How can I count the numbers of $5$ digits such that at least one of the digits appears more than one time?
My thoughts are:
I count all the possible numbers of $5$ digits: $10^5 = 100000$. Then, I subtract the numbers that don't have repeated digits, which I calculate this way: $10*9*8*7*6$ $= 30240 $. Thus, I have $100000 - 30240 = 69760 $ numbers that have at least one digit repeated more than one time.
Is this correct?
It is helpful to look at the negation of your requirement.
No digit is ever repeated for $10 \cdot 9 \cdot 8 \cdot 7 \cdot 6=30240$ numbers.
The total number of $5$ digit numbers is $10^5=100000$.
So the number of $5$ digits numbers with at least $1$ digit repeating more than once is: $$10^5 - 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 69760$$ So, yes you are correct. NOW I noticed I have just repeated your correct arguments. Down vote for it to disappear! :-)