Values
Symmetry
Addition
$$ \left(\begin{matrix}n\\k\end{matrix}\right)
=\frac{n!}{k!(n-k)!}
=n\frac{(n-1)!}{k!(n-k)!}
=k\frac{(n-1)!}{k!(n-k)!}+(n-k)\frac{(n-1)!}{k!(n-k)!}
=\frac{(n-1)!}{(k-1)!(n-k)!}+\frac{(n-1)!}{k!(n-k-1)!}
= \left(\begin{matrix}n-1\\k-1\end{matrix}\right) +
\left(\begin{matrix}n-1\\k\end{matrix}\right)$$
For a set $N$ with $|N|=n$, we can consider the set $P$ of all
subsets of size $k$, and the set $Q$ of all subsets of size
$n-k$, so
$|P|=\left(\begin{matrix}n\\k\end{matrix}\right)$ and $|Q|=\left(\begin{matrix}n\\n-k\end{matrix}\right)$. The map $A\mapsto N\setminus A$ gives a bijection between $P$ and $Q$, so $|P|=|Q|$, so $\left(\begin{matrix}n\\k\end{matrix}\right)= \left(\begin{matrix}n\\n-k\end{matrix}\right)$.
$|P|=\left(\begin{matrix}n\\k\end{matrix}\right)$ and $|Q|=\left(\begin{matrix}n\\n-k\end{matrix}\right)$. The map $A\mapsto N\setminus A$ gives a bijection between $P$ and $Q$, so $|P|=|Q|$, so $\left(\begin{matrix}n\\k\end{matrix}\right)= \left(\begin{matrix}n\\n-k\end{matrix}\right)$.
