The value of (30/0)(30/10) - (30/1)(30/11) + (30/2)(30/12) - .... + (30/20)(30/30) is (a) (30/10) (b

Q: Why did we change $\binom{30}{r+10}$ to $\binom{30}{20-r}$? Isn't that just making it more complicated?

That is a fantastic question! It might feel like extra work, but it is actually the 'Golden Key' to the problem. We need the sum of the lower indices to be constant to use the coefficient extraction method. By transforming $\binom{30}{r+10}$ into $\binom{30}{20-r}$, we turn a difference of $10$ into a constant sum of $20$. It is not complication; it is strategic simplification!

Q: How do I know when to multiply two binomial expansions?

Whenever you see a product of two binomial coefficients $\binom{n}{r}\binom{m}{k}$ where the sum $r+k$ is constant, that is your signal! It is a classic JEE pattern. You are essentially looking for the coefficient of $x^{\text{constant}}$ in the product of two polynomials. If you see this structure, immediately think about multiplying two series.

Q: Why did we pick $(1-x)^{30}$ and $(1+x)^{30}$ specifically?

Look at the alternating sign $(-1)^r$ in your series. That sign is the mathematical fingerprint of $(1-x)^n$. If the series had all positive signs, we would have used $(1+x)^n$ and $(1+x)^n$. The alternating sign forces us to use $(1-x)^n$ to generate that $(-1)^r$ term, while the other term remains positive to keep the structure intact.

Question

The value of $\binom{30}{0}\binom{30}{10} - \binom{30}{1}\binom{30}{11} + \binom{30}{2}\binom{30}{12} - .... + \binom{30}{20}\binom{30}{30}$ is

Options:

(A) $\binom{30}{10}$

(B) $\binom{30}{15}$

(C) $\binom{60}{30}$

(D) $\binom{31}{10}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are going to dismantle a problem that looks intimidating at first glance but is actually a beautifully orchestrated dance of binomial coefficients. We are looking at the series:

$$S = \binom{30}{0}\binom{30}{10} - \binom{30}{1}\binom{30}{11} + \dots + \binom{30}{20}\binom{30}{30}$$

When you see a long, alternating series like this, do not panic. Instead, put on your detective hat. Notice two things: the alternating signs ($+$, $-$, $+$, $-$) and the indices.

The lower indices are $0$ and $10$, $1$ and $11$, $2$ and $12$. The difference is constant! In the world of JEE Advanced, a constant difference or a constant sum in binomial indices is a massive, flashing neon sign telling you exactly what to do.

### The Symmetry Transformation

We want to manipulate this series into a form where we can apply the 'Coefficient Extraction' method. To do that, we need the sum of the lower indices to be constant, not the difference. Here, we invoke our ultimate weapon: the symmetry property of binomial coefficients, which states that $\binom{n}{k} = \binom{n}{n-k}$.

Let us apply this to the second term, $\binom{30}{r+10}$. Using our property, this becomes $\binom{30}{30 - (r+10)}$, which simplifies beautifully to $\binom{30}{20-r}$.

Now, look at our series again. We have $\binom{30}{r}$ multiplied by $\binom{30}{20-r}$. If we add the lower indices, $r + (20-r)$, we get $20$. A constant! This is the breakthrough we needed.

### The Grand Strategy: Multiplying Expansions

Now that we have a constant sum of $20$, we know we are looking for the coefficient of $x^{20}$ in the product of two binomial expansions. But which ones?

Because our series has an alternating sign $(-1)^r$, we know one of the expansions must be $(1-x)^{30}$, because its general term involves $(-1)^r x^r$. The other expansion must be $(1+x)^{30}$, which provides the $x^k$ term.

When we multiply these two, $(1-x)^{30}(1+x)^{30}$, the powers of $x$ combine as $x^r \cdot x^k = x^{r+k}$. We are specifically hunting for the coefficient where $r+k = 20$.

