The domain of the function $\mathrm{f}\left(\mathrm{x}\right){=}^{24-\mathrm{x}}{\mathrm{C}}_{3\mathrm{x}-1}{+}^{40-6\mathrm{x}}{\mathrm{C}}_{8\mathrm{x}-10} \mathrm{is},$

${ }^{24-\mathrm{x}}{\mathrm{C}}_{3\mathrm{x}-1}$ is defined if,
$$\begin{gathered} 24-\mathrm{x}>0,3\mathrm{x}-1\ge 0\text{ and }24-\mathrm{x}\ge 3\mathrm{x}-1 \\ \Rightarrow \mathrm{x}<24,\mathrm{x}\ge \frac{1}{3}\text{ and }\mathrm{x}\le \frac{25}{4} \\ \Rightarrow \frac{1}{3}\le \mathrm{x}\le \frac{25}{4} .........\left(1\right) \end{gathered}$$
${ }^{40-6\mathrm{x}}{\mathrm{C}}_{8\mathrm{x}-10}$ is defined if
$$\begin{gathered} 40-6\mathrm{x}>0,8\mathrm{x}-10\ge 0\text{ and }40-6\mathrm{x}\ge 8\mathrm{x}-10 \\ \begin{array}{l}\Rightarrow \mathrm{x}<\frac{20}{3},\mathrm{x}\ge \frac{5}{4}\text{ and }\mathrm{x}\le \frac{25}{7}\\ \Rightarrow \frac{5}{4}\le \mathrm{x}\le \frac{25}{7} .............\left(2\right)\end{array} \end{gathered}$$
From Eq. (1) and (2), we get
$\frac{5}{4}\le \mathrm{x}\le \frac{25}{7}$
$\mathrm{But}, 24-\mathrm{x}\in \mathrm{N}$
x must be an integer,
x = 2, 3.
Hence, domain (f) = {2, 3}.