If $\sum _{\mathrm{n}=3}^{\infty }\frac{{}^{\mathrm{n}}\mathrm{C}_3}{\mathrm{n}!}=\mathrm{a}·{\mathrm{e}}^{\mathrm{b}}+\mathrm{c}$ then descending order of $\mathrm{a}, \mathrm{b}, \mathrm{c}$ is

$\begin{array}{rcl}\sum _{\mathrm{n}=3}^{\infty } \frac{{}^{\mathrm{n}}\mathrm{C}_3}{\overline{)\underline{\mathrm{n}}}}& =& \sum _{\mathrm{n}=3}^{\infty } \frac{\mathrm{n}(\mathrm{n}-10 (\mathrm{n}-2)}{\overline{)\underline{3} } \overline{)\underline{\mathrm{n}}}}\\ & & \sum _{\mathrm{n}=3}^{\infty } \frac{1}{6 \overline{)\underline{\mathrm{n}-3}}}\\ & =& \frac{1}{6}\left(\mathrm{e}\right)\\ \mathrm{a}& =& \frac{1}{6}, \mathrm{b}=1, \mathrm{c}=0\end{array}$

descending order =b, a, c