Talk:Ordered partition of a set

Name of the "opposite concept"?

Is there a name for partitions in which the order of the subsets does not matter, but the order of elements among each subset does? In other words: {1, 2} {3, 4} {5} and {3, 4} {1, 2} {5} would be equivalent, but would both differ from {1, 2} {4, 3} {5}. 164.15.127.77 (talk) 09:47, 29 April 2008 (UTC)[reply]

I've added a reference to set partitions, which are what you want. 142.177.72.83 (talk) 01:32, 26 May 2011 (UTC)[reply]

Typo?

The lead contains this phrase: "with union is S". I don't understand what this means. Is it a typo? Pburka (talk) 02:07, 16 April 2012 (UTC)[reply]

Can we have a reference on

The number of ordered partitions T_n of { 1, 2, ..., n } can be found by the formula:

T_{n}=\sum _{i=1}^{n}{{n-1} \choose {i-1}}=2^{n-1}

This does not seems to be evident for me. Especially since for $n=2$ the partitions can be {1,2},({1},{2}) or({2},{1}) so there seems to be 3 ordered partitions, which is not $2^{n-1}=2$ .

May be we should have list have an explanation of what are the $i-1$ things we choose, and why we choose between </math>n-1</math> and not $n$ .

Arthur MILCHIOR (talk)