Answer: there are 27,216 5 digit numbers such that there are no repeated digits at all and the leading digit is not zero.
Explanation:
We can think of a five-digit number as:
a-b-c-d-e
Where each letter is a digit.
We want to find the number of combinations such that the digits can not be repeated, and that a can not be zero.
So let's count the number of options for each digit.
a can be {1, 2, 3, 4, 5, 6, 7, 8, 9}
So we have 9 options here.
for the second digit, b, we also have 9 options, because we took one in the previous selection, but now we also can have the number 0 as an option.
For the third digit, c, we have 8 options, because we already selected two.
For the fourth digit, d, we have 7 options, because we already selected 3.
For the last digit, e, we have 6 options, because we already selected 4.
Then the total number of combinations will be equal to the product between the number of options for each selection, this is:
C = 9*9*8*7*6 = 27,216
So there are 27,216 5 digit numbers such that there are no repeated digits at all and the leading digit is not zero.