Animated Solution for Mathematics - Permutations and Combinations: ${}^{n-1}{C}_r=(k^2-8{)}^n{C}_{r+1}$ if and only if :

Welcome, future engineers! Today, we are going to dissect a problem that perfectly encapsulates the spirit of JEE Advanced. It is not about brute-forcing calculations; it is about recognizing the hidden structure within the symbols.

When you look at an equation like ${}^{n-1}{C}_r=(k^2-8{)}^n{C}_{r+1}$, your first instinct might be to expand the factorials. Stop! Take a breath. If you start writing out

you are walking into a trap. Let us look for the elegance instead.

At first glance, this equation looks like a messy algebraic relationship involving combinations. But look closer at the relationship between ${}^{n-1}{C}_r$ and ${}^n{C}_{r+1}$. These are not random terms; they are neighbors in the Pascal triangle.

We need a bridge. The identity

is our bridge. Why does this work? Think about the definition of combinations:

If you manipulate this, you will see it naturally reduces to

By substituting this, we transform our equation into

Now, notice how the complexity collapses. We have ${}^{n-1}{C}_r$ on both sides. Since we are dealing with valid combinations, we know $^{n-1}C_r eq 0$.

We can safely divide both sides by this term. Suddenly, the equation becomes much friendlier:

Our goal is to isolate the term containing $k$. Rearranging this, we get

This is the turning point. The problem is no longer about combinations; it is about the range of a variable. We have successfully reduced a complex combinatorial equation into a simple algebraic inequality.

This is where the JEE Advanced examiners test your maturity. We have the expression

What are the bounds for this? We know that for ${}^{n-1}{C}_r$ to be defined, $r$ must satisfy the condition $0\le r\le n-1$.

This is the constraint that many students miss. By manipulating this inequality, we add 1 to all parts to get $1\le r+1\le n$. Then, dividing by $n$, we obtain

Since $n$ is a positive integer, $\frac{1}{n}$ is always greater than 0. Therefore, we have the strict inequality $0<\frac{r+1}{n}\le 1$.

Now, we substitute this back into our expression for $k$. We have

Adding 8 to all parts, we get

Finally, taking the square root—and keeping in mind that we are looking for positive values of $k$—we arrive at $\sqrt{8}<k\le \sqrt{9}$, which simplifies beautifully to

This is the beauty of mathematics. We started with a daunting equation, used a fundamental identity to simplify it, respected the domain constraints, and arrived at a clean, elegant solution. Keep this mindset: look for the identity, respect the constraints, and the answer will reveal itself.