### The Final Victory

This is where the math becomes elegant. We are looking for the coefficient of $x^{20}$ in the product $(1-x)^{30}(1+x)^{30}$. Using the identity $(a-b)(a+b) = a^2 - b^2$, this product simplifies to:

$$((1-x)(1+x))^{30} = (1-x^2)^{30}$$

Now, the problem is trivial. We need the coefficient of $x^{20}$ in the expansion of $(1-x^2)^{30}$. The general term is:

$$\binom{30}{m} (-x^2)^m = \binom{30}{m} (-1)^m x^{2m}$$

To get $x^{20}$, we set $2m = 20$, which gives us $m = 10$. Substituting $m=10$ into our general term, we get $\binom{30}{10} (-1)^{10}$. Since $10$ is even, $(-1)^{10} = 1$.

And there it is: our final answer is **$\binom{30}{10}$**.

Remember, the complexity of a problem is often just a mask. Once you peel it back with the right properties—symmetry and expansion multiplication—the solution reveals itself. Keep practicing, keep questioning, and keep falling in love with the logic!

Animated Solution for Mathematics - Binomial Theorem: The value of $(0 30) (10 30) - (1 30) (11 30) + (2 30) (12 30) - .... + (20 30) (30 30)$ is

Select Answer:

Visualized Solution

Analyzing the Given Series

Converting to Sigma Notation

The Summation Form

The Symmetry Property of Binomial Coefficients

Transforming the Second Term

The Updated Summation

The Method of Comparing Coefficients

Setting Up the Expansions

Multiplying the Series

Identifying the Target Coefficient

Simplifying the Algebraic Expression

General Term of the Simplified Expansion

Finding the Specific Term

The Final Calculation

Conclusion and Key Takeaway

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Symmetry Transformation

The Grand Strategy

The Final Victory

Similar Questions

For $2 \leq r \leq n$ , $(r n) + 2 (r - 1 n) + (r - 2 n) =$

The value of $(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + (^{21} C_{4} -^{10} C_{4}) + \dots + (^{21} C_{10} -^{10} C_{10})$ is:

Prove that $2^{k} (0 n) (k n) - 2^{k - 1} (1 n) (k - 1 n - 1) + 2^{k - 2} (2 n) (k - 2 n - 2) - .... + (- 1)^{k} (k n) (0 n - k) = (k n)$

The value of $\sum_{r=0}^{20} ^{50-r}C_6$ is equal to:

The value of $\frac{1}{1 ! 50 !} + \frac{1}{3 ! 48 !} + \frac{1}{5 ! 46 !} + \dots + \frac{1}{49 ! 2 !} + \frac{1}{51 ! 1 !}$ is

For $r = 0, 1, ..., 10$ , let $A_{r}, B_{r}$ and $C_{r}$ denote, respectively, the coefficient of $x^{r}$ in the expansions of $(1 + x)^{10}, (1 + x)^{20}$ and $(1 + x)^{30}$ . Then $\sum_{r = 1}^{10} A_{r} (B_{10} B_{r} - C_{10} A_{r})$ is equal to

If $(^{30} C_{0})^{2} + 2 (^{30} C_{1})^{2} + 3 (^{30} C_{2})^{2} + \dots + 31 (^{30} C_{30})^{2} = \frac{α 60 !}{( 30 ! ) ^{2}}$ , then $α$ is equal to

The sum $\sum_{i = 0}^{m} (i 10) (m - i 20)$ , (where $(q p) = 0$ if $p < q$ ) is maximum when $m$ is

$\sum_{i, j = 0 i \neq = j}^{n}^{n} C_{i}^{n} C_{j}$ is equal to

Let $m, n \in N$ and gcd $(2, n) = 1$ . If $30 (0 30) + 29 (1 30) + .... + 2 (28 30) + 1 (29 30) = n . 2^{m}$ , then $n + m$ is equal to

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 (030​)(1030​)−(130​)(1130​)+(230​)(1230​)−....+(2030​)(3030​) is