Let $0 \leq \mathrm{r} \leq \mathrm{n}$. If ${ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}+1}:{ }^n \mathrm{C}_{\mathrm{r}}:{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}=55: 35: 21$, then $2 \mathrm{n}+5 \mathrm{r}$ is equal to:

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Aicharya · Accepted Answer

$\begin{aligned} & \frac{{ }^{n+1} C_r}{{ }^n C_r}=\frac{55}{35} \ & \frac{(n+1) !}{(r+1) !(n-r)} ! \frac{r !(n-r) !}{n !}=\frac{11}{7} \ & \frac{(n+1)}{r+1}=\frac{11}{7}\end{aligned}$ $\begin{aligned} & 7 \mathrm{n}=4+11 \mathrm{r} \ & \frac{{ }^n C_r}{{ }^{n-1} C_{r-1}}=\frac{35}{21} \ & \frac{n !}{r !(n-r) !}=\frac{(r-1) !(n-r) !}{(n-1) !}=\frac{5}{3} \ & \frac{n}{r}=\frac{5}{3} \ & 3 n=5 r \ & 	ext { By solving } r=6 \quad n=10 \ & 2 n+5 r=50\end{aligned}$

Let $0 \leq \mathrm{r} \leq \mathrm{n}$. If ${ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}+1}:{ }^n…

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