Find the number of positive integers that satisfy both the following conditions:
* Each digit is a $1$ or a $2$ or a $3$
* The sum of the digits is $12$
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.
To find the number of positive integers that satisfy both conditions, we will use a method called generating functions.
Let's represent each digit as a term in a polynomial.
The polynomial $(x+x^2+x^3)$ represents the possible choices for each digit, where $x^1$ represents $1$, $x^2$ represents $2$, and $x^3$ represents $3$.
To represent the sum of the digits as $12$, we can multiply the polynomial $(x+x^2+x^3)$ by itself 12 times, since the sum of these terms will give us all combinations of digits adding up to $12$.
Therefore, the generating function for the sum of digits being $12$ is:
\[(x+x^2+x^3)^{12}\]
Now, we need to find the coefficient of $x^{12}$ in this expanded expression. The coefficient will give us the number of positive integers that satisfy both conditions.
To find the coefficient, we can use the binomial theorem or a combinatorial approach.
Using the binomial theorem, we can expand the generating function as:
\[(x+x^2+x^3)^{12} = \binom{12}{0}x^{0}(x^1+x^2+x^3)^{12}+\binom{12}{1}x^{1}(x^1+x^2+x^3)^{11}+\binom{12}{2}x^{2}(x^1+x^2+x^3)^{10}+...+\binom{12}{12}x^{12}(x^1+x^2+x^3)^{0}\]
Considering only the terms with $x^{12}$:
\[\binom{12}{0}x^{0}(x^1+x^2+x^3)^{12} = (1)(1)^{12} = 1\]
\[\binom{12}{1}x^{1}(x^1+x^2+x^3)^{11} = (12)(1)(1)^{11} = 12\]
\[\binom{12}{2}x^{2}(x^1+x^2+x^3)^{10} = (66)(1)(1)^{10} = 66\]
Therefore, the coefficient of $x^{12}$ in the expanded expression is $1+12+66=79$.
Hence, there are $79$ positive integers that satisfy both conditions.
To find the number of positive integers that satisfy both conditions, we can use the concept of generating functions.
In this case, we want to find the number of ways we can represent the number $12$ as a sum of $1$s, $2$s, and $3$s.
Let's define three variables:
$x$ represents the number of $1$s we choose.
$y$ represents the number of $2$s we choose.
$z$ represents the number of $3$s we choose.
We need to find the coefficient of the term $x^{12}$ in the expansion of the polynomial $(x + x^2 + x^3)(y + y^2 + y^3)(z + z^2 + z^3)$.
To simplify this expression, we can notice that each term in the expansion represents choosing a power of $x$, $y$, and $z$ from either $1$, $2$, or $3$. So, we can rewrite the expression as $(x + x^2 + x^3)^3$.
Expanding this expression, we get:
$(x + x^2 + x^3)^3 = x^3 + 3x^4 + 6x^5 + 7x^6 + 6x^7 + 3x^8 + x^9$
Now, we need to find the coefficient of $x^{12}$. From the expression, we can see that there is no term with $x^{12}$, which means there are no positive integers that satisfy both conditions.
Therefore, the number of positive integers that satisfy both conditions is $0$.