Find the number of functions which satisfy given conditions If f :{1,2,3,4} → {1,2,3,4} and y = f(x) be a function defined such that |f(α) – α| ≤ 1, for α ∈ {1,2,3,4} and m be the number of such functions, then m/5 is equal
 A: If we think a function as a permutation of $(1,2,3,4)$, the only functions that satisfy the condition are

$$(1,2,3,4,5),\quad(1,2,4,3)\quad(1,3,2,4)\quad(2,1,3,4)\quad(2,1,4,3)$$

Thus, $m=5$ and $m/5=1$.
A: Okay.
$|f(a) - a| \le 1$ so $-1 \le f(a)-a < 1$ so $a-1 \le f(a) \le a+1$.
so $f(a)4$ may be one of three values $f(a)$ could be $a-1$ if $a-1$ is in range; that is if $a-1\ge 1$ or $a \ge 2$.
Or $f(a)$ could be $a$.
Or $f(a)$ could be $a+1$ if $a+1$ is in range; that is if $a+1 \le 4$ of $a \le 3$.
So
For $a = 1$ there are $2$ things $f(1)$ could be.  $f(1)=1$ is a possibility and $f(1) =2$ are a possiblity.
Of $ a= 2,3$ there are three things $f(a)$ could be.  $a-1, a, a+1$ are the three possibilities.
And for $ a=4$ there are $2$ things for $f(4)$ to be.
So there there are $2$ things $f(1)$ can be and $3$ things $f(2)$ can be and $3$ things $f(3)$ can be and $2$ things $f(4)$ can be.
So how many combinations of ways can $(f(1),f(2),f(3),f(4))$ can be?

 $2\times 3\times 3 \times 2= 36$ possible combinations of value so $f(1),f(2),f(3),f(4)$ so $36$ possible functions.

Um..... why does anyone care what $\frac m5$ is?
