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August 23[edit]

How many partitions into sets of size at least two?[edit]

Formula for partitions without partition of size one? (ways to do "Secret Santa" with no one getting a gift for themselves). So for 1,2,3,4,5,6,7, I think it would be 0,1,1,2 (4, 2|2), 2 (5, 3|2), 3 (3|3,4|2,|2|2|2), 3 (4|3,5|2,3|2|2). Based on Conjugate partitions, this is the same formula for partitions where the two (or more) largest partitions are the same size. I don't see any help at Partition (number theory).Naraht (talk) 13:02, 23 August 2021 (UTC)[reply]

In the 6 and 7 cases, you have left out the trivial partition (with only one part). By searching OEIS for the corrected values, I found that this is A002865. The linked OEIS entry contains several formulae. --116.86.4.41 (talk) 13:51, 23 August 2021 (UTC)[reply]

Why are you counting partitions like these for Secret Santa? Wouldn't you want to count derangements instead?--2406:E003:849:5501:6079:448F:E598:9EDC (talk) 14:20, 23 August 2021 (UTC)[reply]

I was about to ask that; so instead, I added the link. —Tamfang (talk) 00:18, 24 August 2021 (UTC)[reply]

You are right, I did mess up. Thanks for the link to A002865. It does have to do with derangements, but more exactly the possible types of derangements for secret santa. A derangement for 6 items will either be two 3 cycles, (a 4 cycle and a 2 cycle) or three 2 cycles.Naraht (talk) 14:49, 24 August 2021 (UTC)[reply]