Exclamation Marks Everywhere!

$a!b!c!d!e! = 400!$
$397 | RHS \to 397 | LHS$
$397 | a!b!c!d!e! \to$
Therefore one of the (a,b,c,d,e) has to be greater than or equal to $397$ since 397 is the largest prime smaller than 400.

Assume a = 397,

$b!c!d!e! = 398 \cdot 399 \cdot 400$

$19 | 399 \to 19 | RHS \to 19 | b!c!d!e!$

Thus one of (b,c,d,e) to be greater than or equal to 19. Let b = 19

$18!c!de! = 398 \cdot 21 \cdot 400$ .
The minimum value of LHS = $18! > RHS$ .
Thus there is no solution for this case.

Similar reasoning goes for $a = 398$

For a = 399,

$b!c!d!e! = 400$
Since 3 does not divide RHS , $(b,c,d,e) \leq 2$ which does not lead to a solution.

a = 400,
$b!c!d!e! = 1$ .

Thus $(b,c,d,e) = [0,1]$

Thus $(a,b,c,d,e) = (400,1,1,1,1) , (400,1,1,1,0) , (400,1,1,0,0) , (400,1,0,0,0) , (400,0,0,0,0)$
The total ways this can happen $2\left( \dfrac{5!}{4!} + \dfrac{5!}{3!}\right) + \dfrac{5!}{2!2!} = 80$

My method starts similarly to Vighnesh's, but then goes off on a different path. A factorial necessarily contains every prime factor up to its argument. The largest prime $\leq 400$ is $397$ . Obviously since $397$ is prime, $397!$ is the smallest factorial to contain it as a factor, therefore at least one of $(a, b, c, d, e) \geq 397$ . If we assume for the moment that $a \geq 397$ , then we have four cases to consider (prime factorisations given): $a = 400, b!c!d!e! = 1 \\ a = 399, b!c!d!e! = 400 = 2^4 \cdot 5^2 \\ a = 398, b!c!d!e! = 399 \cdot 400 = 2^4 \cdot 3 \cdot 5^2 \cdot 7 \cdot 19 \\ a = 397, b!c!d!e! = 398 \cdot 399 \cdot 400 = 2^5 \cdot 3 \cdot 5^2 \cdot 7 \cdot 19 \cdot 199$ However, a factorial containing a given prime factor must obviously contain any smaller primer factors. The second case contains a factor of $5$ in $b!c!d!e!$ . Obviously the smallest factorial containing this factor is $5!$ ; this or any larger factorial also contains a factor or $3$ , but there is no factor of $3$ in $2^4 \cdot 5^2$ . Therefore this case is impossible. Likewise, the third contains no factor of $11$ , $13$ or $17$ , and the last one is even worse. As such, only the first case is possible.

So we have $b!c!d!e! = 1$ . Since the factorial of a non-negative integer is always a positive integer, we know that $b! = c! = d! = e! = 1$ . Consequently $\{b, c, d, e\} \subseteq \{0, 1\}$ .

Now, since we want the number of ordered quadruples, we note that

• there are five possible positions for the $400$ among $a, b, c, d, e$
• for each possible position of the $400$ , there are $2^4 = 16$ possible ways that the other variables can be assigned the values $0$ and $1$ .

Thus there are $5 \cdot 16 = \boxed{80}$ possible solutions.