I was wondering about these properties of the Sterling numbers of the second kind:
$ S(n,k)=\{n,k\}=$number of partitions of a set of n elements into k subsets.
These properties are: a) $S(n,2)=2^{n-1}-1$
b) $S(n,n-1)=\binom{n}{2}$
c) $S(n,n-2)=\binom{n}{3}+\frac{1}{2}\binom{n}{2}\binom{n-2}{2}$
I can prove a) by using the known formula $$ \displaystyle S(n,k) = \frac{1}{k!} \sum_{j = 0}^{k} (-1)^{k-j} \binom{k}{j} j^n. $$
For b) my thinking is combinatorial: we are to find how many partitions of ${1,2,\ldots,n}$ contain $n-1$ elements. Say we are to choose a partition ${A_1,\ldots,A_{n-1}}$. Then $|A_k|\geq 1$ which means $\sum_k|A_k|\geq n-1$, so there must be one element $A_k$ of the partition to have 2 elements. There are n-1 ways to choose any of the $A_k$ , and n elements to choose from as the second elements. So there are n(n-1) partitions. Is this right?
For c), I am at a loss. Any help would be appreciated. Thanks in advance...
Your reasoning in b) is incorrect. Your reasoning is correct up to the point where you conclude that one of the sets has two elements, and the others have one element. After that, you go astray. The order of the sets in the partition makes no difference, so there is no "choosing one of the $A_k.$" Also, you get the wrong answer. You get $n(n-1),$ whereas the correct answer is $\binom{n}{n-1}=\frac{n(n-1)}{2}.$
The correct reasoning is simple: $\binom{n}{2}$ is the nuimber of ways to choose 2 elements from a set of $n$.
Part c) is answered elsewhere on this site, as Dietrich Burde noted in a comment.