If the coefficients of three consecutive terms in the expansion of ${(1 + x)}^n$ are 45, 120 and 210 then the value of n is

$$\begin{gathered} {}^{\mathrm{n}}{\mathrm{c}}_{\mathrm{r}-1},=45, {}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}=120, {}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}+1}=210 \\ \frac{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}-1}}=\frac{120}{45} \Rightarrow \frac{\mathrm{n}-\mathrm{r}+1}{\mathrm{r}}=\frac{8}{3} \\ 3\mathrm{m}-3\mathrm{r}+3=8\mathrm{r} \\ 3\mathrm{n}-11\mathrm{r}+3=0 -\left(1\right) \\ \frac{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}+1}}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}}=\frac{210}{120} \Rightarrow \frac{\mathrm{n}-\mathrm{r}}{\mathrm{r}+1}=\frac{7}{4} \\ 4\mathrm{n}-4\mathrm{r}=7\mathrm{r}+7 \\ 4\mathrm{n}-11\mathrm{r}-7=0 -\left(2\right) \end{gathered}$$

Solving (1) & (2) we get $\mathrm{n}=10$