JEE Main 2024 — 6 April, Shift 2 — Question 7

$\frac{{ }^{n+1} C_{r}}{{ }^{n} C_{r}}=\frac{55}{35}$

$\frac{(\mathrm{n}+1)!}{(\mathrm{r}+1)!(\mathrm{n}-\mathrm{r})}!\frac{\mathrm{r}!(\mathrm{n}-\mathrm{r})!}{\mathrm{n}!}=\frac{11}{7}$

$\frac{(\mathrm{n}+1)}{\mathrm{r}+1}=\frac{11}{7}$

$7 \mathrm{n}=4+11 \mathrm{r}$

$\frac{{ }^{n} C_{r}}{{ }^{n-1} C_{r-1}}=\frac{35}{21}$

$\frac{n!}{r!(n-r)!}=\frac{(r-1)!(n-r)!}{(n-1)!}=\frac{5}{3}$

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

$3 \mathrm{n}=5 \mathrm{r}$

By solving $r=6, \quad n=10$

$2 \mathrm{n}+5 \mathrm{r}=50$