JEE Main 2024 (Online) 6th April Evening ShiftQuestion 1
Let $0\le r\le n$. If ${}^{n+1}C_{r+1}:{}^{n}C_r:{}^{n-1}C_{r-1}=55:35:21$, then $2n+5r$ is equal to :
Permutation and Combination
Solution
$\frac{{}^{n+1}C_{r+1}}{{}^nC_r}=\frac{n+1}{r+1}=\frac{55}{35}=\frac{11}{7}$ and $\frac{{}^nC_r}{{}^{n-1}C_{r-1}}=\frac nr=\frac{35}{21}=\frac53$. So $3n=5r$ and $7n+7=11r+11$. Then $n=\frac{5r}{3}$: $\frac{35r}{3}+7=11r+11\Rightarrow 2r/3=4\Rightarrow r=6,n=10$. $2n+5r=20+30=50$. Option A.