Exponential generating function in the case of a singular symbol

Question

Suppose I have the following set {p, q, s, 0, 1, 2, 3} where we have $l_n$ is the number of strings with the symbols in this set of length $n$. However, there is exactly one number in each string that is a digit.

To find the generating function for this I started off by treating the alphabets as a ternary string and the digits 1 unit. Essentially we have our exponential generating function $F(x) = x\sum _{n=0}^{\infty }\:3^nx^n\frac{1}{n!}$. But I feel like my approach so far is wrong (it seems too simple) and this for the case where we only have 1 digit. I don't really know how to factor in the other 3 digits.

Answer 1

There are $3^{n-1}$ strings of length $n-1$ with no digit; every string of length $n$ with exactly one digit is uniquely obtained by choosing one of these $3^{n-1}$ strings, choosing one of the $4$ digits, and inserting it in one of the $n$ possible positions in the chosen string. For $n\ge 1$ this can be done in $4n\cdot 3^{n-1}$ ways, so $\ell_0=0$, and $\ell_n=4n\cdot 3^{n-1}$ for $n\ge 1$. The exponential generating function must therefore be

$$\begin{align*} \sum_{n\ge 0}\ell_n\frac{x^n}{n!}&=\sum_{n\ge 1}\frac{4n}{n!}3^{n-1}x^n\\ &=4x\sum_{n\ge 1}\frac{3^{n-1}x^{n-1}}{(n-1)!}\\ &=4x\sum_{n\ge 0}\frac{3^nx^n}{n!}\,, \end{align*}$$

for which you should fairly easily be able to find a closed form.

Answer 2

Summary: you were only off by a factor of four. You would be correct if you replaced $x$ with $4x$.

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The EGF for choosing the digits is $4x^1$, since there can only be one digit, and there are $4$ choices for it.

The EGF for choosing the letters is $\sum_{n\ge 0} 3^nx^n/n!=e^{3x}$, since there can be any number $n$ of letters, and when there are $n$ letters, there are $3^n$ ways to choose the letters.

Therefore, the EGF for choosing the entire string (digits and numbers) is the product of these two EGF's. $$ (4x)\cdot (e^{3x}) $$ To finish, you then need to extract the coefficient of $x^n$ and multiply by $n!$.