Counting number of solutions for $x = (a-1)(b-2)(c-3)(d-4)(e-5)$ 
If $x = (a-1)(b-2)(c-3)(d-4)(e-5)$, where $a,b,c,d,e \in \mathbb{N}$ are distinct natural numbers less than 6. If x is a non zero integer, then the how to count the no of sets of possible values of $(a,b,c,d,e)$? 

which is the fastest (pencil-paper) solution of this problem?
$ \mathbb{N} = 1,2,3,\cdots $
 A: I'll take "natural number" to mean at least 1 (so, in particular, not 0). 
You've got five distinct natural numbers less than 6, that means you have to use each of the numbers 1, 2, 3, 4, 5 exactly once. So you have a permutation of the set $\lbrace1,2,3,4,5\rbrace$. The non-zero condition on the product says that this permutation can't have any fixed points. So the question is a disguised way of asking you about the number of derangements of 5 objects. 
Now any good combinatorics text will tell you all about counting derangements; alternatively, just type the word into your favorite search engine and see what comes up. 
A: do it manualy
here is all the possible values of (abcde) tht respect your constraints
a=  2,3,4,5,6
b=1,  3,4,5,6
c=1,2,  4,5,6
d=1,2,3,  5,6
e=1,2,3,4,  6
there is that 5^5 caseq to studi
and less that 5^5 possible values of x
A: for x to be a non zero solution, (a-1), (b-2).... can't be zero
so possible values of a 2,3,4,5
similarly b= 1,3,4,5
and so on
so possible combinations are 4*4*4*4*4
i.e. 4^5 
A: As a,b,c,d,e can be natural and less then 6{1,2,3,4,5} but x cant be 0
Its a classical example of dearrangements.
A cant take 1
B cant take 2
C cant take 3
D cant take 4
E cant take 5
So its like filling 5 letters in 5 envelopes such that no letter goes to the designated envelope.
Dearrangements of 5 = 5!{1 - 1/1! + 1/2! - 1/3! + 1/4! - 1/5!} = 44.
A: 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 $f_5(x)$.

