JEE 2023 — Mathematics PYQ

Q: How do I solve JEE 2023 Mathematics questions?

Practice JEE 2023 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of JEE Mathematics questions?

JEE Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are JEE previous year papers available for free?

Yes. Mathem Solvex provides free access to JEE previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

\sum_{K=0}^{6} {^{51-K}C_{3}} \text{ is equal to}

Choose the correct answer:

Explanation

Solving

1. Expanding the summation

Let the sum be $S$ :

S = \sum_{K=0}^{6} {^{51-K}C_{3}}

S = ^{51}C_3 + ^{50}C_3 + ^{49}C_3 + ^{48}C_3 + ^{47}C_3 + ^{46}C_3 + ^{45}C_3

2. Using the property of combinations

We use the standard identity:

\sum_{r=k}^{n} {^rC_k} = ^{n+1}C_{k+1}

3. Applying the identity to our series

To use this identity, the sum must start from $r = k = 3$ . We can rewrite our sum by considering the sum up to $n=51$ and subtracting the terms that are not present:

S = \left( \sum_{r=3}^{51} {^rC_3} \right) - \left( \sum_{r=3}^{44} {^rC_3} \right)

Applying the identity to both parts:

For the first part ( $n=51, k=3$ ): $\sum_{r=3}^{51} {^rC_3} = ^{51+1}C_{3+1} = ^{52}C_4$
For the second part ( $n=44, k=3$ ): $\sum_{r=3}^{44} {^rC_3} = ^{44+1}C_{3+1} = ^{45}C_4$

4. Final Expression

S = ^{52}C_4 - ^{45}C_4

Correct Option: (3)