$\large a! \cdot b! \cdot c! \cdot d! \cdot e! = 400!$

Find the number of ordered quintuples of
**
non-negative integers
**
$(a,b,c,d,e)$
satisfying the above equation.

The answer is 80.

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$a!b!c!d!e! = 400!$

$397 | RHS \to 397 | LHS$

$397 | a!b!c!d!e! \to$

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 $2\left( \dfrac{5!}{4!} + \dfrac{5!}{3!}\right) + \dfrac{5!}{2!2!} = 80$