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The total number of 9 digit numbers which have all different digits is

$\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$9\times 9!$
Hence (C) is the correct answer.
answered May 16, 2014 by sreemathi.v

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