Combinatorial number I have some questions about numbers and I think about The "Inclusion-Exclusion Principle" too... but I am not sure.
I. In how many ways can I choose a five digit number from the number 0 - 9 , and be sure that one digit is performing twice at the number?
II. How many numbers do we have between 3000 to 8000 such that all of their digits are different?
III. Is there any connection between those questions to The "Inclusion-Exclusion Principle"?
Any help will be appreciated!
 A: For number 1, I have a few questions:


*

*are leading $0$s allowed in your five-digit number?

*do you mean that the digit has to appear at least twice, or exactly twice?

*can more than one digit appear twice?


If one or more digits have to appear at least twice, then the problem is straightforward without inclusion-exclusion.  (Simply subtract the number of numbers that use each digit exactly once from the total number of five-digit numbers.)
If exactly one digit has to appear exactly twice, the question is also straightforward without inclusion-exclusion.  (Choose which digit is to appear twice and which digits are to appear once, and permute in all possible ways.)
A: So The final answer for question one is:
The number of ways to set the number: $$10^5$$
The first number must be non zero so we have 9 options to choose,in total: $$9*10^4$$
What is next....
How can I be sure that I have 2 same numbers?
final answer for question two is:
We have five ways to choose the first digit in the number (3-7),for the second number we have nine options because we can choose all the digits exapt the digit that was picked for the first place,and so on until the fourth digit, so in total we have: $$5*9*8*7 =  2520$$
