Count Me

If you have access to a computer algebra system or Wolfram Alpha you can work this out as the coefficient of $x^{25}$ in the expansion of $(\frac{x^{10} - 1}{x-1})^4(\frac{x^{10}-1}{x-1} - 1)$

This is because we have five digits. The first digit must be an integer from 1 to 9 and the last 4 digits must be integers from 0 to 9. We can therefore represent the first digit using the generating function $x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9$ which simplifies to $\frac{x^{10}-1}{x-1} - 1$ .

Each of the other four digits can be represented as choosing a term from the function $1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9$ which can be simplified to $\frac{x^{10}-1}{x-1}$ .

Therefore we get that the number of ways to make a digit sum of 25 is equal to the coefficient of $x^{25}$ in the expansion of $(\frac{x^{10} - 1}{x-1})^4(\frac{x^{10}-1}{x-1} - 1)$ .