<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[12pt,reqno]{article}

\usepackage[usenames]{color}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{amscd}
\usepackage{graphicx}

\usepackage[colorlinks=true,
linkcolor=webgreen,
filecolor=webbrown,
citecolor=webgreen]{hyperref}

\definecolor{webgreen}{rgb}{0,.5,0}
\definecolor{webbrown}{rgb}{.6,0,0}

\usepackage{color}
\usepackage{fullpage}
\usepackage{float}

\usepackage{psfig}
\usepackage{graphics}
\usepackage{latexsym}
\usepackage{epsf}
\usepackage{breakurl}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{.1in}
\setlength{\evensidemargin}{.1in}
\setlength{\topmargin}{-.1in}
\setlength{\textheight}{8.4in}

\newcommand{\seqnum}[1]{\href{https://oeis.org/#1}{\underline{#1}}}

\begin{document}

\begin{center}
\epsfxsize=4in
\leavevmode\epsffile{logo129.eps}
\end{center}

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}

\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{conjecture}[theorem]{Conjecture}

\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}

\begin{center}
\vskip 1cm{\LARGE\bf  Two New Identities Involving the Catalan 
Numbers and  Sign-Reversing Involutions  \\
\vskip 1cm}
\large
Jovan Miki\'{c}\\
J.U. S\v{S}C ``Jovan Cviji\'{c}''\\
74480 Modri\v{c}a\\
Republic of Srpska\\
\href{mailto:jnmikic@gmail.com}{\tt jnmikic@gmail.com} \\
\end{center}
\vskip .2 in

\begin{abstract}
We give a combinatorial proof of a known sum concerning the product of a binomial coefficient with two central binomial coefficients. The method of description, involution, and exception is used. The same combinatorial argument also proves the ``$-1$ shifted version'' of this sum. As a consequence, two new binomial coefficient identities with the Catalan numbers are derived.
\end{abstract}

\section{Introduction}

Let $n$ be a non-negative integer. The Catalan numbers are the famous sequence
\[C_n=\frac{1}{n+1}\binom{2n}{n}\text{.}\]
In this paper, we derive two new (to our knowledge) 
binomial coefficient identities with the Catalan numbers. 
\begin{theorem}\label{t:1}
For non-negative integers $n$, we have
\begin{align}
\sum_{k=0}^{n}(-1)^k\binom{n}{k}C_k\binom{2n-2k}{n-k}&amp;={\binom{n}{\lfloor \frac{n}{2}\rfloor}}^2\text{,}\label{eq:1}\\
\sum_{k=0}^{2n}(-1)^k\binom{2n}{k}C_k C_{2n-k}&amp;=C_n\binom{2n}{n}\text{.}\label{eq:2}
\end{align}
\end{theorem}
In order to prove Theorem \ref{t:1}, we consider a known combinatorial sum
\begin{equation}\label{eq:3}
\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2k}{k}\binom{2n-2k}{n-k}=\begin{cases}0,&amp; \text{if } n \text{ is odd;}\\
{\binom{n}{\frac{n}{2}}}^2,&amp; \text{if } n \text{ is even.}\end{cases} 
\end{equation}
Eq.~(\ref{eq:3}) appears several times in the literature;
see, for example,
\cite[Eq.~(6.12), p.~52]{Gould}, \cite[Eq.~(6.61), p.~29]{GouldVol4},
and \cite [Example 3.6.2, p.~45]{MPWZ}.
There are two binomial coefficient identities 
(\cite[Eq.~(6.10), p.~23; Eq.~(7.3), p.~34]{GouldVol4}) similar to 
Eq.~(\ref{eq:3}).
They are proved combinatorially, for example, in \cite{Spivey}. 
See \cite{Nagy} and \cite[Conclusions]{Mikic} for the connection between
these identities and Shapiro's formula \cite[Ex.~(6.C.18), p.~41]{Stanley} and Segner's recurrence relation \cite[Eq.~(5.6), p.~117]{Koshy} respectively.

