JEE Advanced 2018

LEVELJEE Main

Animated Solution for Mathematics - Binomial Theorem: Let $X = (^{10} C_{1})^{2} + 2 (^{10} C_{2})^{2} + 3 (^{10} C_{3})^{2} + \dots + 10 (^{10} C_{10})^{2}$ , where $^{10} C_{r}, r \in {1, 2, \dots, 10}$ denote binomial coefficients. Then the value of $\frac{1}{1430} X$ is ________.

Q: Why did we flip the second binomial coefficient using symmetry? Is that allowed?

Absolutely! The symmetry property $^nC_r = ^nC_{n-r}$ is one of the most powerful tools in combinatorics. We flipped it because the identity requires the sum of the lower indices to be constant. By flipping $^nC_r$ to $^nC_{n-r}$, the lower indices became $(r-1) + (n-r) = n-1$. Now, the sum is independent of $r$, which is exactly what we needed to apply the identity.

Enter Numerical Value:

Progressive Hints

Hint 1

Look at the structure of the series. You have a sum of squares, but that index

r

is the 'villain' preventing a direct application of standard identities. Ask yourself: how can I transform

r \cdot (^{n} C_{r})^{2}

into something where the

r

is absorbed?

Hint 2

Recall the Absorption Property:

r \cdot^{n} C_{r} = n \cdot^{n - 1} C_{r - 1}

. This is your key. Once you apply this, you will have a product of two binomial coefficients. But wait—to use Vandermonde's Identity, the sum of the lower indices must be constant. How can you use the symmetry property

^{n} C_{r} =^{n} C_{n - r}

to fix this?

Hint 3

Your summation should now look like

S = n \sum_{r=1}^{n} ^{n-1}C_{r-1} \cdot ^{n}C_{n-r}

. Recognize this as the combinatorial essence of Vandermonde's Identity: the sum of the product of two binomial coefficients where the lower indices sum to a constant. This simplifies to

n \cdot^{2 n - 1} C_{n - 1}

Visualized Solution

$X = \sum r (^{10} C_{r})^{2}$

Given expression: $X = (^{10} C_{1})^{2} + 2 (^{10} C_{2})^{2} + \dots + 10 (^{10} C_{10})^{2}$
We need to find the value of $\frac{X}{1430}$ .
Let's express $X$ in a compact summation form: $X = \sum_{r = 1}^{10} r (^{10} C_{r})^{2}$ .

The Absorption Property

The general term is $r \cdot (^{n} C_{r})^{2}$ .
We need to eliminate the $r$ multiplying the binomial coefficient.
Recall the absorption property: $r \cdot^{n} C_{r} = n \cdot^{n - 1} C_{r - 1}$ .

Applying the Property

Write $r (^{n} C_{r})^{2}$ as $(r \cdot^{n} C_{r}) \cdot^{n} C_{r}$ .
Substitute the property: $(n \cdot^{n - 1} C_{r - 1}) \cdot^{n} C_{r}$ .
The sum becomes: $S = \sum_{r = 1}^{n} n \cdot^{n - 1} C_{r - 1} \cdot^{n} C_{r}$ .

Factoring out $n$ & Symmetry

Since $n$ is constant with respect to $r$ , take it outside: $S = n \sum_{r=1}^{n} ^{n-1}C_{r-1} \cdot ^{n}C_{r}$ .
To apply standard identities, the sum of lower indices must be constant.
Currently: $(r - 1) + r = 2 r - 1$ (depends on $r$ ).
Use symmetry: $^{n} C_{r} =^{n} C_{n - r}$ .
New sum: $S = n \sum_{r=1}^{n} ^{n-1}C_{r-1} \cdot ^{n}C_{n-r}$ .

Vandermonde's Identity Setup

We now have $S = n \sum_{r=1}^{n} ^{n-1}C_{r-1} \cdot ^{n}C_{n-r}$ .
This perfectly matches Vandermonde's Identity.
Imagine two groups of items. Group 1 has size $n - 1$ , and Group 2 has size $n$ .

