A number theory problem by Priyanshu Mishra

We have $\left(1,1,2,2\right), \left(1,2,1,2\right)$ and $\left(2,1,1,2\right)$ as solutions.

Let's prove that these are the only solutions.

$$

Case 1, at least one of $a,b,c$ is $1$ :

$$

If only one of $a,b,c$ is $1$ or all of them are $1$ then LHS is odd and RHS is even which is impossible.
If two of $a,b,c$ are $1$ , e.g. $b$ and $c$ , we have
$a! + 2 = 2^{d!} \implies \dfrac{a!}2 + 1 = 2^{d!-1} \implies a = 2$ , which gives us the given solutions, when we permutate the values of $a,b,c$ .
( $a = 2$ because $\dfrac {a!}2 + 1 > 1$ , so it must be even, and that forces $a = 2$ or $a = 3$ but only with $a = 2$ we have a solution because $d!-1 = 2$ has no solutions.)

$$

There are no additional solutions in this case.

$$

Case 2, $a,b,c > 1$ :

$$

If $a,b,c > 2$ then LHS is divisible by $3$ which is impossible since RHS is not.

Hence, at least one of $a,b,c$ is $2$ .

If two of them are $2$ we have $a! + 4 = 2^{d!} \implies \dfrac{a!}4 + 1 = 2^{d!-2}$ , this has no solutions since $\dfrac{a!}4 + 1 > 1$ which means that it is even, but there exists no such $a$ that $4$ divides $a!$ but $8$ does not.

If only one of them is $2$ then $a! + b! + 2 = 2^{d!} \implies \dfrac{a!}2 + \dfrac{b!}2 + 1 = 2^{d!-1}$ , which again means that $\dfrac{a!}2 + \dfrac{b!}2 + 1 > 1$ is even, which implies that either $a$ or $b$ is $3$ . ( $2$ is not allowed because we can only have one equal to $2$ here.)

Now $a! + 8 = 2^{d!} \implies \dfrac{a!}8+ 1 = 2^{d!-3}$ , which again means that $\dfrac{a!}8 + 1 > 1$ is even so, $a$ can be either $4$ or $5$ , which yields, $32 = 2^{d!} \implies d! = 5$ , which has no solutions and $128 = 2^{d!} \implies d! = 7$ which has no solutions.

$$

Hence, there are no solutions in this case.

$$

Hence, the only solutions are the solutions given in the beginning, which means that the number solutions is $\boxed{3}$ .

Do you mean positive integral solutions? Because I can show you 12 with integral, 9 of them including 0. Like a,b,c being 0,0,2 will have 3 combinations. Being 1,1,2 will have 3 combinations as well. Being 1,0,2 will have 6 combinations. And d will be 2 in each case. So answer should be 12 if integral solutions are to be considered.