The value of r for which 20Cr·20C0 + 20Cr-1·20C1 + .... 20C0·20Cr is maximum, is (A) 20 (B) 15 (C) 1

Q: Wait, why does the sum of indices being constant matter? How does that lead to multiplication?

That is the heart of the problem! When you multiply two polynomials, say $P(x) = \sum a_i x^i$ and $Q(x) = \sum b_j x^j$, the coefficient of $x^r$ in the product is the sum of all products $a_i b_j$ where $i+j=r$. By noticing that the sum of the lower indices is always $r$, you are essentially identifying the coefficients of the product polynomial. It is the bridge between summation and polynomial multiplication.

Q: Is it always $(1+x)^{40}$? What if the upper indices were different?

Excellent observation. It is only $(1+x)^{40}$ because both upper indices were $20$. If the problem had been ${}^{20}C_r \cdot {}^{30}C_0 + \dots$, you would be looking at the coefficient of $x^r$ in $(1+x)^{20} \cdot (1+x)^{30}$, which would be $(1+x)^{50}$. Always check the upper indices to determine the total power!

Q: Why is the maximum at $r=20$? Is there a formula for this?

Yes, there is! For any binomial coefficient ${}^{n}C_r$, the values increase as $r$ goes from $0$ to $n/2$ and then decrease. If $n$ is even, the maximum is at the exact middle, $r = n/2$. If $n$ is odd, there are two maximums at $r = (n-1)/2$ and $r = (n+1)/2$. Since $n=40$ is even, $40/2 = 20$ is our peak.

Question

The value of $r$ for which ${}^{20}C_r \cdot {}^{20}C_0 + {}^{20}C_{r-1} \cdot {}^{20}C_1 + \dots + {}^{20}C_0 \cdot {}^{20}C_r$ is maximum, is

Options:

(A) 20

(B) 15

(C) 11

(D) 10

Sigma Percentile · Accepted Answer

### The Illusion of Complexity

Welcome, future engineer. Take a deep breath. I know that when you first looked at this problem, your eyes probably darted across the page, trying to process that long, intimidating string of terms:

$${}^{20}C_r \cdot {}^{20}C_0 + {}^{20}C_{r-1} \cdot {}^{20}C_1 + \dots + {}^{20}C_0 \cdot {}^{20}C_r$$

It looks like a monster, doesn't it? A summation that seems to demand brute force calculation.

But here is the secret of JEE Advanced: whenever you see a complex summation of products, you are rarely expected to calculate it term by term. You are being invited to see a pattern. You are being invited to see the 'soul' of the expression.

### The Signature of a Convolution

Let us play detective. Look at the lower indices. In the first term, we have $r$ and $0$. Their sum is $r$.

In the second term, we have $r-1$ and $1$. Their sum is $r$. In the last term, we have $0$ and $r$. Their sum is $r$.

Do you see it? The sum of the lower indices is constant. This is the 'Aha!' moment.

In the world of combinatorics, this constant sum is the signature of a convolution. It is the mathematical equivalent of a fingerprint. It tells us that this sum is not just a random collection of numbers; it is the coefficient of a specific power of $x$ in the product of two polynomials.

### The Binomial Bridge

We know the binomial expansion:

$$(1+x)^n = {}^{n}C_0 + {}^{n}C_1 x + {}^{n}C_2 x^2 + \dots + {}^{n}C_n x^n$$

Our problem involves ${}^{20}C_k$ terms, so let us bring in the expansion of $(1+x)^{20}$. Now, imagine we have two of these expansions. We multiply $(1+x)^{20}$ by itself.

When we multiply these two polynomials, how do we get the term containing $x^r$? We take the $x^r$ term from the first bracket and multiply it by the constant term from the second. Then we take the $x^{r-1}$ term from the first and multiply it by the $x^1$ term from the second.

We continue this until we take the constant term from the first and the $x^r$ term from the second. Does that look familiar? It is exactly the sum given in our problem!

### The Elegance of Reduction

By the laws of exponents, we know that:

$$(1+x)^{20} \cdot (1+x)^{20} = (1+x)^{40}$$

Suddenly, the monster has been tamed. The entire, terrifying summation is nothing more than the coefficient of $x^r$ in the expansion of $(1+x)^{40}$.

And what is the coefficient of $x^r$ in $(1+x)^{40}$? It is simply ${}^{40}C_r$. We have reduced a complex series to a single, elegant binomial coefficient. This is the power of mathematical thinking.

### The Peak of the Mountain

Now, the final step is almost trivial. We need to maximize ${}^{40}C_r$. Think of Pascal's Triangle.

The values of binomial coefficients start small, grow as they approach the middle, and then shrink back down. They form a beautiful, symmetric bell curve.

For any $n$, the maximum value occurs at the middle. Since our $n$ is $40$, which is an even number, the peak is exactly at the center. We calculate:

$$r = \frac{40}{2} = 20$$

And there you have it. The value of $r$ that maximizes the sum is **$20$**. You didn't just solve a problem; you navigated through the logic of combinatorics. Keep this perspective—always look for the underlying structure, and you will find that even the most intimidating problems are just puzzles waiting to be solved.

Animated Solution for Mathematics - Binomial Theorem: The value of $r$ for which $^{20} C_{r} \cdot^{20} C_{0} +^{20} C_{r - 1} \cdot^{20} C_{1} + \dots +^{20} C_{0} \cdot^{20} C_{r}$ is maximum, is

Select Answer:

Visualized Solution

The Given Series

Pattern in Indices

The Binomial Tool

Multiplying Expansions

Coefficient of $x^{r}$

Forming the Sum $S$

Combining the Powers

The Value of $S$

Condition for Maximum

Finding the Peak

The Final Answer

The Sigma Insight: Properties of Binomial Coefficients

The Illusion of Complexity

The Signature of a Convolution

The Binomial Bridge

The Elegance of Reduction

The Peak of the Mountain

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Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of r for which 20Cr​⋅20C0​+20Cr−1​⋅20C1​+⋯+20C0​⋅20Cr​ is maximum, is