Problem source

How many integers between $10\,000$ and $100\,000$ (inclusive) contain exactly two different digits? ($23\,332$ contains exactly two different digits but neither of $33\,333$ and $12\,331$ does.)

Solution source

First consider $5$ digit numbers containing at most $2$ non-zero digits. Then there are $\binom{9}{2}$ ways to choose the two digits, and $2^{5}-2$ different ways to arrange them, removing the ones which are all the same.

Considering all the pairs including zero, there are $9$ ways to choose the non-zero (first) digit. There are $2^4-1$ remaining digits where not all the numbers are the same.

Finally we must not forget $100\,000$.

Therefore there are $\binom{9}{2}(2^5-2) +9\cdot(2^4-1) + 1 = 1216$