# Proving an identity with unsigned sterling numbers of first kind

• May 11th 2014, 09:52 PM
evg952
Proving an identity with unsigned sterling numbers of first kind
How do I prove $c(n+1, m+1) = \sum_{k=0}^{n} c(n, k) \binom{k}{m}$
• May 11th 2014, 11:43 PM
romsek
Re: Proving an identity with unsigned sterling numbers of first kind
have you tried induction?
• May 12th 2014, 12:05 AM
evg952
Re: Proving an identity with unsigned sterling numbers of first kind
The base case is easy to show, but for the induction step do I assume $c(n, m+1) = \sum_{k=0}^{n-1} c(n-1, k) \binom{k}{m}$ is true? and then try to get to $c(n+1, m+1) = \sum_{k=0}^{n} c(n, k) \binom{k}{m}$?

Edit: I am assuming that we want to induct on n.
• May 12th 2014, 11:14 AM
romsek
Re: Proving an identity with unsigned sterling numbers of first kind
Quote:

Originally Posted by evg952
The base case is easy to show, but for the induction step do I assume $c(n, m+1) = \sum_{k=0}^{n-1} c(n-1, k) \binom{k}{m}$ is true? and then try to get to $c(n+1, m+1) = \sum_{k=0}^{n} c(n, k) \binom{k}{m}$?

Edit: I am assuming that we want to induct on n.

err what's the difference between $c(m,n)$ and $\begin{pmatrix}m \\n\end{pmatrix}$ ?