Question

The total number of 9 digit numbers which have all different digits is
(a) 10!
(b) 9!
(c) 9 × 9!
(d) 10 × 10!

Answer 1

Out of 10 digit 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 _ _ _ _ _ _ _ _
First place has 9 options since 0 can not be place for number to be 9 digit.
Second place has 9 options since out of 10 are is already choosen.
Similarly third place has 8 options and so on...
∴ Number of 9 digit number with different digit is
$9\times 9\times 8\times 7\times 6\times \frac{5\times 4\times 3}{4\times 4\times 3}2\times 1$
i.e 9 × 9!

Hence, the correct answer is option C.