How many $14$-digit even numbers can be formed using $0,1,1,2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5$? What is a good way to approach this question?

How many $14$-digit even numbers can be formed using $0,1,1,2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5$?

I know that I can list the possible outcomes with different ending numbers and then add up all the outcomes. However, is there a way to solve the question without categorizing adding different types together?
 A: The answer can be written as
$${6\cdot13!-5\cdot12!\over1!2!2!3!3!3!}$$
which computes to $40{,}471{,}200$, as in EuxhenH's answer. The logic here is to start by making the $14$ digits all distinct by temporarily adding subscripts (e.g., the three $5$'s become $5_1$, $5_2$, and $5_3$). The $6\cdot13!$ counts the number of permutations that end with one of the $6$ even digits ($0$, $2_1$, $2_2$, $4_1$, $4_2$, and $4_3$), but this includes permutations that start with the $0$, so we subtract the $5\cdot12!$ permutations that start with the $0$ and end with one of the $5$ other even digits. Finally, we divide by the rearrangements of the copies of the digits, so as to remove the subscripts that we attached to make them distinct.
A: Case 1: The number ends in $0$. Considering all possible orderings of the other $13$ numbers we get $$\frac{13!}{2!2!3!3!3!}$$ possible outcomes (since we have two of $1$ and $2$ each and three of $3,4,5$ each, we divide by $2!2!3!3!3!$) to remove duplicates.
Case 2: The number does not end in $0$. We have $5$ possible choices for the last digit, which leaves us with $12$ possible choices for the leading digit, $12$ possible choices for the second digit (include zero), $11$ choices for the third and so on. Basically,
$$\frac{5\cdot 12\cdot 12!}{2!2!3!3!3!}.$$
Again, we divide to avoid duplicates.
Hence, we get a total of $$\frac{13! + 5\cdot 12\cdot 12!}{2!2!3!3!3!} = 40471200.$$
A: There are three even numbers, so they make the three cases.
Case 1. Last digit is $0$:
$$\frac{13!}{(2!)^2(3!)^3}$$
Case 2. Last digit is $2$:
$$\frac{13!-12!}{2!(3!)^3}$$
Case 3. Last digit is 4:
$$\frac{13!-12!}{(2!)^3(3!)^2}$$
Hence, adding the three results in $40{,}471{,}200$.
