Given that, the digits $:1,2,3,4,5,6,7,8,9 $ We know that, the number of ways to pick $k$ unordered elements from an $n$ element set $ = \;^{n}C_{k} = \frac{n!}{k!(n-k)!} $ After selecting the number from the given digits, there is only one way to arrange it. So, the total number of ways $ = \;^{9} C_{2} + \;^{9} C_{3} + \;^{9} C_{4} + \dots + \;^{9} C_{9} \quad \longrightarrow (1)$
We know that, $^{n} C_{0} + \;^{n} C_{1} + \;^{n} C_{2} + \dots + \;^{9} C_{n} = 2^{n}$
Here, $^{9}C_{0} + \;^{9}C_{1} + \;^{9} C_{2} + \;^{9} C_{3} + \;^{9} C_{4} + \dots + \;^{9} C_{9} = 2^{9}$ $\Rightarrow \; ^{9} C_{2} + \;^{9} C_{3} + \;^{9} C_{4} + \dots + \;^{9} C_{9} = 2^{9} – \;^{9}C_{0} - \;^{9}C_{1}$ From the equation $(1),$ we get $\therefore$ The total number of ways $ = 2^{9} – \;^{9}C_{0} – \;^{9}C_{1}$ $\quad = 512 – \frac{9!}{0! \cdot 9!} – \frac{9!}{1! \cdot 8!} $ $\quad = 512 – 1 – \frac{9 \times 8!}{1! \cdot 8!} $ $\quad = 512 – 1 – 9 $ $\quad = 512 – 10 $ $\quad = 502 \; \text{ways}.$ Correct Answer $: 502$ $ \textbf{PS}:\text{Important Properties:}$