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Binomial Theorem: JEE Main Mathematics Question with Solution

The value of

\Sigma_{r=0}^{22}{ }^{22} \mathrm{C}_r{ }^{23} \mathrm{C}_r

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{ }^{44} \mathrm{C}_{23}

{ }^{45} \mathrm{C}_{23}

{ }^{44} \mathrm{C}_{22}

{ }^{45} \mathrm{C}_{24}

Step-by-step solutionView

Correct answer

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

Option analysis

Why each option works or fails

A ·

{ }^{44} \mathrm{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 $\sum { }^n\mathrm{C}_r { }^m\mathrm{C}_{k-r} = { }^{n+m}\mathrm{C}_k$ ; here $n+m = 22 + 23 = 45$ , not $44$ .

B ·

{ }^{45} \mathrm{C}_{23}

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

C ·

{ }^{44} \mathrm{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} \mathrm{C}_{24}

An off-by-one index error when rewriting ${ }^{23}\mathrm{C}_r$ in terms of its complement, yielding lower index $24$ instead of $23$ . Since ${ }^{23}\mathrm{C}_r = { }^{23}\mathrm{C}_{23-r}$ , 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 $S = \sum_{r=0}^{22} {}^{22}\mathrm{C}_r \cdot {}^{23}\mathrm{C}_r$ .

02goal

Evaluate the sum $S$ in terms of a single binomial coefficient ${}^n\mathrm{C}_k$ .

03approach

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

04execute

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

✓verify

Using ${}^n\mathrm{C}_r = {}^n\mathrm{C}_{n-r}$ , ${}^{45}\mathrm{C}_{23} = {}^{45}\mathrm{C}_{22}$ . For a small scale check, with upper limits 1 and 2: $\sum_{r=0}^1 {}^1\mathrm{C}_r {}^2\mathrm{C}_r = {}^1\mathrm{C}_0{}^2\mathrm{C}_0 + {}^1\mathrm{C}_1{}^2\mathrm{C}_1 = 1\cdot 1 + 1\cdot 2 = 3 = {}^3\mathrm{C}_2$ , which confirms formula ${}^{n+m}\mathrm{C}_m$ .

Hints that build this answer step by step

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

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why do we replace

{}^{23}\mathrm{C}_r

with

{}^{23}\mathrm{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 ${ }^{23}\mathrm{C}_r = { }^{23}\mathrm{C}_{23-r}$ , the sum becomes $\sum_{r=0}^{22} { }^{22}\mathrm{C}_r { }^{23}\mathrm{C}_{23-r}$ , which by Vandermonde's identity equals ${ }^{45}\mathrm{C}_{23}$ .

Why each option works or fails

A: ${ }^{44} \mathrm{C}_{23}$ - 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 $\sum { }^n\mathrm{C}_r { }^m\mathrm{C}_{k-r} = { }^{n+m}\mathrm{C}_k$ ; here $n+m = 22 + 23 = 45$ , not $44$ .
B · correct: ${ }^{45} \mathrm{C}_{23}$ - Correct option: Rewrite ${ }^{23}\mathrm{C}_r$ as ${ }^{23}\mathrm{C}_{23-r}$ , so the sum corresponds to the coefficient of $x^{23}$ in $(1+x)^{22}(1+x)^{23} = (1+x)^{45}$ , giving ${ }^{45}\mathrm{C}_{23}$ . This is the correct result.
C: ${ }^{44} \mathrm{C}_{22}$ - 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: ${ }^{45} \mathrm{C}_{24}$ - An off-by-one index error when rewriting ${ }^{23}\mathrm{C}_r$ in terms of its complement, yielding lower index $24$ instead of $23$ . Since ${ }^{23}\mathrm{C}_r = { }^{23}\mathrm{C}_{23-r}$ , 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 $S = \sum_{r=0}^{22} {}^{22}\mathrm{C}_r \cdot {}^{23}\mathrm{C}_r$ .
goal: Evaluate the sum $S$ in terms of a single binomial coefficient ${}^n\mathrm{C}_k$ .
approach: Use the symmetry property of binomial coefficients ${}^{n}\mathrm{C}_r = {}^{n}\mathrm{C}_{n-r}$ on ${}^{23}\mathrm{C}_r$ to rewrite the sum with constant indices, then apply Vandermonde's convolution identity $\sum_{r=0}^n {}^n\mathrm{C}_r {}^m\mathrm{C}_{k-r} = {}^{n+m}\mathrm{C}_k$ .
execute: Rewrite ${}^{23}\mathrm{C}_r$ as ${}^{23}\mathrm{C}_{23-r}$ . Then the sum becomes: $S = \sum_{r=0}^{22} {}^{22}\mathrm{C}_r \cdot {}^{23}\mathrm{C}_{23-r}$ This represents the coefficient of $x^{23}$ in the expansion of $(1+x)^{22}(1+x)^{23} = (1+x)^{45}$ . Therefore, $S = {}^{45}\mathrm{C}_{23}$ .
verify: Using ${}^n\mathrm{C}_r = {}^n\mathrm{C}_{n-r}$ , ${}^{45}\mathrm{C}_{23} = {}^{45}\mathrm{C}_{22}$ . For a small scale check, with upper limits 1 and 2: $\sum_{r=0}^1 {}^1\mathrm{C}_r {}^2\mathrm{C}_r = {}^1\mathrm{C}_0{}^2\mathrm{C}_0 + {}^1\mathrm{C}_1{}^2\mathrm{C}_1 = 1\cdot 1 + 1\cdot 2 = 3 = {}^3\mathrm{C}_2$ , which confirms formula ${}^{n+m}\mathrm{C}_m$ .

Binomial Theorem: JEE Main Mathematics Question with Solution

What feels right?

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

Why each option works or fails

Solution

01given

02goal

03approach

04execute

✓verify

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