Let $A,B$ be some sets such that $|A|=a, |B|=b$. Prove $\binom{a}{b}=|P_b(A)|$ is a well defined expression. 
Let $A,B$ be some sets such that $|A|=a, |B|=b$. Prove $\binom{a}{b}=|P_b(A)|$(=the set of all subsets of A of cardinality $b$) is a well defined expression. 

Hello all. In this question I need to show that the equality $\binom{a}{b}=|P_b(A)|$ holds for all sets $A\neq A' \land |A|=|A'|=a, B\neq B'\land |B|=|B'|=b $. I don't really know how to show it, I thought about doing some tricks with the binomial theorem but I guess it has nothing to do with that. Would love to get your help... thanks in advance :)

I think that what is meant is something closer to the following:
Suppose that $A$ and $A'$ have the same cardinality and $B$ and $B'$ have the same cardinality.
I'm going to change your notation slightly and try to define
$$
\binom{a}{b}=\left|P_B(A)\right|
$$
where $P_B(A)$ is the set of all subsets of $A$ whose cardinality is the same as $B$.
The goal is to show that 
$$\left|P_B(A)\right|=\left|P_{B'}(A')\right|.$$
Since $A$ and $A'$ have the same cardinality, there is some bijection $f:A\rightarrow A'$.  Similarly, there is a bijection $g:B\rightarrow B'$.
To show that this notion is well-defined, we construct a bijection between $P_B(A)$ and $P_{B'}(A')$.  Suppose that $C$ is a subset of $A$ with the same cardinality as $B$.  Then, we show the following:


*

*$f(C)$ is a subset of $A'$.

*$f(C)$ has the same cardinality as $C$.

*If $f(C)=f(C')$ then $C=C'$ (injectivity).
The first statement is easy to prove since $f(C)=\{f(c):c\in C\}$ and the codomain of $f$ is $A'$, so $f(c)\in A'$.  
Since $f$ is a bijection, the restriction of $f$ to $C$, denoted $f|_C$, is also a bijection onto its image.  In fact, $f|_C$ is injective because $f$ is injective (skipping details).  On the other hand, maps are always surjective onto their images.  Therefore $f|_C:C\rightarrow f(C)$ is a bijection, since $B$ and $B'$ have the same cardinality, if $f(C)$ and $B'$ have the same cardinality if and only if $C$ and $B$ have the same cardinality.
Finally, we need to show injectivity.  If $C\not=C'$, then there is some $x\in C\setminus C'$ (or vice versa).  Due to injectivity of $f$, $f(x)\in f(C)$ but $f(x)\not\in f(C')$ (skipping details).  Therefore, we have an injective map from subsets of $A$ of cardinality equal to that of $B$ to subsets of $A'$ with cardinality equal to that of $B'$.
By inserting the details and using that the argument for $f^{-1}$ is identical, you get that the identification is well-defined.
