Find $\binom{n}{0} \binom{2n}{n}-\binom{n}{1} \binom{2n-2}{n}+\binom{n}{2} \binom{2n-4}{n}+\cdots$ Find $$\binom{n}{0} \binom{2n}{n}-\binom{n}{1} \binom{2n-2}{n}+\binom{n}{2} \binom{2n-4}{n}+\cdots$$
I have taken $r$th term and modified as follows:
$$T_r =(-1)^r \binom{n}{r} \binom{2n-2r}{n}=(-1)^r \frac{n!}{(n-r)!r!} \times \frac{ (2n-2r)!}{n! (n-2r)!}=(-1)^r \frac{(2n-2r)!}{(n-r)! r!(n-2r)!}$$
Can we continue from here?
 A: Here is an answer based upon generating functions. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write e.g.
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{k=0}^n(-1)^k}&\color{blue}{\binom{n}{k}\binom{2n-2k}{n}}\\
&=\sum_{k=0}^\infty (-1)^k[z^k](1+z)^n[u^n](1+u)^{2n-2k}\tag{1}\\
&=[u^n](1+u)^{2n}\sum_{k=0}^\infty\left(-\frac{1}{(1+u)^2}\right)^k[z^k](1+z)^n\tag{2}\\
&=[u^n](1+u)^{2n}\left(1-\frac{1}{(1+u)^2}\right)^n\tag{3}\\
&=[u^n]u^n(2+u)^n\tag{4}\\
&=[u^0](2+u)^n\tag{5}\\
&\color{blue}{=2^n}\tag{6}
\end{align*}

Comment:


*

*In (1) we apply the coefficient of operator twice. We also set the limit to $\infty$ without changing anything since we are adding zeros only.

*In (2) we use the linearity of the coefficient of operator and do some rearrangements as preparation for the next step.

*In (3) we apply the substitution rule of the coefficient of operator with $z:=-\frac{1}{(1+u)^2}$
\begin{align*}
A(u)=\sum_{k=0}^\infty a_k u^k=\sum_{k=0}^\infty u^k [z^k]A(z)
\end{align*}

*In (4) we do some simplifications.

*In (5) we apply the rule
\begin{align*}
[u^{p-q}]A(u)=[u^p]u^{q}A(u)
\end{align*}

*In (6) we select the coefficient of $[u^0]$.
A: A successful application of  Euler´s Finite Difference Theorem

Given $f(x) = \sum_{j=0}^ra_jx^j$, Euler´s Finite Fifference Theorem statest that:

$$\sum_{k=0}^{n}(-1)^k\binom{n}{k}f(k) = (-1)^n \Delta_1^nf(x)\big|_{x=0}=\left\{
                \begin{array}{ll}
 0, & 0 \leq r<n\\
 (-1)^na_nn!,& r = n
                \end{array}
              \right.
$$



We have
\begin{align*}
\tag1\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{an-ak}{n} &= \sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{a(n-k)}{n}\\
\tag2&=\sum_{k=0}^{n}(-1)^{(n-k)}\binom{n}{(n-k)}\binom{a(n-(n-k))}{n}\\
\tag3&=\sum_{k=0}^{n}(-1)^{(n-k)}\binom{n}{n-k}\binom{ak}{n}\\
\tag4&=\sum_{k=0}^{n}(-1)^{(n-k)}\binom{n}{k}\binom{ak}{n}\\
\tag5&=(-1)^{n}\sum_{k=0}^{n}(-1)^{-k}\binom{n}{k}\binom{ak}{n}\\
\tag6&=(-1)^{n}\sum_{k=0}^{n}(-1)^{k}\binom{n}{k}\binom{ak}{n}\\
\end{align*}
Let $f(k) = \binom{ak}{n}$ where $a$ is any nonzero complex number. The definition of the binomial coefficient implies that $f(k)$ is a polynomial of degree $n$ in $k$, i.e. $\binom{ak}{n} = \sum_{j=0}^na_jk^j$. 
The coefficient of $k^n$ is $\frac{a^n}{n!}$ sience 
\begin{align*}
f(k)=\binom{ak}{n} = \tfrac{(ak)(ak-1)(ak-2)\cdots(ak-n+1)}{n!}=\sum_{j=0}^na_jk^j
\end{align*}
By definition
\begin{align*}
\tag7(-1)^n\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{ak}{n}
&=(-1)^n\sum_{k=0}^{n}(-1)^k\binom{n}{k}f(k)\\
\tag8&=(-1)^n\sum_{k=0}^{n}(-1)^k\binom{n}{k}\sum_{j=0}^na_jk^j\\
\tag9&=(-1)^n\underbrace{\sum_{j=0}^na_j\sum_{k=0}^{n}(-1)^k\binom{n}{k}k^j}_{(-1)^n \Delta_1^nf(x)\big|_{x=0}}=\left\{\begin{array}{ll}
0, &  0\leq j <n\\
a_n n!,& j = n
\end{array}\right.
\end{align*}

Because here the coefficient of $k^n$ is $\frac{a^n}{n!}$ and $j=n$, implies that 
\begin{align*}
\tag{10} (-1)^n\sum_{j=n}^na_j\sum_{k=0}^{n}(-1)^k\binom{n}{k}k^j&=(-1)^na_n 
 \sum_{k=0}^{n}(-1)^k\binom{n}{k}k^n\\
\tag{11} &=(-1)^n\big[a_n(-1)^nn!\big]\\
\tag{12} &= \frac{a^n}{n!}n! \\
\tag{13} &= a^n\\
&&\Box
\end{align*}

In this way we have for $a=2$:

$$\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2n-2k}{n} = 2^n$$


Note In $(10)$: When we study Stirling numbers of the second kind, $S(n, k)$, we will discover that
$\sum_{k=0}^{n}(-1)^k\binom{n}{k}k^j= (-1)^nn!S(n, k),\quad j\leq n$

Bibliographic references:
Gould, H. W. (Oktober 2015). Combinatorial Identities for Stirling Numbers, 68, 69, 70.
