How many numbers between 4000 and 9999 have sum of digits equal to ten (why is exponential generating functions not correct)? How many numbers in between 4000 and 9999 have sum of digits equal to ten?
My attempt
My thinking was to use exponential generating functions because 4510 is not equal to 5401 so the order should matter, which would imply (from what I have learned) that exponential generating functions is the correct one.
So I set it up as
$(\frac{x^4}{4!}+\frac{x^5}{5!} \dots \frac{x^9}{9!})*(\frac{1}{0!} + \frac{x}{1!} \dots \frac{x^9}{9!})^3$
But based on How many numbers between $100$ and $900$ have sum of their digits equal to $15$? and my professor as well, one should use ordinary generating functions, i.e.
$(x^4 + x^5 \dots x^9)*(1 + x \dots x^9)^3$
Why is it not correct to use exponential generating functions for this problem, and why is it correct to use ordinary generating functions? In this case, what would be question that the exponential generating function answers vs. what would be the question that the ordinary generating function answers? I question in a similar vein is also Kind of basic combinatorical problems and (exponential) generating functions
 A: Multiplying $n!$ times the coefficient of $x^n$ in
$$ \left(\frac{x^4}{4!} + \frac{x^5}{5!} + \cdots + \frac{x^9}{9!}\right) \left(\frac{1}{0!} + \frac{x}{1!} + \cdots + \frac{x^9}{9!}\right)^3 $$
gives the number of ways to put $n$ labeled balls in $4$ labeled bins, so that the first bin has at least $4$ balls and no bin has more than $9$ balls.
The coefficient of $x^n$ in
$$ (x^4 + x^5 + \cdots x^9)(1 + x + \cdots + x^9)^3 $$
gives the number of ways to put $n$ identical balls in $4$ labeled bins, so that the first bin has at least $4$ balls and no bin has more than $9$ balls.
So which is relevant to this problem? If we imagine putting balls into bins to represent 4-digit numbers, we'll find it makes sense to assign digits based on the count of balls in specific bins left-to-right, not caring about any color or numbers on the balls themselves, so the ordinary generating function helps.
The difference between the two balls-and-bins problems is whether the order or identity of the "balls" matters, but the fact that 4501 and 5401 are different outcomes relates to the order/identity of the "bins". In general the "balls" are the things being counted/added, and the "bins" are the categories or outcomes for groups of balls. For the four-digit number, the "balls" are sort of the imagined $1$ values repeated in each digit value, such as $3 = 1+1+1$.
