Let X and Y two sets with X=Y=m in N. So exist m! diferents bijectives functions between X and Y. I have this solution but I don't know if is right
Let be bijective functions $f_{i}$ with i=1,2,...,k and $f_{i}:X\mapsto Y$
Now $f(x_{1})=y_{1},y_{2},..,y_{m}$ and this take the value of m.
But $$ f(X_{1})\neq f(X_{2})\neq ...\neq f(X_{k})$$
We know that they aren´t equal so $$f(X_{1})=m\\
f(X_{2})=m-1\\
\vdots \\
f(X_{k})=m-(m-1)$$
so $(m)(m-1)...(m-(m-1))$
 A: (You mean that |X|= |Y|= m or cardinality of X= cardinality of Y= m. X and Y themselves are sets so not equal to a number.)
I don't know what you mean by "$f(x_1)= y_1, y_2,..., y_m$".  $f(x_1)$ is a single element of X (I suppose- you don't actually say what these x's or y's are) not all members of Y.
You then talk about $f(X_1)$, $f(X_2)$, ...  without any previous mention of $X_1$, etc.  Do you mean $x_1$, etc.?
Finally, and most egregious, you write "f(X_1)= m", "f(X_2)= m- 1", "f(X_2)= m- (m-1)", etc. which make no sense at all.  You are told that f is a function from set X to set Y.   Nothing at all is said about Y being a set of numbers or any set on which "1" or subtraction are defined!
I would do this: Since X is finite we can call its cardinality "n" and order the members of X calling them $x_1$, $x_2$, ..., $x_n$.  Since Y has the same cardinality as X we can call its members $y_1$, $y_2$, ..., $y_n$.       Now there exist n! possible permutationa of Y. And each bijective function can be written as mapping $\{x_1, x_2, ..., x_n\}$  to the "y"s in one of those permutations.
A: You're trying to say that there are $m!$ bijections $f:X \to Y$ if $\vert X \vert = \vert Y \vert = m$.  Prove this by induction on $\vert X \vert$.
If $\vert X \vert = \vert Y \vert =1$, then there is only $1$ bijective function $f:X \to Y$.
Now assume we know the result for $\vert X \vert = \vert Y \vert = n$. Assume $X = \{ x_1, x_2, \ldots, x_{n+1} \}, Y= \{ y_1, y_2, \ldots, y_{n+1} \}$ and let $f:X \to Y$ be a bijection.  Then there are $n+1$ choices for $f(x_{n+1})$, and because $f$ is a bijection from $X$ to $Y$, $f$ restricted to $X \setminus \{x_1 \}$ is a bijection to $Y \setminus \{f(x_1) \}$.
By our inductive hypothesis, there are $n!$ choices for the restricted map for each choice of $f(x_1)$, so since there are $n+1$ choices for $f(x_1)$, there are a total of $(n+1)!$ bijections from $X$ to $Y$.
