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 ?

Better version.
Notice that $3 \mid x!$ for $x \geq 3$ and $3 \not \mid 2^d$. Therefore, at least one of $a, b, c \leq 2$. WLOG let $a\leq b\leq c$.
If $c\leq 1$, $a\leq b\leq c\leq 1, a!=b!=c!=1$ which gives no solution.
If $c=2$, $a!+b!+2=2^d$. $a,b\in \{0,1\}$ gives $4$ solutions, while $(a,b)=(1,2)$ and $(2,2)$ doesn't give solution. 
For the cases below, $c\geq 3$.
If $b\leq 1$, $2+c!=2^d$. Notice that $2^2|c!$ for $c \geq 4$, so $c=3$. This gives $(a,b,c,d)=(a,b,3,3) \forall a,b\in \{0,1\}$ (Extreme laziness)
If $b=2$, $a!+2+c!=2^d$. Note that $a!=1$ doesn't give any solution (parity), so $a=b=2$. $4+c!=2^d$. Noticing that $2^3\mid c!$ for $c \geq 4$, $c=3$. $(a,b,c)=(2,2,3)$ doesn't give a solution.
For the cases below, $c \geq b \geq 3$. $2\mid b!+c!$.
Note that $a!=1$ doesn't give a solution. Therefore, $a=2$.
$2+b!+c!=2^d$. If $c \geq b \geq 4$, $2^3 \mid b!+c!$. Therefore $b=3$ gives $8+c!=2^d$. Note that $c\geq 6$ means $2^4 \mid c!$. Therefore, $c=4$ or $c=5$. Checking shows both of them work.
Therefore all the solutions:
$(a,b,c)=(0,0,2),(0,1,2),(1,1,2),(0,0,3),(0,1,3),(1,1,3),(2,3,4),(2,3,5)$ , up to permutations.
