Let 0 ? r ? n . If C r + 1 n + 1 : C r : C r - 1 n - 1 n = 55 : 35 : 21 , then 2 n + 5 r is equal to [2024]

Question

Let $0 \leq r \leq n$ . If ${}^{n + 1}{C_{r + 1}} : {}^{n}{C_{r} : {}^{n - 1}{C_{r - 1}}} = 55 : 35 : 21,$ then $2 n + 5 r$ is equal to [2024]

Smart Achievers · Accepted Answer

(2)

We have, ${}^{n + 1}{C_{r + 1} : {}^{n}{C_{r}}} = 55 : 35$

$\Rightarrow \frac{(n + 1)!}{(r + 1)! (n - r)!} \times \frac{(n - r)! r!}{n!} = \frac{55}{35}$

$\Rightarrow \frac{n + 1}{r + 1} = \frac{11}{7}$

$\Rightarrow 7 n = 11 r + 4$ ...(i)

Also, ${}^{n}{C_{r}} : {}^{n - 1}{C_{r - 1}} = 35 : 21$

$\Rightarrow \frac{n!}{r! (n - r)!} \times \frac{(r - 1)! (n - r)!}{(n - 1)!} = \frac{35}{21}$

$\Rightarrow \frac{n}{r} = \frac{5}{3}$

$\Rightarrow 3 n = 5 r$ ...(ii)

Solving (i) and (ii), we get

$\Rightarrow 7 (\frac{5 r}{3}) = 11 r + 4$

$\Rightarrow 35 r - 33 r = 12$

$\Rightarrow r = 6$ and $n = 10$

So, $2 n + 5 r = 20 + 30 = 50$

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