1. Tardigrade
  2. Question
  3. Mathematics
  4. If 2n+3C2n - 2n+2C2n-1 = 15 (2n+1), then n=

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If $ ^{2n+3}C_{2n} - \,{}^{2n+2}C_{2n-1} = 15 \left(2n+1\right)$, then $n= $

Permutations and Combinations

Solution:

$\frac{\lfloor2n+3 }{\lfloor2n \lfloor3} -\frac{ \lfloor2n+2}{\lfloor2n-1\lfloor3} = 15\left(2n+1\right) $
$ \Rightarrow \frac{\left(2n+2\right)\left(2n+1\right)}{2} = 15\left(2n+1\right) $
$ \Rightarrow n+1=15$
$\Rightarrow n=14.$