If N(n) = n prod_{r=1}^{2023} (n^2 - r^2) for | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $N(n) = n \prod_{r=1}^{2023} (n^2 - r^2) = n \cdot \left[ \prod_{r=1}^{2023} (n-r) \right] \left[ \prod_{r=1}^{2023} (n+r) \right]$.For $n =

Vedclass · Accepted Answer

$(4047)!$

Vedclass · Answer

(A) Given $N(n) = n \prod_{r=1}^{2023} (n^2 - r^2) = n \cdot \left[ \prod_{r=1}^{2023} (n-r) \right] \left[ \prod_{r=1}^{2023} (n+r) \right]$.For $n = 2024$,we have:$N(2024) = 2024 \cdot [(2023)(2022) \dots (1)] \cdot [(2025)(2026) \dots (4047)]$.Rearranging the terms,we get:$N(2024) = (4047)(4046) \dots (2025)(2024)(2023) \dots (1) = (4047)!$.We need to find ${}^{N}C_{N-1}$ where $N = N(2024) = (4047)!$.Using the property ${}^{n}C_{r} = {}^{n}C_{n-r}$,we have ${}^{N}C_{N-1} = {}^{N}C_{1} = N$.Therefore,${}^{N}C_{N-1} = (4047)!$.

If $N(n) = n \prod_{r=1}^{2023} (n^2 - r^2)$ for $n > 2023$,then find the value of ${}^{N}C_{N-1}$ when $n = 2024$.

Similar Questions

$\binom{15}{8} + \binom{15}{9} - \binom{15}{6} - \binom{15}{7} = \dots$

If ${}^nC_{r-1}=330$,${}^nC_r=462$,and ${}^nC_{r+1}=462$,then $r$ is equal to

If $\binom{a^2+a}{3} = \binom{a^2+a}{9}$,then $a = \dots$

If $^{n}C_{9} = ^{n}C_{8}$,find $^{n}C_{17}$.

The number of ways of choosing $10$ objects out of $31$ objects,of which $10$ are identical and the remaining $21$ are distinct,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module