Interpret the equation

The given equation is $^{635}C_r = (k^2-3)\,^{636}C_{r+1},$ where $k$ is an integer.

Use the relation between consecutive binomial coefficients

We know $^{n}C_m = \frac{n!}{m!(n-m)!}.$

So, $\frac{^{635}C_r}{^{636}C_{r+1}} = \frac{\dfrac{635!}{r!(635-r)!}}{\dfrac{636!}{(r+1)!(635-r)!}}.$

Simplifying, $\frac{^{635}C_r}{^{636}C_{r+1}} = \frac{635!}{r!(635-r)!}\cdot \frac{(r+1)!(635-r)!}{636!}$ $= \frac{r+1}{636}.$

Hence, $^{635}C_r = \frac{r+1}{636}\,^{636}C_{r+1}.$

Comparing with the given equation, $(k^2-3) = \frac{r+1}{636}.$

Apply integer constraints

Since $r$ is a valid binomial index for $^{635}C_r$, we must have $0 \le r \le 635.$

Therefore, $1 \le r+1 \le 636,$ so $\frac{r+1}{636} \in \left\{\frac{1}{636}, \frac{2}{636}, \dots, 1\right\}.$

Thus, $0 < k^2 - 3 \le 1.$

Find integer $k$ such that $k^2-3$ lies in this range

Since $k$ is an integer, $k^2$ can be $0,1,4,9,\dots$

Then

• if $k=0$, $k^2-3=-3$
• if $k=\pm 1$, $k^2-3=-2$
• if $k=\pm 2$, $k^2-3=1$
• if $|k|\ge 3$, $k^2-3\ge 6$

The only possible value satisfying $0 < k^2-3 \le 1$ is $k^2-3=1 \implies k=\pm 2.$

Find corresponding $r$

From $k^2-3 = \frac{r+1}{636},$ substituting $k^2-3=1$ gives $\frac{r+1}{636}=1 \implies r+1=636 \implies r=635.$

List ordered pairs

For $r=635$, the possible integer values of $k$ are $2$ and $-2$.

So the ordered pairs are $(635,2), \quad (635,-2).$

Therefore, the number of ordered pairs is $2.$

Compare with stored answer

The derived answer is Option C: 2.

The stored correct answer is D: 4, which does not match.

So the stored answer appears to be incorrect.