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

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(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 =

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$(4047)!$

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(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 $n = 2024$.

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