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Given that, $\dfrac{1}{a}+\dfrac{1}{b} = \dfrac{1}{9}\;; \;a,b\in \mathbb{Z}^{+} $ and $a\leq b$

$\Rightarrow \dfrac{a+b}{ab} = \dfrac{1}{9}$

$\Rightarrow 9a+9b = ab$

$\Rightarrow 9a+9b-ab = 0$

$\Rightarrow ab-9a-9b = 0$

$\Rightarrow ab-9a-9b+81 = 81$

$\Rightarrow (a-9)(b-9) = 81$

We can factorize $81$ such that $a-9\leq b-9 \Rightarrow \boxed{a\leq b}$

$ \qquad \qquad \begin{array} {ccc} \underline{a-9}& \leq & \underline{b-9} \\ 1 & & 81 \\ 3 & & 27 \\ 9 & & 9\end{array}$

$\therefore$ Only three pairs are possible.

Correct Answer $:\text{A}$

$\Rightarrow \dfrac{a+b}{ab} = \dfrac{1}{9}$

$\Rightarrow 9a+9b = ab$

$\Rightarrow 9a+9b-ab = 0$

$\Rightarrow ab-9a-9b = 0$

$\Rightarrow ab-9a-9b+81 = 81$

$\Rightarrow (a-9)(b-9) = 81$

We can factorize $81$ such that $a-9\leq b-9 \Rightarrow \boxed{a\leq b}$

$ \qquad \qquad \begin{array} {ccc} \underline{a-9}& \leq & \underline{b-9} \\ 1 & & 81 \\ 3 & & 27 \\ 9 & & 9\end{array}$

$\therefore$ Only three pairs are possible.

Correct Answer $:\text{A}$