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

Welcome, student. Today, we are going to dissect a problem that, at first glance, looks like a daunting wall of combinatorics. You see the expression $6\cdot {}^{35}{C}_r=(k^2-3)\cdot {}^{36}{C}_{r+1}$ and your instinct might be to panic.

You might think, "Do I need to expand these factorials? Do I need to calculate massive numbers?" The answer is a resounding no.

In the world of JEE Advanced, the most complex-looking problems often hide the most elegant, simple structures. Our goal today is to peel back the layers of this equation and reveal the symmetry underneath.

The Algebraic Simplification

Let us start by grouping the binomial coefficients. We have $6\cdot {}^{35}{C}_r=(k^2-3)\cdot {}^{36}{C}_{r+1}$. If we divide both sides by ${}^{36}{C}_{r+1}$, we get:

Now, look at that ratio on the left. It is a classic pattern. Whenever you see a ratio of consecutive binomial coefficients, your mind should immediately jump to the identity:

In our specific case, $n=35$. So, the denominator ${}^{n+1}{C}_{r+1}$ becomes ${}^{36}{C}_{r+1}$. Applying the identity, the ratio $\frac{{}^{35}{C}_r}{{}^{36}{C}_{r+1}}$ simplifies beautifully to $\frac{r+1}{36}$.

Just like that, the complex combinations vanish, replaced by a simple linear fraction. Our equation now reads:

The Integer Constraint

We are left with $\frac{r+1}{6}=k^2-3$, or more cleanly, $k^2=3+\frac{r+1}{6}$. This is where the "Number Theory" part of the problem kicks in.

We are told that $k$ is an integer. This is a massive hint. If $k$ is an integer, then $k^2$ must be a perfect square integer.

Since $3$ is an integer, the term $\frac{r+1}{6}$ must also be an integer. This means $r+1$ must be a multiple of $6$.

But we are not done with constraints. We must respect the definition of the binomial coefficient ${}^{36}{C}_{r+1}$. For this to be defined, the lower index $r+1$ must satisfy $0\le r+1\le 36$.

Combining these two facts, we know that $r+1$ must be a multiple of $6$ within the range $[0,36]$. The possible values for $r+1$ are therefore $6,12,18,24,30,36$.

The Systematic Search

Now, we simply test these values. This is the "detective work" phase of the problem. We plug each candidate into our equation $k^2=3+\frac{r+1}{6}$ and check if the result is a perfect square.

1. If $r+1=6$: $k^2=3+\frac{6}{6}=4$. This gives $k=\pm 2$. That is two valid pairs: $(5,2)$ and $(5,-2)$.

2. If $r+1=12$: $k^2=3+\frac{12}{6}=5$. Not a perfect square.

3. If $r+1=18$: $k^2=3+\frac{18}{6}=6$. Not a perfect square.

4. If $r+1=24$: $k^2=3+\frac{24}{6}=7$. Not a perfect square.

5. If $r+1=30$: $k^2=3+\frac{30}{6}=8$. Not a perfect square.

6. If $r+1=36$: $k^2=3+\frac{36}{6}=9$. This gives $k=\pm 3$. That is two more valid pairs: $(35,3)$ and $(35,-3)$.

The Final Count

By systematically checking our candidates, we have found exactly four ordered pairs $(r,k)$ that satisfy the original equation: $(5,2),(5,-2),(35,3),$ and $(35,-3)$.

Do you see the beauty here? We started with a terrifying expression involving binomial coefficients, and through the power of a single identity and a bit of logical constraint analysis, we reduced it to a simple search.

This is the essence of JEE Advanced mathematics. It is not about brute force; it is about finding the right tool to simplify the chaos. Keep this mindset, and you will find that no problem is truly insurmountable.