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Q.
Total number of eight digit numbers in which all digits are different is
Permutations and Combinations
Solution:
There are ten digits $0,1,2, \ldots \ldots \ldots .9 .$ Permutations of these digits taken eight at a time $={ }^{10} P _{8}$ which includes permutations having 0 at the first. When 0 is fixed at the first place,
then number of such permutations = ${ }^{9} P _{7}$
So, required number $={ }^{10} P _{8}-{ }^{9} P _{7}$
$=\frac{10 !}{2}-\frac{9 !}{2}=\frac{9.9 !}{2}$