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Let $0 \leq \mathrm{r} \leq \mathrm{n}$. If ${ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}+1}:{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}:{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}=55: 35: 21$, then $2 n+5 r$ is equal to:
$60$
$62$
$50$
$55$
Solution
$ \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}$
$ 7 \mathrm{n}=4+11 \mathrm{r} $
$ \frac{{ }^n C_r}{{ }^{n-1} C_{r-1}}=\frac{35}{21} $
$ \frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}=\frac{(\mathrm{r}-1) !(\mathrm{n}-\mathrm{r}) !}{(\mathrm{n}-1) !}=\frac{5}{3} $
$ \frac{\mathrm{n}}{\mathrm{r}}=\frac{5}{3} $
$ 3 \mathrm{n}=5 \mathrm{r} $
By solving $ \mathrm{r}=6 \quad \mathrm{n}=10 $
$ 2 \mathrm{n}+5 \mathrm{r}=50$
