ice cream cones can a customer order at this shop?
Your answer is actually too small!
Since you say that the order of the scoops matters, the correct answer is
$$2(31+31^2+31^3)=61566$$
Think of it this way. The store owner starts by asking what kind of cone you want. There are $2$ choices. Then he asks you how many scoops you want, and you tell him either one, two, or three. He then asks what flavor you want for your first scoop; there are $31$ choices. If you had said you wanted two or three scoops, he keeps going, asking what you want for the second scoop, and then for the third. There are $31$ choices each time. In other words, the number of one-scoop cones is $2\times31$, the number of two-scoop cones is $2\times31\times31$, and the number of three-scoop cones is $2\times31\times31\times31$.
It's unclear what you meant when you said that two scoops of the same kind of ice cream are "indistinguishable." It doesn't seem to have any relevance for the counting problem.