We give a proof of  Eq.~(\ref{eq:3}) by using the method of ``description, involution, and exception'' \cite{BenjaminQuinn}.

By using the same idea, we derive the ``$-1$ shifted version'' of  Eq.~(\ref{eq:3}).
We assert that
\begin{equation}\label{eq:4}
\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2k}{k-1}\binom{2n-2k}{n-k}=\begin{cases}
-{\binom{n}{\lfloor\frac{n}{2}\rfloor}}^2, &amp;\text{if } n \text{ is odd;}\\
0, &amp;\text{if } n \text{ is even.}\end{cases} 
\end{equation}


Clearly, subtracting  Eq.~(\ref{eq:4}) from Eq.~(\ref{eq:3}) and by using the well-known relation  $C_k=\binom{2k}{k}-\binom{2k}{k-1}$, we get Eq.~(\ref{eq:1}). Furthermore, it can be shown that Eq.~(\ref{eq:2}) is a consequence of  Eq.~(\ref{eq:1}).

Throughout the paper, $[n]$ denotes the set $\{1,2,\ldots,n\}$, if $n$ is a positive integer; and $[0]$ denotes the empty set $\emptyset$.

Let $A$ and $B$ be sets. Then $|A|$ denotes the cardinality of the set $A$, and $A\backslash B$ denotes the set difference: $\{x:x \in A,x\notin B\}$.

We end this paper with the generalization of Eqns.~(\ref{eq:3}) and (\ref{eq:4}).

\section{Definitions}

Let $n$ be a fixed non-negative integer.
\begin{definition}\label{def:1} 
For $A\subset [n]$, we define
\begin{equation}\notag
A^t = \begin{cases}\emptyset,&amp;\text{ if }A=\emptyset \text{;}\\
\{ x+n:x\in A\} ,&amp;\text{ if }A\neq \emptyset\text{.}
\end{cases}
\end{equation}
\end{definition}
Obviously, if $A\subset [n]$, then $A^t\subset[2n]\backslash[n]$.

\begin{definition}\label{def:2}
We define the function $\varphi:[2n]\to [2n]$, as follows:
\begin{equation}\notag
\varphi(x)=\begin{cases}x+n,&amp;\text{ if }x\in [n]\text{;}\\
x-n,&amp;\text{ if }x \in [2n]\backslash [n]\text{.}
\end{cases}
\end{equation}
\end{definition}

\begin{definition}\label{def:3}
Let $S\subset [2n]$. The set $S$ is \textit{balanced} if $S=(S\cap [n])\cup (S\cap [n])^t$.
Otherwise, $S$ is a \textit{unbalanced} set.
\end{definition}

In other words,  set $S$ is balanced if $\forall x\,(x\in S \Leftrightarrow \varphi (x) \in S)$.
Clearly, if set $S$ is balanced, then $|S|$ must be even.

The union of two balanced sets is a balanced set. Note that the converse  does not  hold even if two sets are disjoint. For example, unbalanced sets $[n]$ and $[2n]\backslash [n]$ are disjoint, but their union is a  balanced set $[2n]$.

However, if $S_1$  and $S_2$ are disjoint sets such that  $ S_1\cup \varphi(S_1)$  and  $S_2 \cup \varphi(S_2)$ are disjoint sets, then $S_1 \cup S_2$ is  a balanced set if and only if both sets  $S_1$  and $S_2$  are balanced.

\section{Proof of Eq.~(\ref{eq:3})}
\begin{proof}
Let $n$ be a fixed non-negative integer.
We use the method of description,
involution, and exception, as discussed in \cite{BenjaminQuinn}.

