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?

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

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