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

$\begin{array}{1 1}(A)\;10!\\(B)\;9!\\(C)\;9.9!\\(D)\;10.10!\end{array}$

Can you answer this question?

Toolbox:
• $n!=n(n-1)(n-2)(n-3)......(3)(2)(1)$
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