I need to show
$\displaystyle S(n,n-2) = \binom{n}{3} + 3\binom{n}{4}$
I understand how to get $\displaystyle \binom{n}{3}$ but the 3$\displaystyle \binom{n}{4}$ is throwing me off a little bit, can anyone help?
The $\displaystyle \binom{n}{3}$ comes from partitioning the n elements into 1 set of size 3 and n - 3 sets of size 1.
Then you have to add the number of ways of partitioning the n elements into 2 sets of size 2 and n - 4 sets of size 1. This is equal to (the number of ways of choosing 4 elements from the n elements) times (number of ways of choosing 2 elements from 4)/2:
$\displaystyle {n \choose 4} \, \frac{{4 \choose 2}}{2}$.
You divide by 2 because otherwise you've double counted .... Think about it.
First, these must be Stirling Numbers of the second kind.
S(n,k) is the number of ways to partition a set of n individuals into k non-empty cells.
Therefore, in this problem we must have $\displaystyle n \ge 4$.
So if $\displaystyle n=4$ then how many ways can one partition the set in two non-empty subsets?
It is clear that we can have one cell with three elements and one with one element: $\displaystyle \binom {4}{3}$ ways.
Or we can have two cells with two elements each: $\displaystyle \frac {\binom {4}{2}} {2}$.
Can you continue?
There's no problem with n = 2 or n = 3 ....
n = 2: S(2, 0) = 0, since there's no way to partition 2 elements into zero sets .... (note that in general, $\displaystyle S(n, 0) = \delta_{n\, 0}$, where $\displaystyle \delta_{n\, 0}$ is the Kronecker delta).
n = 3: S(3, 1) = 1, since there's only one way to partition 3 elements into one set ....
However ..... these values of n cannot be used in the formula (since the combinatorials are undefined for n < 4), which is what I think Plato was refering to.
One might try to argue a case that $\displaystyle {3 \choose 4} = 0$, $\displaystyle {2 \choose 3} = 0$ and $\displaystyle {2 \choose 4} = 0$ ......