The number of nine digit numbers that can be formed with different digits is

9. 8!

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8.9!

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9. 9!

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10!

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Solution

The correct option is A $9. 9!$
The first digit is one of $(1, 2, 3, 4, 5, 6, 7, 8, 9)$ (i.e., all expect $0$ ).
So, the no. of ways $= 9$
The remaining $8$ digits are to be taken from the remaining $9$ numbers.
So, the no. of ways $=^{9} P_{8}$
So, total no. of ways $= 9 \times^{9} P_{8}$
$= 9 \times 9!$

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The number of nine digit numbers that can be formed with different digits is

The correct option is A 9.9!The first digit is one of (1,2,3,4,5,6,7,8,9) (i.e., all expect 0).So, the no. of ways =9The remaining 8 digits are to be taken from the remaining 9 numbers.So, the no. of ways =9P8So, total no. of ways =9×9P8=9×9!