Error in code or need more runtime?

edit

@Nathann's answer explains that with your code, Sage is not calculating the number but enumerating all submultisets, which in this case just takes too long.

For your particular counting problem, you can use compositions (also known as ordered partitions). You can iterate through compositions of 14 with maximum part 2, (there are 610 of them), and count each of them with an appropriate weight (some binomial coefficient) to account for the number of ways to choose which numbers the parts refer to.

sage: C = Compositions(14,outer=[2]*53) 
sage: C.cardinality()
610
sage: sum(binomial(53,len(c)) for c in C)
48204860086530

You can kill the computation, for Sage is apparently trying to build all of these sets in order to count them. This is the default behaviour (and not always a very good idea) when no .cardinality() method has been implemented specifically for this problem.

Here, the number of submultiset of [0..52] containing at most twice the same element is at least the number of subsets of 53, i.e. at least 2^53. That is already too much.

If you know how this value should be computed, however, you are welcome to add this missing feature into Sage !

http://www.sagemath.org/doc/developer/