It's hard to tell exactly what the OP is asking for, but I interpret the question to be asking about a function I'll call $f_5(x)$, which counts the number of permutations $(a,b,c,d,e)$ of $(1,2,3,4,5)$ for which $x=(a-1)(b-2)(c-3)(d-4)(e-5)$. It's rather cumbersome to compute the answer to this (especially if it's not what the OP means), so instead I'll give the answer for $f_4(x)$, defined in the analogous way:
$$f_4(-3)=2,\quad f_4(1)=1,\quad f_4(4)=2,\quad f_4(9)=1,\quad f_4(12)=2,\quad f_4(16)=1 $$
and $f_4(x)=0$ for all other non-zero $x$. If you like, we also have $f_4(0)=15$.
Just to round things out, we also have
$$f_2(-1)=1,\quad f_3(-2)=1,\quad f_3(2)=1$$
Any guesses as to the number of non-zero $x$'s for which $f_5(x)$ takes a non-zero value?
Sorry, the answer is not $24$. If I did everything correctly, there are $12$ values for which $f_5(x)=1$ and $16$ values for which $f_5(x)=2$, for a total of $28$ non-zero $x$s with a non-zero values of $f_5(x)$.