# Sum over k of Stirling Numbers of the Second Kind of k with m by n choose k

## Theorem

Let $m, n \in \Z_{\ge 0}$.

$\displaystyle \sum_k \left\{ {k \atop m}\right\} \binom n k = \left\{ {n + 1 \atop m + 1}\right\}$

where:

$\displaystyle \left\{ {k \atop m}\right\}$ denotes a Stirling number of the second kind
$\dbinom n k$ denotes a binomial coefficient.

## Proof

The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:

$\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left\{ {k \atop m}\right\} \binom n k = \left\{ {n + 1 \atop m + 1}\right\}$

### Basis for the Induction

$P \left({0}\right)$ is the case:

 $\displaystyle \sum_k \left\{ {k \atop m}\right\} \binom 0 k$ $=$ $\displaystyle \sum_k \left\{ {k \atop m}\right\} \delta_{0 k}$ Zero Choose n $\displaystyle$ $=$ $\displaystyle \left\{ {0 \atop m}\right\}$ all terms vanish except for $k = 0$ $\displaystyle$ $=$ $\displaystyle \delta_{0 m}$ Stirling Number of the Second Kind of 0 $\displaystyle$ $=$ $\displaystyle \delta_{1 \left({m + 1}\right)}$ Definition of Kronecker Delta $\displaystyle$ $=$ $\displaystyle \left\{ {1 \atop m + 1}\right\}$ Stirling Number of the Second Kind of 1 $\displaystyle$ $=$ $\displaystyle \left\{ {0 + 1 \atop m + 1}\right\}$

So $P \left({0}\right)$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 1$, then it logically follows that $P \left({r + 1}\right)$ is true.

So this is the induction hypothesis:

$\displaystyle \sum_k \left\{ {k \atop m}\right\} \binom r k = \left\{ {r + 1 \atop m + 1}\right\}$

from which it is to be shown that:

$\displaystyle \sum_k \left\{ {k \atop m}\right\} \binom {r + 1} k = \left\{ {r + 2 \atop m + 1}\right\}$

### Induction Step

This is the induction step:

 $\displaystyle$  $\displaystyle \sum_k \left\{ {k \atop m}\right\} \binom {r + 1} k$ $\displaystyle$ $=$ $\displaystyle \sum_k \left\{ {k \atop m}\right\} \left({\binom r k + \binom r {k - 1} }\right)$ Pascal's Rule $\displaystyle$ $=$ $\displaystyle \sum_k \left\{ {k \atop m}\right\} \binom r k + \sum_k \left\{ {k \atop m}\right\} \binom r {k - 1}$ $\displaystyle$ $=$ $\displaystyle \left\{ {r + 1 \atop m + 1}\right\} + \sum_k \left\{ {k \atop m}\right\} \binom r {k - 1}$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \left\{ {r + 1 \atop m + 1}\right\} + \sum_k \left({m \left\{ {k - 1 \atop m}\right\} + \left\{ {k - 1 \atop m - 1}\right\} }\right) \binom r {k - 1}$ Definition of Stirling Numbers of the Second Kind $\displaystyle$ $=$ $\displaystyle \left\{ {r + 1 \atop m + 1}\right\} + m \sum_k \left\{ {k - 1 \atop m}\right\} \binom r {k - 1} + \sum_k \left\{ {k - 1 \atop m - 1}\right\} \binom r {k - 1}$ $\displaystyle$ $=$ $\displaystyle \left\{ {r + 1 \atop m + 1}\right\} + m \sum_k \left\{ {k \atop m}\right\} \binom r k + \sum_k \left\{ {k \atop m - 1}\right\} \binom r k$ Translation of Index Variable of Summation $\displaystyle$ $=$ $\displaystyle \left\{ {r + 1 \atop m + 1}\right\} + m \left\{ {r + 1 \atop m + 1}\right\} + \left\{ {r + 1 \atop m}\right\}$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \left({m + 1}\right) \left\{ {r + 1 \atop m + 1}\right\} + \left\{ {r + 1 \atop m}\right\}$ $\displaystyle$ $=$ $\displaystyle \left\{ {r + 2 \atop m + 1}\right\}$ Definition of Stirling Numbers of the Second Kind

So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \left\{ {k \atop m}\right\} \binom n k = \left\{ {n + 1 \atop m + 1}\right\}$

$\blacksquare$