Rule of sum problem I am supposed to solve the following problem:

How many different 4-digit numbers exist, which sum is 9 and do not
  contain the number 0.

My solution: 
$\frac{4!}{3!}+\frac{4!}{3!}+\frac{4!}{2!}+\frac{4!}{2!}+\frac{4!}{2!}=44$
My approach: I tried to write all possible combinations:
$$6 1 1 1=\frac{4!}{3!}\\5211=\frac{4!}{2!}\\4311=\frac{4!}{2!}\\4221=\frac{4!}{2!}\\3222=\frac{4!}{3!}$$
 A: If the four-digit number is written like $ABCD$, then you are looking for the number of ways to solve the equation $A+B+C+D=9$ in positive integers.  Because it's easier to work with non-negative numbers (for reasons we'll see shortly), let's set $A'=A-1$, $B'=B-1$, $C'=C-1$, $D'=D-1$.  Now we need to find the number of solutions to $$A'+B'+C'+D'=5$$ in non-negative integers.
This is easily addressed with the stars-and-bars strategy.  It's worth reading to get all the detail, but essentially there is a one-to-one correspondence between solutions to that equation and ways to arrange five * and three | in a line.  For instance, the solution $(A',B',C',D')=(3,0,2,0)$ corresponds to the arrangement * * * | | * * | and vice-versa.  However, the number of arrangements of those characters is easily seen to be $\binom83=56$ since you just need to choose which of the 8 characters in the string need to be filled with the bars.

Your solution  strategy would have been fine if you had included all the possible digit combinations.  You're missing 3321, which accounts for why you had 12 fewer solutions than the correct answer.
A: If you can find the number of tuples $(a,b,c,d)$ where $a,b,c,d$ are positive integers that satisfy: $$a+b+c+d=9$$ then you are ready. 
This because then automatically $a,b,c,d\in\{1,2,3,4,5,6,7,8,9\}$ so that string $abcd$ can be recognized as a $4$-digit number that does not contain digit $0$.
Use stars and bars for this.