\medskip

\noindent \textbf{Description:}

\medskip

Let $X$ denote the set \[\{(A,B,C):A\subset [n],B\subset A \cup A^t, |A|=|B|, C\subset [2n]\backslash (A \cup A^t),|A|+|C|=n \}\text{.}\]

For integers $k$, where $0\leq k\leq n$, we define the sets $X_k$, as follows:
\[X_k=\{(A,B,C)\in X:|A|=k\}\text{.}\]
Obviously, $X=\bigcup_{k=0}^{n}X_k$ and $|X_k|=\binom{n}{k}\binom{2k}{k}\binom{2n-2k}{n-k}$.
We have 
\begin{equation}\label{eq:5}
\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2k}{k}\binom{2n-2k}{n-k}=\sum_{k=0}^{n}(-1)^k|X_k|=|\mathcal{E}|-|\mathcal{O}|\text{;}
\end{equation}
where 
\begin{equation}\notag
\mathcal{E}=\{(A,B,C)\in X: |A| \text{ is even}\} \text{ and } \mathcal{O}=\{(A,B,C)\in X: |A| \text{ is odd}\} \text{.}
\end{equation}

\noindent \textbf{Involution:}

\medskip

Let us define sets $D$ and $E$, as follows:
\begin{align}
D&amp;=\{(A,B,C)\in X: B\cup C \text{ is an unbalanced set}\}\label{eq:6}\\
E&amp;=\{(A,B,C)\in X: B\cup C \text{ is balanced set}\}\label{eq:7}
\end{align}
Obviously, $D$ and $E$ are disjoint sets and $X=D\cup E$.

Let $(A,B,C)\in D$. We let $d_{B,C}$ denote $\min\{x\in B\cup C: \varphi(x) \notin B \cup C\}$.
The integer $d_{B,C}$ is well-defined because $B \cup C$ is an unbalanced set.

Let us define the function $\varPsi:D \to D$, as follows:
\begin{equation}\label{eq:8}
\varPsi((A,B,C))=\begin{cases}(A\backslash\{d_{B,C}\},B\backslash\{d_{B,C}\},C\cup \{d_{B,C}\}), &amp;\text{ if } d_{B,C}\in B \cap [n]\text{;}\\
(A\backslash\{d_{B,C}-n\},B\backslash\{d_{B,C}\},C\cup \{d_{B,C}\}), &amp;\text{ if } d_{B,C}\in B \cap [n]^t\text{;}\\
(A\cup\{d_{B,C}\},B\cup\{d_{B,C}\},C\backslash \{d_{B,C}\}), &amp;\text{ if } d_{B,C}\in C \cap [n]\text{;}\\
(A\cup\{d_{B,C}-n\},B\cup\{d_{B,C}\},C\backslash \{d_{B,C}\}), &amp;\text{ if } d_{B,C}\in C \cap [n]^t\text{.}
\end{cases}
\end{equation}

The function $\varPsi$ is well-defined and an involution on $D$.
Moreover, if $(A,B,C)\in D\cap \mathcal{E}$, then $\varPsi((A,B,C))\in D\cap \mathcal{O}$; and vice versa.
Therefore, we may conclude that $|D\cap \mathcal{E}|=|D\cap \mathcal{O}|$.

We have 
\begin{align*}
|\mathcal{E}|-|\mathcal{O}|&amp;=|\mathcal{E}\cap X|-|\mathcal{O}\cap X|\\
&amp;=|\mathcal{E}\cap D|+|\mathcal{E}\cap E|-(|\mathcal{O}\cap D|+|\mathcal{O}\cap E|)&amp;&amp; (X=D\cup E)\\
&amp;=|\mathcal{E}\cap E|-|\mathcal{O}\cap E|&amp;&amp;( \text{because }|\mathcal{E}\cap D |=|\mathcal{O}\cap D |)\text{.}
\end{align*}
Therefore, we obtain \begin{equation}
|\mathcal{E}|-|\mathcal{O}|=|\mathcal{E}\cap E|-|\mathcal{O}\cap E|\text{.}\label{eq:9} 
\end{equation}

From Eqns.~(\ref{eq:5}) and (\ref{eq:9}), it follows that
\begin{equation}
\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2k}{k}\binom{2n-2k}{n-k}=|\mathcal{E}\cap E|-|\mathcal{O}\cap E|\text{.}\label{eq:10} .
\end{equation}

\noindent\textbf{Exception:}

\medskip

Let $(A,B,C)\in E$.

