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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)\;9.9!\\(D)\;10.10!\end{array} $

1 Answer

  • $n!=n(n-1)(n-2)(n-3)......(3)(2)(1)$
No of ways of filling first place =9
No of ways of filling remaining 8 place =$9\times 8\times 7\times 6\times 5\times 4\times 3\times 2$
$\Rightarrow 9!$
Total no of numbers =$9\times 9!$
Hence (C) is the correct answer.
answered May 19, 2014 by sreemathi.v

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