Solving a multi-variable function Let $a$ be a real number, and define function $f$ on $\mathbb{N}$ $\cup$ {$0$} $\times \mathbb{N}$ $\cup$ {$0$} as follows:
$f(0,0) = 1$
$f(m,0)=f(0,n)=1$
$f(m,n)=af(m,n-1)+(1-a)f(m-1,n-1)$
where $m,n$ are positive integers.
$(a)$ Find all $a$ such that $|f(m,n)| \leq 1989, \forall m,n \in \mathbb{N}$.
$(b)$ Find the explicit expression of $f(m,n)$
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\mrm{f}\pars{m,n} =
a\,\mrm{f}\pars{m,n - 1} + \pars{1 - a}\,\mrm{f}\pars{m - 1,n - 1}\,,\quad
\left\{\begin{array}{lclcl}
\mrm{f}\pars{0,0} & \ds{=} & \ds{1}&&
\\[1mm]
\ds{\mrm{f}\pars{m,0}} & \ds{=} & \ds{\mrm{f}\pars{0,n}} & \ds{=} & \ds{1}
\end{array}\right.}$

  $\ds{m, n \in \mathbb{Z}_{\ \geq\ 0}\,}$.

$$
\begin{array}{|l|}\hline\mbox{}\\
\quad\mbox{As already pointed out, in an above comment, by}\
\color{#44f}{\texttt{@Bettybel}};
\quad
\\[1mm]
\quad\ds{\mrm{f}\pars{m,n} = 1\,,\ \forall\ m,n\ \in\ \mathbb{Z}_{\ \geq 0}}\ 
\mbox{satisfies the above}\ \ds{\mrm{f}\pars{m,n}}\ \mbox{conditions. However,}
\quad
\\[1mm]
\quad\mbox{it would be interesting to illustrate a general procedure which, in the present case,}\quad
\\[1mm]
\quad\mbox{can be considered an}\ overkill.\quad
\\ \mbox{}\\ \hline
\end{array}
$$
$$\bbx{%
\mbox{Lets}\quad\mc{F}\pars{x,y} \equiv
\sum_{m = 0}^{\infty}\sum_{n = 0}^{\infty}\mrm{f}\pars{m,n}x^{m}\,y^{n}\quad
\mbox{such that}\quad\bbx{\mrm{f}\pars{m,n} = \bracks{x^{m}\,y^{n}}\mc{F}\pars{x,y}}}
$$
Then,
\begin{align}
&\!\!\!\!\!\!\!\!\!\!\!\!\!
\sum_{m = 1}^{\infty}\sum_{n = 1}^{\infty}\mrm{f}\pars{m,n}x^{m}y^{n} =
a\sum_{m = 1}^{\infty}\sum_{n = 1}^{\infty}\mrm{f}\pars{m,n - 1}x^{m}y^{n} +
\pars{1 - a}\sum_{m = 1}^{\infty}\sum_{n = 1}^{\infty}
\mrm{f}\pars{m - 1,n - 1}x^{m}y^{n}\label{1}\tag{1}
\end{align}


*

*$\large\mathsf{Left\ Hand\ Side}$:
\begin{align}
&\sum_{m = 1}^{\infty}x^{m}\bracks{\sum_{n = 0}^{\infty}\mrm{f}\pars{m,n}y^{n} -
\mrm{f}\pars{m,0}} =
\sum_{n = 0}^{\infty}y^{n}\sum_{m = 1}^{\infty}\mrm{f}\pars{m,n}x^{m} -
\sum_{m = 1}^{\infty}x^{m}
\\[5mm] = &\
\sum_{n = 0}^{\infty}y^{n}\bracks{\sum_{m = 0}^{\infty}\mrm{f}\pars{m,n}x^{m}
-\mrm{f}\pars{0,n}} - \sum_{m = 1}^{\infty}x^{m} =
\bbx{\mc{F}\pars{x,y} - {x \over 1 - x} - {1 \over 1 - y}}\label{2}\tag{2}
\end{align}

*$\large\mathsf{Right\ Hand\ Side\ \underline{First\ Term}}$:
\begin{align}
&a\sum_{m = 1}^{\infty}\sum_{n = 0}^{\infty}\mrm{f}\pars{m,n}x^{m}y^{n + 1} =
ay\sum_{n = 0}^{\infty}y^{n}\bracks{\sum_{m = 0}^{\infty}\mrm{f}\pars{m,n}x^{m}
- \mrm{f}\pars{m,0}}
\\[5mm] = &\
\bbx{ay\,\mc{F}\pars{x,y} - {ay \over 1 - y}}
\end{align}

*$\large\mathsf{Right\ Hand\ Side\ \underline{Second\ Term}}$:
\begin{align}
&\pars{1 - a}\sum_{m = 0}^{\infty}\sum_{n = 0}^{\infty}
\mrm{f}\pars{m,n}x^{m + 1}y^{n + 1} =
\bbx{\pars{1- a}xy\,\mc{F}\pars{x,y}}\label{3}\tag{3}
\end{align}

With \eqref{1}, \eqref{2} and \eqref{3}:
\begin{align}
&\mc{F}\pars{x,y} - {x \over 1 - x} - {1 \over 1 - y} =
\bracks{ay\,\mc{F}\pars{x,y} - {ay \over 1 - y}} +
\bracks{\pars{1- a}xy\,\mc{F}\pars{x,y}}
\\[5mm] & \implies
\mc{F}\pars{x,y} = {1 \over 1 - x}\,{1 \over 1 - y} =
\sum_{m = 0}^{\infty}\sum_{n = 0}^{\infty}x^{m}y^{n} 
\implies
\bbox[15px,#ffe,border:1px dotted navy]{\ds{\mrm{f}\pars{m,n} = 1\,,\
\forall\ m,n\ \in\ \mathbb{Z}_{\, \geq\ 0}}}
\end{align} 

