How many solution does $1/a+1/b+1/c+1/d=1$ have? From my friend, he gives me a competition question:
"How many solution $(a,b,c,d)$ does $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1$ have where $a,b,c,d$ are positive integers? (the size of $a,b,c,d$ doesn't matter, either one can be the biggest or smallest, and they are not necessarily distinct)"
I want to ask if there is any solution shorter than mine? I think mine is too long, and maybe yields a wrong answer.

My solution:
WLOG, let $a\leq b\leq c\leq d$
$$1=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\leq \frac{4}{a}$$
$$a\leq4$$
Because $a=1$ yields no solution, so consider $a=2,3,4$

Case 1:$a=2$, then $\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{1}{2}$
Do that again: $\frac{1}{2}=\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\leq\frac{3}{b}$, so $b\leq 6$. 
Let $b=6$,  then $\frac{1}{c}+\frac{1}{d}=\frac{1}{3}$, going to $$(c-3)(d-3)=9$$ $$(c,d)=(4,12),(6,6)$$ so in the case have: $(a,b,c,d)=(2,6,4,12),(2,6,6,6)$ then eliminate some case not satisfy $a\leq b\leq c\leq d$
Then going through when $b=5$,$b=4$,$b=3$... yields $184$ distinct solutions.
Case 2: Following the same procedure as Case 1... yields $18$ solutions.
Case 3: As above... yields only a solution which is $(4,4,4,4)$
Conclude it, the equation has $203$ solutions.

That is my solution, I wrote it using one and a half piece of A4 paper, I have recently tried $abcd=abc+abd+acd+bcd$ but don't know how to continue, or should I use Vieta theorem?
---After first edit---
According to Robert Z, I had miscount quadruplet $(3,4,4,6)$ which add up the count to $215$ solutions.

-- After last edit --
Seems like there is no faster solution, I will close this question and marked as solved. Thanks to everyone who spend effort to my question.
 A: In my opinion your approach is fine and I am not aware of a faster method. 
However I got a different number of solutions. 
Assuming that $a\leq b\leq c\leq d$ then
1) if $a=2$ and $b=3$ then $(c,d)$ can be
$$(7,42),\;(8,24),\;(9,18),\;(10,15),\;(12,12).$$
2) if $a=2$ and $b=4$ then $(c,d)$ can be
$$(5,20),\;(6,12),\;(8,8).$$
3) if $a=2$ and $b\geq 5$ then $(b,c,d)$ can be
$$(5,5,10),\;(6,6,6).$$
4) if $a=3$ then $(b,c,d)$ can be
$$(3,4,12),\;(3,6,6),\;(4,4,6).$$
5) if $a=4$ then $(b,c,d)=(4,4,4)$.
Hence rearranging the $14$ ordered solutions we find the total number of solutions:
$$6\cdot 4!+5\cdot \frac{4!}{2}+1\cdot \frac{4!}{2!2!}+1\cdot \frac{4!}{3!}+1=215.$$
A: Above equation shown below has parametric form:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1$
$a=k(3-k)/2k$
$b=k(3-k)/k$
$a=3(k-3)/2k$
$a=3(k-3)/k$
For   $k=-7$ we get $(a,b,c,d)=[5,10,(15/7),(30/7)]$
A: There was a typo in my previous answer yesterday for the below equation:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1$
The variables (a,b,a,a) should read as (a,b,c,d) & is shown below:
$a=k(3-k)/2k$
$b=k(3-k)/k$
$c=3(k-3)/2k$
$d=3(k-3)/k$
Hence for $k=-7$ , we get:
$(1/5)+(1/10)+(7/15)+(7/30)=1$
