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Question

Find the total number of 9 digit numbers which have all different digits.


A

3265920

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B

3267720

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C

9×9!

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D

9×10!

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Solution

The correct options are
A

3265920


C

9×9!


10P99P8=10!1!9!1=9!(101)=9(9!)
=9×(9×8×7×6!)
=81×56×720=3265920.

Alternative method.
The number is to be of 9 digits. The first place can be filled in 9 ways only (as zero cannot be in the first place). Having filled up the first place the remaining 8 places can be filled up by the remaining 9 digits in 9P8=9! ways.
Hence the total is 9×9!.


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