If ${ }^{36}{C}_{r+1}\left(k^2-3\right)=6{\times }^{35}{C}_r$ then the number of ordered pairs $(r, k)$ where$k \in Z,$ is 

We have,

$\begin{array}{l}{ }^{36}{C}_{r+1}\left(k^2-3\right)=6{\times }^{35}{C}_r\\ \Rightarrow {\frac{36}{r+1}}^{35}{C}_r\left(k^2-3\right){=}^{35}{C}_r\times 6\\ \Rightarrow k^2-3=\frac{r+1}{6}\Rightarrow k^2=3+\frac{r+1}{6}\end{array}$

Since $k^2$is an integer and r takes values from Oto 35. Therefore,

$r=5$ or $35$

Putting $r=5$in (i), we obtain

$k^2=3+1=4\Rightarrow k=\pm 2$

Putting $r=35$ in (ii), we obtain

$k^2=3+6=9\Rightarrow k=\pm 3$

Hence, the total number of order pairs $(r, k)$ is 4.