I need to show
I understand how to get but the 3 is throwing me off a little bit, can anyone help?
The 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:
.
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 .
So if 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: ways.
Or we can have two cells with two elements each: .
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, , where 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 , and ......