Number of all ordered lists using characters of a word

Here is a solution using exponential generating functions. This approach is most convenient if you have access to a computer algebra system. I used Mathematica, but I suppose SageMath or Wolfram Alpha should work as well. Readers not familiar with generating functions may find many resources in the answers to this question: How can I learn about generating functions?

The exponential generating function for the number of $n$-letter words taken from MISSISIPPI is $$f(x) = (1+x) \left( 1+x+\frac{1}{2!} x^2 \right) \left( 1+x+\frac{1}{2!} x^2 + \frac{1}{3!} x^3 \right) \left( 1+x+\frac{1}{2!} x^2 + \frac{1}{3!} x^3 + \frac{1}{4!} x^4 \right)$$ (Note: the usual spelling is MISSISSIPPI, with four S's. I have stuck with the spelling in the OP for the sake of consistency.) The number of $n$-letter words is the coefficient of $(1/n!) x^n$ when $f(x)$ is expanded: $$f(x) = 1 + 4x + \frac{15}{2!} x^2 + \frac{53}{3!}x^3 + \frac{175}{4!}x^4 + \\\frac{535}{5!}x^5 +\ \frac{1490}{6!}x^6 + \frac{3675}{7!}x^7 + \frac{7700}{8!}x^8 + \\ \frac{12600}{9!}x^9 + \frac{12600}{10!}x^{10}$$ so, for example, the number of $5$-character words is $535$. If we add all these counts together, $1+4+15+ \dots + 12600$, the sum is $38848$. Note that this total includes one zero-length word.

However, there is a shortcut, provided our algebra system supports integration. In general, if $$f(x) = \sum_{n=0}^{\infty} \frac{a_n}{n!} x^n$$ then since $$\int_0^{\infty} e^{-x} x^n \; dx = n!$$ we have $$\sum_{n=0}^{\infty} a_n = \int_0^{\infty} e^{-x} f(x) \; dx$$ Using Mathematica to evaluate $$\int_0^{\infty} e^{-x} f(x) \; dx$$ we find the result is $38848$, agreeing with our previous result.