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Q. The total number of 9-digits numbers which have all different digits is

Permutations and Combinations

Solution:

The total number of 9-digits numbers, having all digit are different $=9{ }^9 P_8$
$=\frac{9 \times 9 !}{1 !}=9 \times 9 !$
$(\because$ In first place, we select anyone of nine numbers except zero. In rest of the eight places, we select any eight numbers from remaining 9 numbers)