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 :)

Using binomial theorem could work. You could also consider the "definition" (or interpretation) of the number $\binom{a}{b}$:

$\binom{a}{b}$ is the number of ways to choose $b$ elements from a set with $a$ elements.

This is, in a very literal sense, the number of subsets of size $b$ from a set of size $a$. 
This is clear if you are willing to accept this interpretation of the binomial coefficient.
On the other hand, if you want to prove it completely what you would need to show is basically that the binomial coefficient represents what it says it does: the number of subsets of size $b$ from a set of size $a$. 
This is a standard reasoning when you're first introduced to the binomial coefficient, and the approach is something as follows (try to think about it first, put the pointer on the space below if you need help):

 To choose $b$ elements out of a set of $a$ elements, first we determine which elements will we choose, in order (i.e. which is the first, which is the second, and so on). For this we have $n\cdot (n-1)\cdots (n-r+1)$ possibilities. Now, since the order doesn't really matter when it comes to choosing a subset, we have to divide by the number of permutations of $r$ elements, which is $r!$. The result is $\frac{n\cdot (n-1)\cdots (n-r+1)}{r!}$, which is the binomial coefficient $\binom{a}{b}$. 

