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

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Text solutionVerified

^{10} P_{9} -^{9} P_{8} = \frac{10 !}{1 !} - \frac{9 !}{1 !} = 9! (10 - 1) = 9 (9!)

= 9 \times (9 \times 8 \times 7 \times 6!)

= 81 \times 56 \times 720 = 3265920

.
Alternative : The number is to be of 9 digits. The first place can be filled in 9 ways only (as zero can not 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

^{9} P_{8} = 9!

ways.
Hence the total is

9 \times 9!

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Question Text	Find the total number of 9 digit numbers which have all different digits.
Answer Type	Text solution:1
Upvotes	150

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

Text solutionVerified

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