By  Eq.~(\ref{eq:7}), the set $B\cup C$ is balanced. Sets $B$ and $C$ are disjoint. Moreover,  $B \cup \varphi(B)$  and $C \cup \varphi(C)$ are disjoint sets too. Then it follows that both sets $B$ and $C$  must be balanced. Hence integers $|B|$ and $|C|$ are even. Since $|A|$=$|B|$ (by the definition of $X$), $|A|$ is even and $(A,B,C)\in \mathcal{E}$. Therefore, $E \subset \mathcal{E}$.

Eq.~(\ref{eq:10}) simplifies to
\begin{equation}\label{eq:11}
\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2k}{k}\binom{2n-2k}{n-k}=|E|\text{.}
\end{equation}

We have two cases:

\medskip

\noindent \textbf{Case (a)}: $n$ is odd.

\medskip

Since $|B\cup C|=n$, it follows that the set $B\cup C$ is unbalanced and $E=\emptyset$.
By Eq.~(\ref{eq:11}), it follows that
\[\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2k}{k}\binom{2n-2k}{n-k}=0,\]
as desired.

\medskip

\noindent \textbf{Case (b)}: $n$ is even.

\medskip

We use the Chu-Vandermonde convolution formula: 
\begin{equation}
\sum_{k=0}^{c}\binom{a}{k}\binom{b}{c-k}=\binom{a+b}{c}\mbox{,}\label{eq:12}
\end{equation}
where $a$, $b$, and $c$ are non-negative integers.

Let us count the number of elements of the set $E$.
The set $E$ is equal to the set
\begin{equation}\notag
\{(A,B_1\cup B_1^t,C_1\cup C_1^t):A\subset [n],|A| \text{ even},B_1\subset A, |B_1|=\frac{|A|}{2},C_1\subset [n]\backslash A, |C_1|+|B_1|=\frac{n}{2}\}\text{.}
\end{equation}
Obviously, it follows that $|E|$ is equal to
\begin{equation}\notag
|\{(A,B_1,C_1):A\subset [n],|A| \text{ even},B_1\subset A, |B_1|=\frac{|A|}{2},C_1\subset [n]\backslash A, |C_1|+|B_1|=\frac{n}{2}\}|\text{.}
\end{equation}
Clearly, there is a one-to-one correspondence between $(A,B_1,C_1)$
and $(B_1\cup C_1,B_1,A\backslash B_1)$.
Therefore, $|E|$ is equal to
\begin{equation}\label{eq:13}
|\{(B_2,B_1,A_1):B_2\subset [n],|B_2|=\frac{n}{2},B_1\subset B_2, A_1\subset[n]\backslash B_2, |A_1|=|B_1|\}|\text{.}
\end{equation}
Let $k=|B_1|$.
By Eq.~(\ref{eq:13}), it follows that
\begin{align*}
|E|&amp;=\binom{n}{\frac{n}{2}}\sum_{k=0}^{\frac{n}{2}}\binom{\frac{n}{2}}{k}\binom{\frac{n}{2}}{k}\\
&amp;=\binom{n}{\frac{n}{2}}\sum_{k=0}^{\frac{n}{2}}\binom{\frac{n}{2}}{k}\binom{\frac{n}{2}}{\frac{n}{2}-k}&amp;&amp;(\text{by symmetry})\\
&amp;=\binom{n}{\frac{n}{2}}^2&amp;&amp;(\text{by Eq.~(\ref{eq:12})}) .
\end{align*}
We obtain \[|E|=\binom{n}{\frac{n}{2}}^2\text{.}\] By Eq.~(\ref{eq:11}), it follows that
\[\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2k}{k}\binom{2n-2k}{n-k}=\binom{n}{\frac{n}{2}}^2,\]
as desired.
This completes the proof of  Eq.~(\ref{eq:3}).
\end{proof}

\section{Proof of Eq.~(\ref{eq:4})}
\begin{proof}
The proof of Eq.~(\ref{eq:4}) is similar to the proof of Eq.~(\ref{eq:3}).

