Thanks and Yes

Is what I am supposed to be showing is bijective.

So let me get this straight (sorry I'm a little slow) you're showing that it's injective by showing that f(A,b) = f(C,D) implies (A,B) = (C,D) (which is the definition of injective, I get that much. I don't see why this implies that AUB = CUD. Also I'm not sure where the conclusion that x is not an element of D comes from. The biggest problem with my understanding, is I'm not sure how it relates to the "power set" function. We were given the hint:

If

and

, we can join them to make the subset

of

. On the other hand, given

, we can break it up into

and

.

This gives a bijection from [tex]{\mathcal{P} (X)\times\mathcal{P} (Y)}\mapsto\mathcal{P}\left(X\cup Y\right)[/tex], namely, f

A,B) =

and

f^-1: (C) =

Now look at sets of a particular size. If |A| = i and |B| =k-i, then |

| =k (A and B are disjoint). Inversely if |C| =k, what can you say about |

| and |

|? Take it from here...

Which is what I tried to do but seem to be wandering in the dark.

On the second part of the question I know that it is basically the cardinality of the first part and that I have to prove that the Cardinality of the RHS (which is a Union) is the cardinality of the LHS (which is a Sum of products) but I'm also lost there...