Animated Solution for Mathematics - Binomial Theorem: The value of the expression ${}^{47}{C}_4+{\sum }_{j=1}^5{}^{52-j}{C}_3$ is equal to

Welcome, fellow traveler on the path to JEE excellence! Today, we are not just solving a problem; we are uncovering a beautiful, rhythmic dance hidden within the world of combinatorics.

When you first look at the expression ${}^{47}{C}_4+{\sum }_{j=1}^5{}^{52-j}{C}_3$, it might look like a daunting wall of symbols. In mathematics, complexity is often just simplicity in disguise, waiting for the right perspective to reveal itself.

Before we dive into the deep end, let us lay out our tools. We have a summation, and the best way to understand a sum is to see its components. Let us expand the term ${}^{52-j}{C}_3$ for $j=1,2,3,4,5$:

For $j=1$, we get ${}^{51}{C}_3$. For $j=2$, we get ${}^{50}{C}_3$. For $j=3$, we get ${}^{49}{C}_3$. For $j=4$, we get ${}^{48}{C}_3$. For $j=5$, we get ${}^{47}{C}_3$.

Now, let us place these back into our original expression. We have ${}^{47}{C}_4+{(}^{51}{C}_3{+}^{50}{C}_3{+}^{49}{C}_3{+}^{48}{C}_3{+}^{47}{C}_3)$.

To see the pattern clearly, let us rearrange them in ascending order of the upper index:

Here is where the magic happens. We invoke the legendary Pascal's Identity:

Think of this as a "merging" rule. It tells us that if we have two groups of combinations with the same upper index $n$, but consecutive lower indices $r$ and $r-1$, they can fuse into a single, larger combination with a higher upper index.

Look at our first two terms: ${}^{47}{C}_4$ and ${}^{47}{C}_3$. They are perfect candidates! Applying the identity, we get:

This is the most thrilling part of the journey. We have just created ${}^{48}{C}_4$. Look at what is waiting next in our sequence: ${}^{48}{C}_3$.

We can see the pattern repeating:

We are climbing a ladder! We take the result, merge it with the next term, and watch the upper index grow. Let us continue this elegant chain:

1. Next, we take our new ${}^{49}{C}_4$ and add it to the next term, ${}^{49}{C}_3$, to get ${}^{50}{C}_4$. 2. Then, we take ${}^{50}{C}_4$ and add it to ${}^{50}{C}_3$, resulting in ${}^{51}{C}_4$. 3. Finally, we reach the last term in our sequence, ${}^{51}{C}_3$. We combine our ${}^{51}{C}_4$ with ${}^{51}{C}_3$ to reach our final destination:

Look at what we have achieved. We started with a complex sum, and through the systematic application of Pascal's Identity, we collapsed it into a single, elegant term: ${}^{52}{C}_4$.

This is the essence of JEE Advanced mathematics—not brute force, but the ability to recognize patterns and apply fundamental laws with precision and grace. You didn't just solve a problem; you performed a mathematical symphony. Keep this clarity of thought, and there is no problem you cannot conquer!