Question: A fleet of nine taxis is to be dispatched to three airports in such a way that three go to airport A, five go to airport B, and one goes to airport C. In how many distinct ways can this be accomplished?
What is the difference between solving this with multinomial-coefficients $\frac{9!}{3!5!1!} = 504$ and permutation $\frac{9!}{(9-3)!} = 504$
If in the end we get the same answer of how many distinct ways the taxis can be dispatched to three airports.
Why does how many taxis go to airport A, B, or C matter in calculating the answer?
Your solution with multinomial coefficients is correct. Your solution with permutations looks like a fortuitous coincidence. Try distributing three taxis each to three different airports. Do you still get the same numbers?
We must choose which three taxis are sent to airport $A$, which can be done in $\binom{9}{3}$ ways; which five of the remaining six taxis are sent to airport $B$, which can be done in $\binom{6}{5}$ ways; and then send the remaining taxi to airport $C$, which can be done in one way. Hence, the number of ways of distributing the taxis is $$\binom{9}{3}\binom{6}{5} = \frac{9!}{3!6!} \cdot \frac{6!}{5!1!} = \frac{9!}{3!5!1!} = \binom{9}{3, 5, 1}$$ One purpose of the multinomial coefficient $$\binom{n}{n_1, n_2, \ldots, n_k}$$ is to count the number of ways of distributing $n = n_1 + n_2 + \cdots + n_k$ objects so that $n_i$ objects are placed in box $i$ for $1 \leq i \leq k$, which is what we are doing here. In this case, the multinomial coefficient $\binom{9}{3, 5, 1}$ counts the number of ways of sending three taxis to airport $A$, $5$ taxis to airport $B$, and $1$ taxi to airport $C$.
What is your rationale for using permutations? The formula $P(n, k)$ is the number of ways we can select $k$ objects from a set of $n$ objects when order matters. However, in this problem, we do not care about the order of selection, just which taxis are sent to which airport. Also, we are selecting all nine objects, not just three of them.