This method is really inefficient. There are b! / (b - n)! ways to choose n of b unique numbers, whereas this solution is O(b^n).

That being said, it's still fast enough for this example.

Also, Python solution along the same lines:

    import itertoolsdef
    def valueOf(*values, base=10):
        v = 0
        for v in itertools.accumulate(values, lambda a,b: a*base+b):
            pass
        return v
    def solve():
        for s,e,n,d,m,o,r,y in itertools.permutations(range(10), r=8):
            if valueOf(s,e,n,d) + valueOf(m,o,r,e) == valueOf(m,o,n,e,y):
                yield (s,e,n,d),(m,o,r,e),(m,o,n,e,y)

(If anyone knows of a better way to do valueOf, let me know)