Choosing from Groups

From Group 1, we choose $r - 1$ items.
From Group 2, we choose $n - r$ items.
The product $^{n - 1} C_{r - 1} \cdot^{n} C_{n - r}$ represents the number of ways to do this for a specific $r$ .

Total Selection

Total items chosen = $(r - 1) + (n - r) = n - 1$ .
Total items available = $(n - 1) + n = 2 n - 1$ .
Summing over all possible values of $r$ gives the total ways to choose $n - 1$ items from $2 n - 1$ items.
Therefore, $\sum_{r=1}^{n} ^{n-1}C_{r-1} \cdot ^{n}C_{n-r} = ^{2n-1}C_{n-1}$ .

General Formula for $S$

Substituting the result back, we get the general formula:
$S = n \cdot^{2 n - 1} C_{n - 1}$ .
For our specific problem, $n = 10$ .
So, $X = 10 \cdot^{2 (10) - 1} C_{10 - 1} = 10 \cdot^{19} C_{9}$ .

Setting up the Final Target

The question asks for the value of $\frac{X}{1430}$ .
Substitute the value of $X$ : $\frac{10 \cdot ^{19} C _{9}}{1430}$ .
Simplifying the fraction gives: $\frac{^{19} C _{9}}{143}$ .

Expanding $^{19} C_{9}$

Expand $^{19} C_{9} = \frac{19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11}{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$ .
Notice that $143 = 11 \times 13$ .
This is perfect because $11$ and $13$ are present in the numerator of $^{19} C_{9}$ .

Cancelling Terms

$\frac{^{19} C _{9}}{143} = \frac{19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 12}{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$ .
Group the denominator to cancel terms in the numerator:
$\frac{18}{9 \times 2} = 1$ , $\frac{15}{5 \times 3} = 1$ .
$\frac{14}{7} = 2$ , $\frac{12}{6} = 2$ , $\frac{16}{8} = 2$ .

Final Calculation

Remaining expression: $\frac{19 \times 17 \times 2 \times 2 \times 2}{4}$ .
Wait, the $4$ in the denominator is still there!
Actually, $2 \times 2 \times 2 = 8$ . So $\frac{8}{4} = 2$ .
Final value: $19 \times 17 \times 2 = 323 \times 2 = 646$ .

00:00 / 00:00

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The series provided is

X = \sum_{r = 1}^{n} r \cdot (^{n} C_{r})^{2}

. At first glance, this appears to be a complex summation that would require brute force to evaluate.

However, in JEE Advanced mathematics, we avoid brute force by identifying the underlying structure. Our goal is to simplify the general term

r \cdot (^{n} C_{r})^{2}

using binomial identities.

The Absorption Property

The primary obstacle is the factor

r

multiplying the binomial coefficient. We utilize the Absorption Property, which states:

r \cdot^{n} C_{r} = n \cdot^{n - 1} C_{r - 1}

By applying this, the

r

is absorbed into the binomial coefficient. The expression transforms into:

X = n \sum_{r=1}^{n} ^{n-1}C_{r-1} \cdot ^nC_r

The Symmetry Shuffle

To apply Vandermonde's Identity, the sum of the lower indices must be a constant. Currently, the indices are

(r - 1)

and

r

, which sum to

2 r - 1

We invoke the Symmetry Property, which states

^{n} C_{r} =^{n} C_{n - r}

. We rewrite the second term:

^{n} C_{r} =^{n} C_{n - r}

Substituting this back into our summation, the indices become

(r - 1) + (n - r) = n - 1

. The variable

r

is now eliminated from the sum of the lower indices.

