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?

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- Apr 16th 2008, 02:36 PMJrb599Stirling numbers
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? - Apr 16th 2008, 03:34 PMmr fantastic
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. - Apr 16th 2008, 03:49 PMPlato
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? - Apr 17th 2008, 06:09 AMJrb599
- Apr 17th 2008, 07:43 PMmr fantastic
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 $\displaystyle {2 \choose 4} = 0$ ......