Re: [math-fun] Counting problem

On Wednesday 07 December 2011 01:05:04 Mike Stay wrote:

I think you take each of the 19 partitions and let a=#1s, b=#2s, c=#3s; the number of ways to order the elements of that partition is (a+b+c-1)! / a! b! c! 2 (Where the -1 comes from cyclic reorderings and the 2 comes from reflections.)

Am I missing something?

Yes, I think so. Consider, for instance, the case a=12, b=0, c=0, where your formula gives 11/24. Or the case a=3, b=0, c=3, where your formula gives 5/3. -- g