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? On Tue, Dec 6, 2011 at 4:17 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Ha -- and in my spare moments today I counted 87 (by hand).
So, can we split the difference and agree on 86 ?
--Dan
Allan wrote:
<< As luck would have it, this afternoon I had a boring staff meeting in which I enumerated all of them ... I think. And the answer is 85 ... I think.
On Tue, Dec 6, 2011 at 12:36 PM, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
This seems to be a particular case of the necklace problem which was solved by Hazel Perfect in Math Gaz, when? (more than half a century ago -- not in MR) R.
On Tue, 6 Dec 2011, Dan Asimov wrote:
What are all the ways that 30-, 60-, and 90-degree angles can be arranged
about the origin in the plane?
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