I'm pretty sure the answer is 32!/(8!)4 which is 99 561 092 450 391 000.
Three numbers can be arranged in 27 different sequences if repetition is allowed, and in 6 different sequences if it's not.
9,999 if zeroes can be used than it would be 10,999
24
The four digits can be used to produce infinitely many different numbers if repetition is permitted. Without repetition, there are 24 possible numbers. A lot more can be produced if the numbers are combined using binary oprations, fore example, 19 * 8/4 = 19*2 = 38.
There are 10 to the 10th power possibilities of ISBN numbers if d represents a digit from 0 to 9 and repetition of digits are allowed. That means there are 10,000,000,000 ISBN numbers possible.
9999 - 102 = 9897
4
Three numbers can be arranged in 27 different sequences if repetition is allowed, and in 6 different sequences if it's not.
9999-1023 = 8976
8! = 40320 ways.
30 without repetition (6P2) 66 with repetition (12C2)
9,999 if zeroes can be used than it would be 10,999
24
The four digits can be used to produce infinitely many different numbers if repetition is permitted. Without repetition, there are 24 possible numbers. A lot more can be produced if the numbers are combined using binary oprations, fore example, 19 * 8/4 = 19*2 = 38.
There are 10 to the 10th power possibilities of ISBN numbers if d represents a digit from 0 to 9 and repetition of digits are allowed. That means there are 10,000,000,000 ISBN numbers possible.
9888
When the repetition is allowed: 4*4*4 = 64 numbers. Without repetition: 4*3*2 = 24 numbers.