The Vandermonde Revelation

We are now evaluating the sum:

X = n \sum_{r=1}^{n} ^{n-1}C_{r-1} \cdot ^nC_{n-r}

According to Vandermonde's Identity, the sum of the product of two binomial coefficients

\sum^{m} C_{k} \cdot^{p} C_{q - k}

is equal to

^{m + p} C_{q}

. Here,

m = n - 1

p = n

, and

q = n - 1

The summation simplifies elegantly:

X = n \cdot^{2 n - 1} C_{n - 1}

Final Calculation

Given

n = 10

, the expression becomes:

X = 10 \cdot^{19} C_{9}

We are asked to calculate the value of

\frac{X}{1430}

\frac{10 \cdot ^{19} C _{9}}{1430} = \frac{^{19} C _{9}}{143}

Expanding

^{19} C_{9}

\frac{19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11}{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 143}

Noting that

143 = 11 \times 13

, we cancel these factors from the numerator. After performing the remaining arithmetic simplification, we arrive at the final result:

646

Common Traps & FAQs

Wait, why can't we just expand the squares and sum them up? Isn't that safer?

That is the classic trap! In JEE Advanced, brute force is rarely the intended path. If you expand

(^{10} C_{r})^{2}

, you get a massive, unmanageable expression. The beauty of this problem lies in recognizing the pattern—the 'Absorption Property' and 'Vandermonde's Identity' are tools designed to collapse complexity. Trust the identities; they are the shortcuts built into the fabric of mathematics.

How did you know to use Vandermonde's Identity? It feels like magic.

It isn't magic; it's pattern recognition. Whenever you see a sum of the product of two binomial coefficients, your brain should immediately scream 'Vandermonde!' The key is the constant sum of the lower indices. Once we used the symmetry property to make the lower indices sum to

n - 1

, the structure of Vandermonde's Identity became undeniable. Keep practicing, and this intuition will become second nature.

Why did we flip the second binomial coefficient using symmetry? Is that allowed?

Absolutely! The symmetry property

^{n} C_{r} =^{n} C_{n - r}

is one of the most powerful tools in combinatorics. We flipped it because the identity requires the sum of the lower indices to be constant. By flipping

^{n} C_{r}

^{n} C_{n - r}

, the lower indices became

(r - 1) + (n - r) = n - 1

. Now, the sum is independent of

r

, which is exactly what we needed to apply the identity.

Animated Solution for Mathematics - Binomial Theorem: Let $X = (^{10} C_{1})^{2} + 2 (^{10} C_{2})^{2} + 3 (^{10} C_{3})^{2} + \dots + 10 (^{10} C_{10})^{2}$ , where $^{10} C_{r}, r \in {1, 2, \dots, 10}$ denote binomial coefficients. Then the value of $\frac{1}{1430} X$ is ________.

Enter Numerical Value:

Visualized Solution

$X = \sum r (^{10} C_{r})^{2}$

The Absorption Property

Applying the Property

Factoring out $n$ & Symmetry

Vandermonde's Identity Setup

Choosing from Groups

Total Selection

General Formula for $S$

Setting up the Final Target

Expanding $^{19} C_{9}$

Cancelling Terms

Final Calculation

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Absorption Property

The Symmetry Shuffle

The Vandermonde Revelation

Final Calculation

Similar Questions

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 value of $(0 30) (10 30) - (1 30) (11 30) + (2 30) (12 30) - .... + (20 30) (30 30)$ is

If $\sum_{k = 1}^{31} (^{31} C_{k}) (^{31} C_{k - 1}) - \sum_{k = 1}^{30} (^{30} C_{k}) (^{30} C_{k - 1}) = \frac{α ( 60 !)}{( 30 !) ( 31 !)}$ , where $α \in R$ , then the value of $16 α$ is equal to

If $a_{r}$ is the coefficient of $x^{10 - r}$ in the Binomial expansion of $(1 + x)^{10}$ , then $\sum_{r = 1}^{10} r^{3} (\frac{a _{r}}{a _{r - 1}})^{2}$ is equal to

Let $^{n} C_{r}$ denote the binomial coefficient of $x^{r}$ in the expansion of $(1 + x)^{n}$ . If $\sum_{k = 0}^{10} (2^{2} + 3 k)^{10} C_{k} = α \cdot 3^{10} + β \cdot 2^{10}, α, β \in R$ , then $α + β$ 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

If $(\frac{1}{^{15} C _{0}} + \frac{1}{^{15} C _{1}}) (\frac{1}{^{15} C _{1}} + \frac{1}{^{15} C _{2}}) \dots (\frac{1}{^{15} C _{12}} + \frac{1}{^{15} C _{13}}) = \frac{α ^{13}}{^{14} C _{0} \cdot ^{14} C _{1} \dots ^{14} C _{12}}$ , then $30 α$ is equal to .........