\medskip

\noindent \textbf{Description:}

\medskip

Let $X$ denote the set \[\{(A,B,C):A\subset [n],B\subset A \cup A^t, |B|=|A|-1, C\subset [2n]\backslash (A \cup A^t),|A|+|C|=n \}\text{.}\]

For integers $k$, where $0\leq k\leq n$, we define the following sets $X_k$, as follows:
\[X_k=\{(A,B,C)\in X:|A|=k\}\text{.}\]

Obviously, $X=\bigcup_{k=0}^{n}X_k$ and $|X_k|=\binom{n}{k}\binom{2k}{k-1}\binom{2n-2k}{n-k}$.
We have 
\begin{equation}\label{eq:14}
\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2k}{k-1}\binom{2n-2k}{n-k}=\sum_{k=0}^{n}(-1)^k|X_k|=|\mathcal{E}|-|\mathcal{O}|\text{;}
\end{equation}
where 
\begin{equation}\notag
\mathcal{E}=\{(A,B,C)\in X: |A| \text{ is even}\} \text{ and } \mathcal{O}=\{(A,B,C)\in X: |A| \text{ is odd}\} \text{.}
\end{equation}

\noindent\textbf{Involution:}

\medskip

Same as in Eq.~(\ref{eq:3}).
Let $D$, $E$, and $\varPsi$ be same as in Eqns. (\ref{eq:6}),(\ref{eq:7}), and (\ref{eq:8}) respectively.
It is readily verified that the function $\varPsi$ is  well-defined and an involution on $D$.
Hence the equation \begin{equation}
\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2k}{k-1}\binom{2n-2k}{n-k}=|\mathcal{E}\cap E|-|\mathcal{O}\cap E|\label{eq:15}
\end{equation}
holds.

\medskip

\noindent\textbf{Exception:}

\medskip

Let $(A,B,C)\in E$.
By Eq.~(\ref{eq:7}), the set $B\cup C$ is balanced. As before, both sets $B$ and $C$ must be balanced. Thus, integers $|B|$ and $|C|$ are even. Since $|A|$=$|B|+1$ (by the new definition of $X$), $|A|$ is odd and $(A,B,C)\in \mathcal{O}$. Therefore, $E \subset \mathcal{O}$.
Eq.~(\ref{eq:15}) simplifies to
\begin{equation}\label{eq:16}
\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2k}{k-1}\binom{2n-2k}{n-k}=-|E|\text{.}
\end{equation}
We have two cases:

\medskip

\noindent\textbf{Case (a)}: $n$ is even.

\medskip

Since $|B\cup C|=n-1$,  $|B\cup C|$ is odd. It follows that the set $B\cup C$ is unbalanced and $E=\emptyset$.
By  Eq.~(\ref{eq:16}), it follows that
\[\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2k}{k-1}\binom{2n-2k}{n-k}=0, \]
as desired.

\medskip

\noindent \textbf{Case (b)}: $n$ is odd.

\medskip

Again, we use Eq.~(\ref{eq:12}).
Let us count the number of elements of the set $E$.

