Let $m\geq n\geq 1$. Prove the following identity using PIE: $\displaystyle\sum_{k=0}^n(-1)^k\left(\begin{array}{c} m-k \\ n \end{array}\right)\left(\begin{array}{c} n \\ k \end{array}\right)=1$.
For $i=1,2,\dots,n$, let $A_i$ be the set of all $(m-n)$-subsets of $\{1,2,\dots,m\}$ that contain $i$. Use this to count the number of $(m-n)$-subsets of $\{1,2,\dots,m\}$ that do not contain any elements of $\{1,2,\dots,n\}$.
This is what I have so far:
Let $A_i=\{(m-n)\subseteq \{1,2,\dots,m\}\ \text{containing }i,\ \text{for }i=1,2,\dots n\}$. For each $A_i$, there are $\left(\begin{array}{c} n \\ 1 \end{array}\right)$ ways of choosing them multiplied $\left(\begin{array}{c} m-1 \\ m-n-1 \end{array}\right)$ elements. Then for each intersection of two $A_i$'s, there are $\left(\begin{array}{c} n \\ 2 \end{array}\right)$ ways of choosing them multipied by $\left(\begin{array}{c} m-2 \\ m-n-2 \end{array}\right)$ elements. Thus for $k$ many intersections of the $A_i$'s, there are $\left(\begin{array}{c} n \\ k \end{array}\right)$ choices multiplied by $\left(\begin{array}{c} m-k \\ m-n-k \end{array}\right)$ elements. Thus \begin{align*} |A_1\cup\cdots\cup A_n|&=\sum_{k=1}^n(-1)^{k+1}\left(\begin{array}{c} m-k \\ m-n-k \end{array}\right)\left(\begin{array}{c} n \\ k \end{array}\right)\\ &=\sum_{k=1}^n(-1)^{k+1}\left(\begin{array}{c} m-k \\ n \end{array}\right)\left(\begin{array}{c} n \\ k \end{array}\right) \end{align*}
define for $i\in${$1,...,n$} $A_i$ to be the set of all subsets of ${1,...,m}$ such that $|A_i|=n$ and $i\notin A_i$ (means all n-subsets of [m] that do not contain i).
Now by PIE we know that $|\bigcap_{k=1}^{n}A_k ^C| =|S|+ \sum_{k=1}^{n}(-1)^k\sum_{I\subseteq [m], |I|=k} |\bigcap_{i\in I}A_i|= \sum_{k=0}^{n}(-1)^k\sum_{I\subseteq [m], |I|=k} |\bigcap_{i\in I}A_i| $ (where S is the set of all n-subsets of [m])
now we see that $\sum_{k=0}^{n}(-1)^k\sum_{I\subseteq [m], |I|=k} |\bigcap_{i\in I}A_i| = \sum_{k=0}^{n}(-1)^k $$ n\choose k $$m-k \choose n$ because for every $I\subseteq[m] $with $|I|=k : |\bigcap_{i\in I}A_i|=$$m-k\choose n$ and there are $n \choose k$ ways to choose such I.
also $\bigcap_{k=1}^{n}A_k ^C =${{$1,2,...,n $}} (singleton containing only [n]) therefor we get: $1 = |\bigcap_{k=1}^{n}A_k ^C| = \sum_{k=0}^{n}(-1)^k $$ n\choose k $$m-k \choose n$