The number of elements in the set S = \ (r, k) : k ∈ ℤ and ^ 36 C_...

Given equation:

$^{36}C_{r+1} = \dfrac{6 \left(^{35}C_r\right)}{k^2 - 3}$

Using the property $^{n}C_r = \dfrac{n}{r} {}^{n-1}C_{r-1}$, we can write:

$^{36}C_{r+1} = \dfrac{36}{r+1} {}^{35}C_r$

Substituting this into the given equation:

$\dfrac{36}{r+1} {}^{35}C_r = \dfrac{6}{k^2 - 3} {}^{35}C_r$

For $^{35}C_r$ to be defined, $r$ must be an integer such that $0 \le r \le 35$. Thus, $^{35}C_r \neq 0$. Dividing both sides by $^{35}C_r$:

$\dfrac{36}{r+1} = \dfrac{6}{k^2 - 3}$

$\dfrac{6}{r+1} = \dfrac{1}{k^2 - 3}$

$k^2 - 3 = \dfrac{r+1}{6}$

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

For $k \in \mathbb{Z}$, $k^2$ must be a perfect square integer. This requires $r+1$ to be a multiple of $6$.

Since $0 \le r \le 35$, we have $1 \le r+1 \le 36$.

Let $r+1 = 6m$, where $m \in \{1, 2, 3, 4, 5, 6\}$.

Then $k^2 = 3 + m$.

Checking the possible values of $m$ for which $k^2$ is a perfect square:

For $m = 1$, $k^2 = 4 \Rightarrow k = \pm 2$. (Here $r+1 = 6 \Rightarrow r = 5$)

For $m = 2$, $k^2 = 5$ (not a perfect square).

For $m = 3$, $k^2 = 6$ (not a perfect square).

For $m = 4$, $k^2 = 7$ (not a perfect square).

For $m = 5$, $k^2 = 8$ (not a perfect square).

For $m = 6$, $k^2 = 9 \Rightarrow k = \pm 3$. (Here $r+1 = 36 \Rightarrow r = 35$)

The possible pairs $(r, k)$ are $(5, 2)$, $(5, -2)$, $(35, 3)$, and $(35, -3)$.

Therefore, the number of elements in the set $S$ is $4$.

Answer: $4$