The set $E$ is equal to the set
\begin{gather*}
\{(A,B_1\cup B_1^t,C_1\cup C_1^t):\\A\subset [n],|A| \text{ odd},B_1\subset A, |B_1|=\frac{|A|-1}{2},C_1\subset [n]\backslash A, |C_1|+|B_1|=\frac{n-1}{2}\}\text{.}
\end{gather*}

Obviously, $|E|$ is equal to
\[|\{(A,B_1,C_1):A\subset [n],|A| \text{ odd},B_1\subset A, |B_1|=\frac{|A|-1}{2},C_1\subset [n]\backslash A, |C_1|+|B_1|=\frac{n-1}{2}\}|\text{.}\]
Clearly, there is one-to-one correspondence between $(A,B_1,C_1)$
and $(B_1\cup C_1,B_1,A\backslash B_1)$.

Therefore, $|E|$ is equal to
\begin{equation}\label{eq:17}
|\{(B_2,B_1,A_1):B_2\subset [n],|B_2|=\frac{n-1}{2},B_1\subset B_2, A_1\subset[n]\backslash B_2, |A_1|=|B_1|+1\}|\text{.}
\end{equation}
Let $k=|B_1|$.
By  Eq.~(\ref{eq:17}), it follows that
\begin{align*}
|E|&amp;=\binom{n}{\frac{n-1}{2}}\sum_{k=0}^{\frac{n-1}{2}}\binom{\frac{n-1}{2}}{k}\binom{\frac{n+1}{2}}{k+1}\\
&amp;=\binom{n}{\frac{n-1}{2}}\sum_{k=0}^{\frac{n-1}{2}}\binom{\frac{n-1}{2}}{k}\binom{\frac{n+1}{2}}{\frac{n-1}{2}-k}&amp;&amp;(\text{by symmetry})\\
&amp;=\binom{n}{\frac{n-1}{2}}^2&amp;&amp;(\text{by Eq.~(\ref{eq:12})}) .
\end{align*}
Thus we have shown
\[|E|=\binom{n}{\frac{n-1}{2}}^2\text{.}\]
By  Eq.~(\ref{eq:16}), it follows that
\[\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2k}{k-1}\binom{2n-2k}{n-k}=-\binom{n}{\frac{n-1}{2}}^2,\]
as desired.
This completes the proof of  Eq.~(\ref{eq:4}).
\end{proof}


\section{Proof of Theorem \ref{t:1}}

\begin{proof}
Eq.~(\ref{eq:1}) directly follows from Eqns.~(\ref{eq:3}),(\ref{eq:4}), and from the relation \cite[p.~106]{Koshy}
$C_k=\binom{2k}{k}-\binom{2k}{k-1}$.

Let us prove Eq.~(\ref{eq:2}).
By  Eq.~(\ref{eq:1}), it follows that
\begin{equation}\notag
\sum_{k=0}^{2n}(-1)^k\binom{2n}{k}C_k\binom{4n-2k}{2n-k}={\binom{2n}{n}}^2\text{.}
\end{equation}
Changing $k$ to $2n-k$, we obtain that
\begin{equation}\notag
\sum_{k=0}^{2n}(-1)^k\binom{2n}{k}C_k\binom{4n-2k}{2n-k}=\sum_{k=0}^{2n}(-1)^k\binom{2n}{k}C_{2n-k}\binom{2k}{k}\text{.}
\end{equation}

Now we use a lesser known identity on the Catalan numbers:
\begin{equation}\label{eq:18}
C_k\binom{4n-2k}{2n-k}+\binom{2k}{k}C_{2n-k}=2(n+1)C_kC_{2n-k}\text{.}
\end{equation}
Eq.~(\ref{eq:18}) is a special
case \cite[p.~8]{Mikic} of the following identity:
\begin{equation}
C_k\binom{2n-2k}{n-k}+\binom{2k}{k}C_{n-k}=(n+2) C_kC_{n-k}\text{.}\notag
\end{equation}
We have 
\begin{align*}
\sum_{k=0}^{2n}(-1)^k\binom{2n}{k}C_k\binom{4n-2k}{2n-k}+\sum_{k=0}^{2n}(-1)^k\binom{2n}{k}C_{2n-k}\binom{2k}{k}&amp;=2{\binom{2n}{n}}^2\text{,}\\
2(n+1)\sum_{k=0}^{2n}(-1)^k\binom{2n}{k}C_k C_{2n-k}&amp;=2{\binom{2n}{n}}^2 \quad (\text{by Eq.~(\ref{eq:18})}) \\
\sum_{k=0}^{2n}(-1)^k\binom{2n}{k}C_k C_{2n-k}&amp;=\frac{1}{n+1}{\binom{2n}{n}}^2\text{.}
\end{align*}
The last equation above proves Eq.~(\ref{eq:2}).
This completes the proof of Theorem \ref{t:1}.
\end{proof}

