Animated Solution for Mathematics - Binomial Theorem: The number of ordered pairs $(r,k)$ for which $6{\cdot }^{35}{C}_r=(k^2-3){\cdot }^{36}{C}_{r+1}$, where $k$ is an integer, is :

The given equation is $6{\cdot }^{35}{C}_r=(k^2-3){\cdot }^{36}{C}_{r+1}$, where $k$ is an integer and $r$ is a non-negative integer. In JEE Advanced problems, we avoid brute-force factorial expansion in favor of recursive identities.

We utilize the identity:

By applying this to the right side of the equation, we transform ${}^{36}{C}_{r+1}$ into:

Substituting this back into our original expression, we obtain:

Since ${}^{35}{C}_r$ is strictly positive for valid $r$, we divide both sides by ${}^{35}{C}_r$ to simplify the expression:

Dividing both sides by 6, we isolate the relationship between $k$ and $r$:

We know the constraint $0\le r\le 35$, which implies $1\le r+1\le 36$. Since $k$ is an integer, $k^2$ must be an integer, forcing $(r+1)$ to be a multiple of 6.

The possible values for $(r+1)$ in the range $[1,36]$ are $\{6,12,18,24,30,36\}$. We test these to find perfect squares for $k^2$:

If $r+1=6$, then $k^2=\frac{6}{6}+3=4$, yielding $k=\pm 2$. If $r+1=12$, then $k^2=2+3=5$ (not a square). If $r+1=18$, then $k^2=3+3=6$ (not a square). If $r+1=24$, then $k^2=4+3=7$ (not a square). If $r+1=30$, then $k^2=5+3=8$ (not a square). If $r+1=36$, then $k^2=6+3=9$, yielding $k=\pm 3$.

We have identified the valid cases for the ordered pairs $(r,k)$:

1. For $r+1=6$ ($r=5$), we have $(5,2)$ and $(5,-2)$. 2. For $r+1=36$ ($r=35$), we have $(35,3)$ and $(35,-3)$.

This results in a total of four valid ordered pairs. Through the application of symmetry and logical constraints, we have successfully navigated the complexity of the equation.