number of one-one function;a set to itself How do you find the number of all one-one function from a set to itself?
If you are asked to find the number of all one-one functions possible from any set A to itself ,how do you do it?The following is the question I found:
Write the number of all one-one function from the set A = {a,b,c} to itself.The answer says six.But how??
 A: For a finite set this is simple. A function between finite sets of the same size is injective if and only if it is surjective if and only if it is bijective. So the set of injective functions is simply the set of permutations, i.e. the ways you can "re-order" $\{a,b,c\}$. For infinite sets this becomes much more difficult.
A: Suppose $f:A\to A$.  There are 3 choices for $f(a)$.  Once one is chosen, there are only two
choices for $f(b)$ since $f(a)\ne f(b)$.  This leaves one choice for $f(c)$. So the total number of possibilities is $3\cdot 2\cdot 1=6$.
A: This is the same problem as chosing elements from a set without repetition.
Let $f:A\rightarrow A$.
Put some order in your elements from $A$ and construct the array:
$$
f(a_1)f(a_2)\dots f(a_m),
$$
where $m = |A|$. Now ask the question: in how many ways the $a_1$ element could be mapped to another element of $A$? Clearly there are $m$-ways to do so. Now, in how many ways the $a_2$ element could be mapped to another element in $A$? The only restriction is that we do not chose the element to which $a_1$ was mapped, so there are $m-1$ options. Continue this line of thinking for the rest of elements. At the end you will have:
$$
m(m-1)\dots(1)=m!,
$$
possible $one-to-one$ mappings from $A$ to itself.
