The number of values of $n\in N$ for which ${ }^{(n+2)}{C}_2:{}^{(n+3)}{C}_1=4:2$

$\frac{(\mathrm{n}+2){\mathrm{C}}_2}{(\mathrm{n}+3){\mathrm{C}}_1}=\frac{4}{2}$

$2\left(\frac{(\mathrm{n}+2) (\mathrm{n}+1)}{1\times 2}\right)=4(\mathrm{n}+3)$

$$\begin{gathered} {\mathrm{n}}^2+3\mathrm{n}+2=4\mathrm{n}+12 \\ \Rightarrow {\mathrm{n}}^2-\mathrm{n}-10=0 \end{gathered}$$

N cant take integer value. Number of values of n=0