Find the number of positive integers that satisfy both the following conditions:

Question

* Each digit is a $1$ or a $2$ or a $3$
* The sum of the digits is $12$

Answer 1

To solve this problem, we will use casework. We start with the number of 1's. Suppose there are $n$ 1's in the number. Then there must be $12-n$ 2's and no 3's. Since each digit is either 1, 2, or 3, there are $n$ digits in total. Thus, we want to find the number of positive integers where the sum of the digits is 12 and only includes the digits 1 and 2. If there are $n$ 1's, then there must be $12-n$ 2's. Since each digit is 1 or 2, we want to find the number of non-negative solutions to $n+(12-n)=12$. Simplifying, we get $2n=12$, so $n=6$. Therefore, there is $\boxed{1}$ possibility where there are 6 1's and 6 2's.

Note: We do not need to consider cases with 3's because a digit of 3 would make the sum of the digits greater than 12.