The problem reads:
How many ways are there to distribute $26$ of $34$ distinguishable balls to $5$ people if Lucy gets at most $4$ balls?
The generating function to distribute distinguishable balls to $5$ people is:
$$ e^{4z}(1+z+z^2/2!+z^3/3!+z⁴/4!) $$
So the answer is the $26th$ coefficient of this expression, multiplied by the ways in which one can choose 26 balls among 34, which give me the answer:
$$ \binom{34}{26}e^{4z}(1+z+z^2/2!+z^3/3!+z⁴/4!) $$
The answer of the book is $\binom{34}{12}e^{4z}(1+z+z^2/2!+z^3/3!+z⁴/4!)$, I want to understand what I am doing wrong.
I can tell you that much, your reasoning seams quite correct. However I am a little bit shocked from such a typo in a book, which one?
Because $e^{4 z}(1 + z + z^2/2! + z^3/3! + z^4/4! )$ is the egf of the number of ways to distribute $n$ distinguishable balls for 5 people where one of them gets at most 4.
I can only imagine that it was an unfortunate attempt to simplify ${34 \choose 26}$