Find all non-negative integers $a, b, c ,d$ such that $a! + b! + c! = 2^d$ 
Find all non-negative integers $a, b, c ,d$ such that $a! + b! + c! = 2^d$.

By trial I found  $a= 2 , b= 3 , c= 5$ and $d= 7$ which is one solution. How to find all the solutions of it ?

Without loss of generality $a\le b\le c$, so $a!|2^d\implies a!\in\{1,\,2\}$.
If $a!=1$, $b!+c!$ is odd so $b!=1$ and $c!=2^d-2$, so $c!\nmid4$ and $c\le3$. This gives the solutions $c=2$ and $c=3$.
If $a!=2$, $b!+c!$ isn't a multiple of $4$, so $b\le3$. In particular, if $a=b-2$ then $c!=2^d-4$ is a multiple of $4$ but not $8$ so $4\le c\le7$, and similarly if $a=2,\,b=3$ then $8\le c\le15$. I'll leave you to check these cases.