If some three consecutive in the binomial expansion of $(x + 1)^{n}$ in powers of $x$ are in the ratio $2 : 15 : 70$ , then the average of these three coefficients is :-

The value of $^{50}C_4 + \sum_{r=1}^6 ^{56-r}C_3$ is

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X=∑r(10Cr​)2

The Absorption Property

Applying the Property

Factoring out .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n & Symmetry

Vandermonde's Identity Setup

Choosing from Groups

Total Selection

General Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Setting up the Final Target

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 19C9​

Cancelling Terms

Final Calculation

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Absorption Property

The Symmetry Shuffle

The Vandermonde Revelation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (30C0​)2+2(30C1​)2+3(30C2​)2+⋯+31(30C30​)2=(30!)2α60!​, then α is equal to

.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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∑k=131​(31Ck​)(31Ck−1​)−∑k=130​(30Ck​)(30Ck−1​)=(30!)(31!)α(60!)​, where α∈R, then the value of 16α is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ar​ is the coefficient of x10−r in the Binomial expansion of (1+x)10, then ∑r=110​r3(ar−1​ar​​)2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let nCr​ denote the binomial coefficient of xr in the expansion of (1+x)n. If ∑k=010​(22+3k)10Ck​=α⋅310+β⋅210,α,β∈R, then α+β is equal to ___

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of 1!50!1​+3!48!1​+5!46!1​+⋯+49!2!1​+51!1!1​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (15C0​1​+15C1​1​)(15C1​1​+15C2​1​)⋯(15C12​1​+15C13​1​)=14C0​⋅14C1​⋯14C12​α13​, then 30α is equal to .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If some three consecutive in the binomial expansion of (x+1)n in powers of x are in the ratio 2:15:70, then the average of these three coefficients is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of ^{50}C_4 + \sum_{r=1}^6 ^{56-r}C_3 is

$X = \sum r (^{10} C_{r})^{2}$

Factoring out $n$ & Symmetry

General Formula for $S$

Expanding $^{19} C_{9}$

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 value of $(0 30) (10 30) - (1 30) (11 30) + (2 30) (12 30) - .... + (20 30) (30 30)$ is

If $\sum_{k = 1}^{31} (^{31} C_{k}) (^{31} C_{k - 1}) - \sum_{k = 1}^{30} (^{30} C_{k}) (^{30} C_{k - 1}) = \frac{α ( 60 !)}{( 30 !) ( 31 !)}$ , where $α \in R$ , then the value of $16 α$ is equal to

If $a_{r}$ is the coefficient of $x^{10 - r}$ in the Binomial expansion of $(1 + x)^{10}$ , then $\sum_{r = 1}^{10} r^{3} (\frac{a _{r}}{a _{r - 1}})^{2}$ is equal to

Let $^{n} C_{r}$ denote the binomial coefficient of $x^{r}$ in the expansion of $(1 + x)^{n}$ . If $\sum_{k = 0}^{10} (2^{2} + 3 k)^{10} C_{k} = α \cdot 3^{10} + β \cdot 2^{10}, α, β \in R$ , then $α + β$ 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

If $(\frac{1}{^{15} C _{0}} + \frac{1}{^{15} C _{1}}) (\frac{1}{^{15} C _{1}} + \frac{1}{^{15} C _{2}}) \dots (\frac{1}{^{15} C _{12}} + \frac{1}{^{15} C _{13}}) = \frac{α ^{13}}{^{14} C _{0} \cdot ^{14} C _{1} \dots ^{14} C _{12}}$ , then $30 α$ is equal to .........

If some three consecutive in the binomial expansion of $(x + 1)^{n}$ in powers of $x$ are in the ratio $2 : 15 : 70$ , then the average of these three coefficients is :-

The value of $^{50}C_4 + \sum_{r=1}^6 ^{56-r}C_3$ is