Animated Solution for Mathematics - Binomial Theorem: Let $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ denotes ${}^n{C}_k$ and $\left[\begin{array}{c}n\\ k\end{array}\right]=\{\begin{array}{ll}\left(\genfrac{}{}{0ex}{}{n}{k}\right),& \text{if }0\le k\le n\\ 0,& \text{otherwise}\end{array}$. If $A_k={\sum }_{i=0}^9\left(\genfrac{}{}{0ex}{}{9}{i}\right)\left[\begin{array}{c}12\\ 12-k+i\end{array}\right]+{\sum }_{i=0}^8\left(\genfrac{}{}{0ex}{}{8}{i}\right)\left[\begin{array}{c}13\\ 13-k+i\end{array}\right]$ and $A_4-A_3=190p$, then $p$ is equal to :

When you first look at this problem, it is designed to intimidate you. That square bracket notation $\left[\begin{array}{c}n\\ k\end{array}\right]$ is a psychological barrier.

In the heat of a JEE Advanced exam, seeing unfamiliar notation can make your heart race. But take a deep breath. As we peel back the layers, you will see that this is just a standard binomial coefficient in disguise, specifically $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$.

The problem is not testing your ability to decipher code; it is testing your ability to see through the noise.

Our first mission is to simplify the summations. We have the expression:

This looks messy. But remember the fundamental symmetry of combinations: $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\left(\genfrac{}{}{0ex}{}{n}{n-r}\right)$.

This is not just a formula; it is a physical reality. Choosing $r$ items to keep is the same as choosing $n-r$ items to discard. When we apply this to $\left(\genfrac{}{}{0ex}{}{12}{12-k+i}\right)$, it transforms into $\left(\genfrac{}{}{0ex}{}{12}{k-i}\right)$.

Suddenly, the expression becomes:

Do you see the beauty emerging? We have a product of two binomial coefficients where the sum of the lower indices is $i+(k-i)=k$. This is the signature of a master theorem.

Vandermonde's Identity is one of the most powerful tools in your combinatorial arsenal. It states:

It essentially says that if you want to choose $k$ items from a combined group of $n+m$ items, you can do it by choosing $r$ from the first group and $k-r$ from the second, and summing over all possible values of $r$.

Applying this to our simplified sums, both the first and second summations collapse beautifully into $\left(\genfrac{}{}{0ex}{}{21}{k}\right)$. The complexity vanishes, leaving us with:

Now, the problem reduces to simple arithmetic. We need to evaluate $A_4-A_3$:

Calculating $\left(\genfrac{}{}{0ex}{}{21}{4}\right)$ gives us $5985$, and $\left(\genfrac{}{}{0ex}{}{21}{3}\right)$ gives us $1330$. The difference is $4655$.

Multiplying by $2$, we get $9310$. Equating this to $190p$, we find:

You have conquered the problem not by brute force, but by recognizing the underlying structure. Keep this clarity of mind, and you will solve any problem JEE throws at you. The final answer is 49.