Properties of Combination, including Yang-Pascal's identity and Zhu-Vandermonde's identity, a polynomials approach

In this article, I will introduce some properties of Combination, including Yang-Pascal's identity and Zhu-Vandermonde's identity, and give proof by polynomials.

Actually, we can define $\begin{array}{c}(\genfrac{}{}{0ex}{}{n}{k})\end{array}$$\binom{n}{k}\\$ as the coefficient of $x^k{y}^{n-k}$$x^k y^{n-k}$ in $(x+y{)}^k$$(x+y)^k$

The first property of $\begin{array}{c}(\genfrac{}{}{0ex}{}{n}{k})\end{array}$$\binom{n}{k}\\$ is $\begin{array}{c}(\genfrac{}{}{0ex}{}{n}{k})=(\genfrac{}{}{0ex}{}{n}{n-k})\end{array}$$\binom{n}{k}=\binom{n}{n-k}\\$

And observe that $(x+y{)}^n=(y+x{)}^n$$(x+y)^n=(y+x)^n$

Thus $\begin{array}{c}(\genfrac{}{}{0ex}{}{n}{k})x^k{y}^{n-k}=(\genfrac{}{}{0ex}{}{n}{n-k})y^{n-k}{x}^k\Rightarrow (\genfrac{}{}{0ex}{}{n}{k})=(\genfrac{}{}{0ex}{}{n}{n-k})\end{array}$$\binom{n}{k}x^k y^{n-k}=\binom{n}{n-k} y^{n-k}x^k\Rightarrow\binom{n}{k}=\binom{n}{n-k}\\$

And we know that Yang-Pascal's identity as follows

$\begin{array}{c}(\genfrac{}{}{0ex}{}{n}{k})=(\genfrac{}{}{0ex}{}{n-1}{k-1})+(\genfrac{}{}{0ex}{}{n-1}{k})\end{array}$$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}\\$

We can consider that $(1+x{)}^n=(1+x{)}^{n-1}(1+x)$$(1+x)^n=(1+x)^{n-1}(1+x)$

And in the LHS, the coefficient of $x^k$$x^k$ is $\begin{array}{c}(\genfrac{}{}{0ex}{}{n}{k})x^k\end{array}$$\binom{n}{k}x^k\\$

In the RHS, the coefficient of $x^k$$x^k$ is $\begin{array}{c}(\genfrac{}{}{0ex}{}{n-1}{k-1})x^{k-1}\cdot x+(\genfrac{}{}{0ex}{}{n-1}{k})x^k\end{array}$$\binom{n-1}{k-1} x^{k-1}\cdot x+\binom{n-1}{k}x^k\\$

Thus $\begin{array}{c}(\genfrac{}{}{0ex}{}{n}{k})=(\genfrac{}{}{0ex}{}{n-1}{k-1})+(\genfrac{}{}{0ex}{}{n-1}{k})\end{array}$$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}\\$

And in general, we can consider $(1+x{)}^{m+n}=(1+x{)}^m(1+x{)}^n$$(1+x)^{m+n}=(1+x)^m(1+x)^n$

In the LHS, $\begin{array}{c}(\genfrac{}{}{0ex}{}{m+n}{k})x^k\end{array}$$\binom{m+n}{k}x^k\\$

In the RHS, $\begin{array}{c}\sum _{i+j=k}(\genfrac{}{}{0ex}{}{m}{i})x^i(\genfrac{}{}{0ex}{}{n}{j})x^j\end{array}$$\sum_{i+j=k}\binom{m}{i}x^i\binom{n}{j}x^j\\$

Thus we get $\begin{array}{c}(\genfrac{}{}{0ex}{}{m+n}{k})=\sum _{i=0}^k(\genfrac{}{}{0ex}{}{m}{i})(\genfrac{}{}{0ex}{}{n}{k-i})\end{array}$$\binom{m+n}{k} = \sum_{i=0}^{k} \binom{m}{i} \binom{n}{k-i}\\$