How many ways can the sum of 4 positive integers equal 12? I am currently doing research in Combinatorial Geometry and I have been able to reduce a quite complicated problem relating to extending the Newton-Gregory problem of kissing spheres to a simple number theory problem and then checking every case to see that a conjecture of mine holds.
In any case, for background reasons which are not necessary for me to get into, I need to determine the explicit sets of 4 positive integers which when summed together give 12. Order does not matter as I will need to permute the set of 4 positive integers in each case to satisfy a different case for verifying my conjecture by exhaustion. So, I am not interested in some abstract "there are this many ways", I am actually interested in generating the explicit sets of numbers. I have been able to come up with the following so far:
$$\{1,1,5,5\},\{2,2,4,4\},\{3,3,3,3\},\{2,2,3,5\},\{1,2,4,5\},\{1,3,3,5\},\{1,3,4,4\},\{2,3,3,4\}$$
Any ideas for how to solve this? I can clarify any ambiguities as needed!
EDIT: I forgot to mention the following important detail:

I only want to consider integers from the set $\{2,3,4,5\}$ in summing to 12, since these correspond to the degree of a vertex and I have proved that for my particular problem that all vertices have either degree 2, 3, 4, or 5.

 A: You want $a+b+c+d = 12$ where $2 \leq a \leq b \leq c \leq d \leq 5$. Let $a_1 = a-2$, $b_1 = b-2$, $c_1 = c-2$ and $d_1 = d-2$. This gives us that $$a_1 + b_1 + c_1 + d_1 = 4$$ where $0 \leq a_1 \leq b_1 \leq c_1 \leq d_1 \leq 3$.
Let $b_1 = a_1 + b_2$, $c_1 = b_1 + c_2$ and $d_1 = c_1 + d_2$. Then we need $$a_1 + (a_1 + b_2) + (a_1 + b_2 + c_2) + (a_1 + b_2 + c_2 + d_2) = 4$$
i.e. $$4a_1 + 3b_2 + 2 c_2 + d_2 = 4$$
where $0 \leq a_1,b_2,c_2,d_2$.
Note that $a = a_1 +2$, $b = a_1 + b_2 + 2$, $c = a_1 + b_2 + c_2 + 2$ and $d = a_1 + b_2 + c_2 + d_2 + 2$
Now all we want is $$4a_1 + 3b_2 + 2 c_2 + d_2 = 4$$
such that $0 \leq a_1 \leq b_1 \leq c_1 \leq d_1 \leq 3$.
This means $a_1 \leq 1$.
If $a_1 = 1$, then $b_2 = c_2 = d_2 = 0$. Hence the solution is $$(a,b,c,d) = (3,3,3,3)$$
If $a_1 = 0$, then $$3b_2 + 2 c_2 + d_2 = 4$$ where $0 \leq b_2,c_2,d_2$.
This means $b_2 \leq 1$.
If $b_2 = 1$, then $c_2 = 0$ and $d_2 = 1$. Hence, the solution is $$(a,b,c,d) = (2,3,3,4)$$
If $b_2 = 0$, then $$2c_2 + d_2 = 4$$ where $0 \leq c_2,d_2$. This gives us $(c_2,d_2) = (2,0)$, $(c_2,d_2) = (1,2)$ and $(c_2,d_2) = (0,4)$. But $d_1 \leq 3$. Hence, the last solution is not possible.
Hence, these now give the solutions
$$(a,b,c,d) = (2,2,4,4)$$
$$(a,b,c,d) = (2,2,3,5)$$
Hence, the only four possible solutions for $a+b+c+d = 12$, with the constraint that $a,b,c,d \in \{2,3,4,5\}$ are
$$(a,b,c,d) = (3,3,3,3)$$
$$(a,b,c,d) = (2,3,3,4)$$
$$(a,b,c,d) = (2,2,4,4)$$
$$(a,b,c,d) = (2,2,3,5)$$
A: Since order does not matter, assume $a \geq b \geq c \geq d \geq 1$ and $a + b + c + d = 12$.  All you have to do is a case-by-case exhaustion of what possible values you can have.  For example, the largest possible value of $a$ is $9$, in which case $b = c = d = 1$.  If $a = 8 \dots$
