Find all positive integers $(a,b,c,n)$ such that $2^n=a!+b!+c!$ I have found the solutions by a little calculation $(2,3,5,7)$ and $(2,3,4,5)$. But I don't know if there's any other solutions or not?
 A: Let's grind it out. We want all ordered quadruples, but to get them it is enough to find all ordered triples $(a,b,c)$ where $a\le b\le c$. Then any triple of this kind we find can be permuted arbitrarily.
There cannot be triples $(a,b,c)$ with $a\le b\le c$ and $a\gt 2$, since for these we get that $3$ divides $a!+b!+c!$. So $a=1$ or $a=2$.
You don't seem to have considered $a=1$. So here goes. If $a=1$ then $b=c=1$ doesn't work. Let $b=1$. Then $c=2$ works, and we get the triple $(1,1,2)$. We also have the triple $(1,1,3)$. There is no triple of the shape $(1,1,k)$ for $k\gt 3$, for then $k!$ is divisible by $4$, so $1!+1!+k!$ cannot be a power of $2$.
Now look for triples of the shape $(1,b,c)$ where $b\ge 2$. There are none, for then $1!+b!+c!$ is odd.
Now look for triples of the form $(2,b,c)$. Try for $b=2$. Note that $c=2,3$  don't work, and $c\gt 3$ doesn't work, for in that case $8$ divides $c!$, so $2!+2!+c!$ cannot be a power of $2$.
So try for $b=3$. For $c$, calculate. We find, as you did, that $c=4$ and $c=5$ work. Nothing bigger than $5$ works, for if $c\gt 5$ then $c!$ is divisible by $16$, so $2!+3!+c!$ cannot be a power of $2$.
Now look at $b\ge 4$. This cannot work for any $c$, because then $2!+b!+c!$  has remainder $2$ on division by $4$.
A: Assume $1\le a\le b\le c$. Then $2^n=a!(1+b\cdots(a+1)+c\cdots(a+1)$.
It follows that either $a=2$ or $a=1$. 
Consider the case $a=2$. Then one of the numbers $b\cdots(a+1)$ and $c\cdots(a+1)$ must be even, and the other odd, and since $b\cdots(a+1)$ divides $c\cdots(a+1)$ it follows that $b\cdots(a+1)$ must be odd, and $c\cdots(a+1)$ must be even. Hence $b=2$ or $b=3$, and $c\ge 4$. So we are looking for $c$ such that either $\displaystyle 2+\frac{c!}2$ is a power of $2$, or $\displaystyle 4+\frac{c!}2$ is a power of $2$. 
These are only ideas how to possibly start (seems right), I will think if I could add more details and a more definite answer. Edit: I added all details below. 
Consider the case $a=1$. Then $2^n=1+b!+c!$. It follows that $b=1$ and $\displaystyle 2^n=2(1+\frac{c!}2)$, so $\displaystyle 2^{n-1}=1+\frac{c!}2$. This works if (and only if) $c=2$ or $c=3$. So we get solutions $(a,b,c)=(1,1,2)$ with $1+1+2=4=2^2$, and $(a,b,c)=(1,1,3)$ with $1+1+6=8=2^3$. It looks like you missed these small integers solutions. 
Back to the case $a=2$. First let $b=2$. Then $\displaystyle 2^n=2(1+1+\frac{c!}2)$, so $\displaystyle 2^{n-2}=1+\frac{c!}4$. No solutions in this case.  
Finally, let $a=2$, $b=3$. Then $\displaystyle 2^n=2(1+3+\frac{c!}2)$, so $\displaystyle 2^{n-3}=1+\frac{c!}8$. Works if $c=4$ or $c=5$. Solutions $(a,b,c)=(2,3,4)$ when $2+6+24=32=2^5$, and $(a,b,c)=(2,3,5)$ when $2+6+120=128=2^7$ (these are the two solutions that you found). This finishes all possible cases, no other solutions. 
A: Let $a \le b \le c$.
If $m \ge 2$ m! is even so if $a = 1$ (odd), $b = 1$ and $c! = 2^n - 2 = 2(2^{n-1} - 1)$ so $c = 2; n=1$ or $c = 3; n = 2$.  Two answers so far. (1,1,2,1)(1,1,3,2)
If $a > 1$ then $a! + b! + c! = a!(1 + b!/a! + c!/a!) = 2^n$ If $a \ge 3$ $3|a!$ and so $3|2^n$ which is impossible so $a = 2$.
So $1 + b!/2 + c!/2 = 2^{n-1}$  $n$ can't be $1$. so $b!/2$ must be odd so $b= 3$.  So $c!/2 = 2^{n -1} - 4= 4(2^{n -3} - 1)$ so $c$ = 4 and $n = 5$. Or 3*5...c = $2^{n-3}-1).$  Possibly, $c=5$ and $15 = $ $2^{n-3} - 1$; $n = 7$.  So two more answers $(2,3,4,5)$ and $(2,3,5,7)$.
Any more will require $3*5...c = 2^{n-3}-1)$ which is odd so $c<6$.  So those $4$ are the only solutions.
