Number of positive unequal integer solutions of $x+y+z+w=20$ What is the number of positive different integer solutions of $x+y+z+w=20$, where $x,y,z,w$ are all different and positive?
It would be nice if coding is not used.  I am given the answer $552$.

Just for comparison with Brian Scott's solution, here is the longhand count solution.
As is observed in several places above, it is sufficient to find the number of solutions with $x<y<z<w$ and multiply by 24, because each such solution can be permuted to give 24 solutions to the original question. 
Next note that $1+2+3+4=10$ is the smallest total for 4 unequal positive integers, so we are not going to get many solutions. Now $4+5+6+7>20$, so $x=1,2$ or 3. Since $3+4+5+6=18$ the only possible solutions for $x=3$ are $(3,4,6,7),(3,4,5,8)$. 
Similarly for $x=2$ we must have $y=3,4$ or 5. $y=5$ gives only $(2,5,6,7)$. $y=4$ gives only $(2,4,6,8),(2,4,5,9)$. $y=3$ gives only $(2,3,4,11),(2,3,5,10),(2,3,6,9),(2,3,7,8)$.
Finally in a similar way $x=1$ implies $y=2,3,4,5$ and gives solutions $(1,2,3,4),(1,2,4,13),(1,2,5,12),(1,2,6,11),(1,2,7,10),(1,2,8,9)$ and $(1,3,4,12),(1,3,5,11),(1,3,6,10),(1,3,7,9)$ and $(1,4,5,10),(1,4,6,9),(1,4,7,8)$, and $(1,5,6,8)$.
Total 23, hence 552 to question.
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I have spelt this out in full detail, but note that since only the count is required setting out all those cases with $z+w=k$ is not really needed. It is enough to say something like:
$x=3$ implies $y=4$ and 2 solutions
$x=2$ implies: $y=3$, 4 solutions; or $y=4$, 2 solutions; or $y=5$, 1 solution
$x=1$ implies: $y=2$, 6 solutions; or $y=3$, 4 solutions; or $y=4$, 3 solutions; or $y=5$, 1 solution.
Total 23 solutions. Times 4! gives 552 to the original question.
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Teachers often give simple questions for you to practice methods like generating functions. But simple questions can often be answered faster by more elementary methods. Often fastest is best!
