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

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

• $n!=n(n-1)(n-2)(n-3)......(3)(2)(1)$
The no of ways is 9.9.8.7.6.5.4.3.2=$9\times 9!$