Hi I've been working on this problem, and don't know how to solve it.
Here it is:
How many positive integers less than 10,000 have at most two different digits.
I think this problem has something to do with the inclusion-exclusion principle, but I'm not entirely sure. Thanks!
Best Answer
For simplicity I ignore the digit $0$. You will have to figure that out yourself. But here is the answer to the question "how many positive integer less than $10000$ have at most two different digits and no $0$ in them":
There are $2^4 = 16$ four-digit numbers containing only the digits $1$ and $2$. The same for all other of the $\binom{9}{2}=\frac{9\cdot8}{2} = 36$ combinations of $2$ of the $9$ possible digits. If you add all those together, you will have counted the number "1111" and "2222" and so on too often. In fact you counted them each $8$ times, so you have to substract 7 again. This is called the inclusion-exclusion-principle.
The answer therefore is $36\cdot16 - 9\cdot 7$.