if n-1cr:ncr:n+1cr=6:9:13 find n and r - Maths - Linear Inequalities

Question

if
n-1cr:ncr:n+1cr=6:9:13
find n and r

Ankush Jain · Accepted Answer

$Given: {}^{n - 1}{C_{r}} : {}^{n}{C_{r}} : {}^{n + 1}{C_{r}} = 6 : 9 : 13 \Rightarrow {}^{n - 1}{C_{r}} : {}^{n}{C_{r}} = 6 : 9 and {}^{n}{C_{r}} : {}^{n + 1}{C_{r}} = 9 : 13$

$\Rightarrow \frac{\frac{(n - 1)!}{(n - 1 - r)! r!}}{\frac{n!}{(n - r)! r!}} = \frac{6}{9} and \frac{\frac{n!}{(n - r)! r!}}{\frac{(n + 1)!}{(n + 1 - r)! r!}} = \frac{9}{13} \Rightarrow \frac{(n - 1)!}{(n - 1 - r)!} \times \frac{(n - r)!}{n!} = \frac{6}{9} and \frac{n! \times (n + 1 - r)!}{(n - r)! \times (n + 1)!} = \frac{9}{13} \Rightarrow \frac{(n - 1)! \times (n - r) (n - r - 1)!}{(n - r - 1)! \times n (n - 1)!} = \frac{6}{9} and \frac{n! \times (n + 1 - r) (n - r)}{(n - r)! \times (n + 1) n!} = \frac{9}{13} \Rightarrow \frac{n - r}{n} = \frac{6}{9} and \frac{n + 1 - r}{n + 1} = \frac{9}{13}$

$\Rightarrow 9 n - 9 r = 6 n and 13 n + 13 - 13 r = 9 n + 9 \Rightarrow 3 n = 9 r and 4 n - 13 r = - 4 \Rightarrow n = 3 r ........ (1) and 4 (3 r) - 13 r = - 4 {using (1)] \Rightarrow 12 r - 13 r = - 4 \Rightarrow - r = - 4 \Rightarrow r = 4$

On putting r in (1), we get

$n = 3 r = 3 \times 4 = 12$