JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Vandermonde's Identity and Binomial Coefficient Sums

The value of Σ_(r=0)^22 ^22 C_r ^23 C_r is

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A^44 C_23B^45 C_23C^44 C_22D^45 C_24

Step-by-step solutionView

Correct answer

Using the symmetry property $^{23} C_{r} =^{23} C_{23 - r}$ , the sum becomes $\sum_{r = 0}^{22}^{22} C_{r}^{23} C_{23 - r}$ , which by Vandermonde's identity equals $^{45} C_{23}$ .

Option analysis

Why each option works or fails

A · ^44 C_23

Adding the lower indices to form $n = 22 + 22 = 44$ instead of properly summing the upper indices $22 + 23 = 45$ . Vandermonde's identity states that $\sum^{n} C_{r}^{m} C_{k - r} =^{n + m} C_{k}$ ; here $n + m = 22 + 23 = 45$ , not $44$ .

B · ^45 C_23

Correct option: Rewrite $^{23} C_{r}$ as $^{23} C_{23 - r}$ , so the sum corresponds to the coefficient of $x^{23}$ in $(1 + x)^{22} (1 + x)^{23} = (1 + x)^{45}$ , giving $^{45} C_{23}$ . This is the correct result.

C · ^44 C_22

Confusing the upper index with $22 + 22$ while also incorrectly setting the bottom parameter to $22$ . The two binomial coefficients have upper parameters $22$ and $23$ , whose sum is $45$ , not $44$ .

D · ^45 C_24

An off-by-one index error when rewriting $^{23} C_{r}$ in terms of its complement, yielding lower index $24$ instead of $23$ . Since $^{23} C_{r} =^{23} C_{23 - r}$ , the sum of the indices $r + (23 - r)$ is $23$ , so the bottom index must be $23$ .

Reviewed route

Solution

StepWorking

01Given

The expression to evaluate is $S = \sum_{r = 0}^{22}^{22} C_{r} \cdot^{23} C_{r}$ .

02Goal

Evaluate the sum $S$ in terms of a single binomial coefficient $^{n} C_{k}$ .

03Approach

Use the symmetry property of binomial coefficients $^{n} C_{r} =^{n} C_{n - r}$ on $^{23} C_{r}$ to rewrite the sum with constant indices, then apply Vandermonde's convolution identity $\sum_{r = 0}^{n}^{n} C_{r}^{m} C_{k - r} =^{n + m} C_{k}$ .

04Execute

Rewrite $^{23} C_{r}$ as $^{23} C_{23 - r}$ . Then the sum becomes: $S = \sum_{r = 0}^{22}^{22} C_{r} \cdot^{23} C_{23 - r}$ This represents the coefficient of $x^{23}$ in the expansion of $(1 + x)^{22} (1 + x)^{23} = (1 + x)^{45}$ . Therefore, $S =^{45} C_{23}$ .

✓Verify

Using $^{n} C_{r} =^{n} C_{n - r}$ , $^{45} C_{23} =^{45} C_{22}$ . For a small scale check, with upper limits 1 and 2: $\sum_{r = 0}^{1}^{1} C_{r}^{2} C_{r} =^{1} C_{0}^{2} C_{0} +^{1} C_{1}^{2} C_{1} = 1 \cdot 1 + 1 \cdot 2 = 3 =^{3} C_{2}$ , which confirms formula $^{n + m} C_{m}$ .

Hints that build this answer step by step

Which symmetry property should we apply to $^{23} C_{r}$ to make the lower indices sum to a constant?
$^{23} C_{r} =^{23} C_{23 - r}$
Using Vandermonde's identity $\sum_{r = 0}^{k}^{n} C_{r}^{m} C_{k - r} =^{n + m} C_{k}$ , what does $\sum_{r = 0}^{22}^{22} C_{r}^{23} C_{23 - r}$ evaluate to?
$^{45} C_{23}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why do we replace

^{23} C_{r}

with

^{23} C_{23 - r}

Because Vandermonde's convolution requires the sum of the lower indices of the two binomial terms to remain constant across all terms ( $r + (23 - r) = 23$ ).

Answer

Using the symmetry property $^{23} C_{r} =^{23} C_{23 - r}$ , the sum becomes $\sum_{r = 0}^{22}^{22} C_{r}^{23} C_{23 - r}$ , which by Vandermonde's identity equals $^{45} C_{23}$ .

Why each option works or fails

A: ^44 C_23 - Adding the lower indices to form $n = 22 + 22 = 44$ instead of properly summing the upper indices $22 + 23 = 45$ . Vandermonde's identity states that $\sum^{n} C_{r}^{m} C_{k - r} =^{n + m} C_{k}$ ; here $n + m = 22 + 23 = 45$ , not $44$ .
B · correct: ^45 C_23 - Correct option: Rewrite $^{23} C_{r}$ as $^{23} C_{23 - r}$ , so the sum corresponds to the coefficient of $x^{23}$ in $(1 + x)^{22} (1 + x)^{23} = (1 + x)^{45}$ , giving $^{45} C_{23}$ . This is the correct result.
C: ^44 C_22 - Confusing the upper index with $22 + 22$ while also incorrectly setting the bottom parameter to $22$ . The two binomial coefficients have upper parameters $22$ and $23$ , whose sum is $45$ , not $44$ .
D: ^45 C_24 - An off-by-one index error when rewriting $^{23} C_{r}$ in terms of its complement, yielding lower index $24$ instead of $23$ . Since $^{23} C_{r} =^{23} C_{23 - r}$ , the sum of the indices $r + (23 - r)$ is $23$ , so the bottom index must be $23$ .

Step-by-step solution

Given: The expression to evaluate is $S = \sum_{r = 0}^{22}^{22} C_{r} \cdot^{23} C_{r}$ .
Goal: Evaluate the sum $S$ in terms of a single binomial coefficient $^{n} C_{k}$ .
Approach: Use the symmetry property of binomial coefficients $^{n} C_{r} =^{n} C_{n - r}$ on $^{23} C_{r}$ to rewrite the sum with constant indices, then apply Vandermonde's convolution identity $\sum_{r = 0}^{n}^{n} C_{r}^{m} C_{k - r} =^{n + m} C_{k}$ .
Execute: Rewrite $^{23} C_{r}$ as $^{23} C_{23 - r}$ . Then the sum becomes: $S = \sum_{r = 0}^{22}^{22} C_{r} \cdot^{23} C_{23 - r}$ This represents the coefficient of $x^{23}$ in the expansion of $(1 + x)^{22} (1 + x)^{23} = (1 + x)^{45}$ . Therefore, $S =^{45} C_{23}$ .
Verify: Using $^{n} C_{r} =^{n} C_{n - r}$ , $^{45} C_{23} =^{45} C_{22}$ . For a small scale check, with upper limits 1 and 2: $\sum_{r = 0}^{1}^{1} C_{r}^{2} C_{r} =^{1} C_{0}^{2} C_{0} +^{1} C_{1}^{2} C_{1} = 1 \cdot 1 + 1 \cdot 2 = 3 =^{3} C_{2}$ , which confirms formula $^{n + m} C_{m}$ .

Vandermonde's Identity and Binomial Coefficient Sums

What feels right?

Using the symmetry property $^{23} C_{r} =^{23} C_{23 - r}$ , the sum becomes $\sum_{r = 0}^{22}^{22} C_{r}^{23} C_{23 - r}$ , which by Vandermonde's identity equals $^{45} C_{23}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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