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

by Kevin   Last Updated September 11, 2019 10:20 AM

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?

Tags :

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 $$\frac{13! + 5\cdot 12\cdot 12!}{2!2!3!3!3!} = 40471200.$$