$\begin{array}{1 1}(A)\;10!\\(B)\;9!\\(C)\;9\times 9!\\(D)\;10\times 10!\end{array} $

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- $n!=n(n-1)(n-2)(n-3)......(3)(2)(1)$

The number is to be of 9 digit,the first place can be filled in 9 ways only as (0 cannot be in the left)

Having filled up the first place the remaining 8 place can be filled by the remaining 9 digits in.

The no of ways is 9.9.8.7.6.5.4.3.2=$9\times 9!$

Hence (C) is the correct answer.

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