\section{Conclusion}

Let $n$, $\alpha$, $\beta$ be non-negative integers. By using the same idea from  proofs of Eqns.~(\ref{eq:3}) and (\ref{eq:4}), we can conclude that
\begin{equation}\label{eq:19}
\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{2k}{k-\alpha}\binom{2n-2k}{n-k-\beta}=\begin{cases}
0,\text{ if  } n\text{ and  }\alpha+\beta  \text{ are of opposite parity;}\\
(-1)^\alpha\binom{n}{\frac{n-\alpha-\beta}{2}}\binom{n}{\frac{n-\alpha+\beta}{2}},\text{ otherwise.}
\end{cases}
\end{equation}

\section{Acknowledgments}

I would like to thank my teachers Vanja Vuji\'{c} and Ivana Bo\v{z}i\v{c}kovi\'{c} for proofreading this paper. Also I would like to thank the referee for useful suggestions.

\begin{thebibliography}{99}

\bibitem{BenjaminQuinn}
A. T. Benjamin and J. J. Quinn, An alternate approach to alternating sums:
a method to DIE for, \textit{College Math.\  J.} \textbf{39} (2008), 191--201.

\bibitem{Gould}
H. W. Gould,
\textit{Combinatorial Identities},
published by the author, revised edition, 1972.

\bibitem{GouldVol4}
H. W. Gould and J. Quaintance, Combinatorial identities: Table I: intermediate techniques for summing finite series, preprint, 2010. Available at \url{https://www.math.wvu.edu/~gould/Vol.4.PDF},

\bibitem{Koshy}
T. Koshy, \textit{Catalan Numbers with Applications}, Oxford University
Press, 2009.

\bibitem{Mikic}
J. Miki\'{c}, A proof of a famous identity concerning the convolution of the central binomial coefficients, \textit{J. Integer Sequences} \textbf{19} (2016), 
\href{https://cs.uwaterloo.ca/journals/JIS/VOL19/Mikic2/mikic15.html}{Article 16.6.6}.

\bibitem{Nagy}
G. V. Nagy, A combinatorial proof of Shapiro's Catalan convolution, \textit{Adv.\  in Appl.\  Math.\ } \textbf{50} (2012) 391--396.

\bibitem{MPWZ}
M. Petkov\v{s}ek, H. Wilf, and D. Zeilberger, \textit{A = B}, A. K. Peters,
1996.

\bibitem{Spivey}
M. Z. Spivey, A combinatorial proof for the alternating convolution of the central
binomial coefficients, \textit{Amer.\  Math.\  Monthly} \textbf{121} (2014), 537--540.

\bibitem{Stanley}
R. P. Stanley, \textit{Catalan Numbers}, Cambridge University Press, 2015.

\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary  05A19 ; Secondary 05A10.

\noindent \emph{Keywords:} Catalan number, central binomial coefficient, combinatorial proof, method of involution, binomial coefficient identity.
\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received May 5 2019;
revised versions received  May 13 2019; May 26 2019; November 9 2019.
Published in {\it Journal of Integer Sequences}, November 11 2019.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


\end{document}

                                                                                


                                                                                
</pre></body></html>