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?

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 $40{,}471{,}200$.
