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How many numbers with two or more digits can be formed with the digits $1,2,3,4,5,6,7,8,9$, so that in every such number, each digit is used at most once and the digits appear in the ascending order?
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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:}$

  • $n! = n(n-1)(n-2) \dots 1 = n(n-1)!$
  • $0! = 1 $
  • $1! = 1 $
  • $ ^{n}C_{n} = \frac{n!}{n! \; 0!} = 1 $
  • $ ^{n}C_{0} = \frac{n!}{0! \; n!} $
  • $ ^{n}C_{1} = \frac{n!}{1!(n-1)!} = \frac{n(n-1)!}{1!(n-1)!} = n $
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