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

Question:

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

(a) 10!

(b) 9!

(c) 9 × 9!

(d) 10 × 10!

Solution:

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.

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