The number of ordered pairs r , k for which 6 . C r 35 = k 2 - 3 . C r + 1 36 , where k is an integer is

Given, $6.{}^{35}{C}_r=\left(k^2-3\right).{}^{36}{C}_{r+1}$

$\Rightarrow \frac{36}{r+1}{\times }^{35}{C}_r\left(k^2-3\right)=6{\times }^{35}{C}_r$, using properties of binomial coefficient.
$\Rightarrow k^2-3=\frac{r+1}{6}$

$\Rightarrow k^2=3+\frac{r+1}{6}$

$r$ can be $\mathrm{5,35}$

For $r=5,k=\pm 2$

$r=35,k=\pm 3$

Hence, number of solution $=4$