Question 1011202: how many number of 9 digits can be formed which have all digits different?
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website!
Question:
how many numbers of 9 digits can be formed which have all digits different?
Solution:
Numbers do not normally have a leading zero, so a nine-digit number cannot have a zero as the first digit.
Examples: 123456789 is a valid 9 digit number with distinct digits, but 012345678 is not considered a valid 9 digit number (for the purpose of this question).
A nine-digit number with distinct digits will have 9 choices for the first digit (1-9) and 9 choices for the second digit (0-9, less the first digit), then 8 choices for the third, etc.
So the total count (N) of numbers with distinct digits is then
N=9*9*8*7*6*5*4*3*2=9*(9!/1!)=3265920
Note:
If the answer is 910, it may refer to a different question, or a question worded differently.
It might help if the question could be posted exactly as it appears.
For example, the answer to the question "how many ways to choose up to 9 different digits out of the digits 0-9" would be 1022.
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