# Sum over k of m-n choose m+k by m+n choose n+k by Unsigned Stirling Number of the First Kind of m+k with k

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## Contents

## Theorem

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

- $\displaystyle \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brack k} = {n \brace n - m}$

where:

- $\dbinom {m - n} {m + k}$ etc. denote binomial coefficients
- $\displaystyle {m + k \brack k}$ denotes an unsigned Stirling number of the first kind
- $\displaystyle {n \brace n - m}$ denotes a Stirling number of the second kind.

## Proof

The proof proceeds by induction on $m$.

For all $m \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

- $\forall n \in \Z_{\ge 0}: \displaystyle \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brack k} = {n \brace n - m}$

### Basis for the Induction

$\map P 0$ is the case:

\(\displaystyle \) | \(\) | \(\displaystyle \sum_k \binom {0 - n} {0 + k} \binom {0 + n} {n + k} {0 + k \brack k}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_k \binom {- n} k \binom n {n + k}\) | Unsigned Stirling Number of the First Kind of Number with Self | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_k \binom {- n} k \delta_{0 k}\) | as $\dbinom n {n + k} = 0$ for $k = 0$, and Binomial Coefficient with Self | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \binom {- n} 0\) | All terms but where $k = 0$ vanish | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1\) | Binomial Coefficient with Zero | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle {n \brace n - 0}\) | Stirling Number of the Second Kind of Number with Self |

So $\map P 0$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

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

So this is the induction hypothesis:

- $\displaystyle \sum_k \binom {r - n} {r + k} \binom {r + n} {n + k} {r + k \brack k} = {n \brace n - r}$

from which it is to be shown that:

- $\displaystyle \sum_k \binom {r + 1 - n} {r + 1 + k} \binom {r + 1 + n} {n + k} {r + 1 + k \brack k} = {n \brace n - r + 1}$

### Induction Step

This is the induction step:

\(\displaystyle \) | \(\) | \(\displaystyle \sum_k \binom {r + 1 - n} {r + 1 + k} \binom {r + 1 + n} {n + k} {r + 1 + k \brack k} = {n \brace n - r + 1}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_k \paren {\binom {r - n} {r + 1 + k} + \binom {r - n} {r + k} } \binom {r + 1 + n} {n + k} {r + 1 + k \brack k} = {n \brace n - r + 1}\) | Pascal's Rule |

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

Therefore:

- $\forall m, n \in \Z_{\ge 0}: \displaystyle \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brack k} = {n \brace n - m}$

$\blacksquare$

## Sources

- 1997: Donald E. Knuth:
*The Art of Computer Programming: Volume 1: Fundamental Algorithms*(3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: $(54)$