Number of Integer Pairs Satisfying Combinatorial Relation

To find the number of elements in the set $S = \left\{ (r, k) : k \in \mathbb{Z} \text{ and } {}^{36}C_{r+1} = \frac{6\left({}^{35}C_r\right)}{k^2-3} \right\}$

we begin by using the standard identity for combinations: ${}^nC_k = \frac{n}{k} {}^{n-1}C_{k-1}$

Applying this to ${}^{36}C_{r+1}$, we get: ${}^{36}C_{r+1} = \frac{36}{r+1} {}^{35}C_r$

Substitute this into the given equation: $\frac{36}{r+1} {}^{35}C_r = \frac{6}{k^2 - 3} {}^{35}C_r$

For the combinations to be well-defined, we must have $0 \le r \le 35$ with $r \in \mathbb{Z}$, which implies ${}^{35}C_r \neq 0$. Dividing both sides by ${}^{35}C_r$, we get: $\frac{36}{r+1} = \frac{6}{k^2 - 3}$

Simplifying by dividing both numerators by $6$: $\frac{6}{r+1} = \frac{1}{k^2 - 3}$

Cross-multiplying yields: $k^2 - 3 = \frac{r+1}{6}$ $k^2 = 3 + \frac{r+1}{6}$

Since $k \in \mathbb{Z}$, $k^2$ must be a perfect square integer. Furthermore, for $\frac{r+1}{6}$ to be an integer, $r+1$ must be a multiple of $6$.

Given $0 \le r \le 35$, the range for $r+1$ is $1 \le r+1 \le 36$. Let $r+1 = 6m$, where $m \in \{1, 2, 3, 4, 5, 6\}$.

Now, we test each possible value of $m$:

1. For $m = 1$: $r+1 = 6 \implies r = 5$ $k^2 = 3 + 1 = 4 \implies k = \pm 2$ This gives $2$ integer pairs: $(5, 2)$ and $(5, -2)$.

2. For $m = 2$: $r+1 = 12 \implies r = 11$ $k^2 = 3 + 2 = 5$ Since $5$ is not a perfect square, there are no integer solutions for $k$.

3. For $m = 3$: $r+1 = 18 \implies r = 17$ $k^2 = 3 + 3 = 6$ No integer solutions for $k$.

4. For $m = 4$: $r+1 = 24 \implies r = 23$ $k^2 = 3 + 4 = 7$ No integer solutions for $k$.

5. For $m = 5$: $r+1 = 30 \implies r = 29$ $k^2 = 3 + 5 = 8$ No integer solutions for $k$.

6. For $m = 6$: $r+1 = 36 \implies r = 35$ $k^2 = 3 + 6 = 9 \implies k = \pm 3$ This gives $2$ integer pairs: $(35, 3)$ and $(35, -3)$.

Thus, the set $S$ contains the elements: $S = \{(5, 2), (5, -2), (35, 3), (35, -3)\}$

The total number of elements in $S$ is $4$.

Correct Option: B