Using the generating functions calculate the constant $C$. Let 
$P (X=k, Y=j) = C \frac{ \binom{k}{j}}{2^{j+k}}$ ,
$k \in \mathbb{N}_{0}, 0\leq j \leq k$ be distribution of random vector $(X,Y)$. Using the generating functions calculate the constant $C$.
I don't know how to start. Any hint helps.
 A: We proceed by finding a bivariate probability generating function for $(X,Y)$.  This is not fundamentally different from the excellent solution already provided by Hayk, but uses a generating function as requested in the OP.
We define $g(x,y)$ by
$$g(x,y) = \sum_k \sum_j P(X=k, Y=j) x^k y^j$$
so
$$\begin{align}
g(x,y) &= \sum_{k=0}^{\infty} \sum_{j=0}^k C \binom{k}{j} 2^{-j-k} x^k y^j \\
&=C \sum_{k=0}^{\infty} 2^{-k} x^k \sum_{j=0}^k \binom{k}{j} 2^{-j}y^j \\
&=C \sum_{k=0}^{\infty} 2^{-k} x^k (1+y/2)^k \tag{1} \\
&= \frac{C}{1-(x/2)(1+y/2)} \tag{2}
\end{align}$$
where we have used the Binomial Theorem at $(1)$ and the formula for the sum of a geometric series at $(2)$.
Finally, using the fact that $g(1,1)=1$ for a probability generating function, we find from $(2)$ that
$$1 = 4C$$
so $$C=\frac{1}{4}$$
A: Since you have the joint distribution of $(X,Y)$ then summing the probabilities over all outcomes of $X$ and $Y$ must equal to $1$, from there you determine the normalizing constant $C$. Namely
$$
C \sum\limits_{k=0}^\infty \sum\limits_{j=0}^k \frac{1}{2^{k + j}}\binom{k}{j} = 1.
$$
For the sum, using Newton's binomial formula we have
$$
\sum\limits_{k=0}^\infty \sum\limits_{j=0}^k \frac{1}{2^{k + j}}\binom{k}{j} = \sum\limits_{k=0}^\infty \frac{1}{2^k} \sum\limits_{j=0}^k \frac{1}{2^j}\binom{k}{j} = \sum\limits_{k = 0}^\infty \frac{1}{2^k} \left( 1 + \frac{1}{2}\right)^k = \sum\limits_{k=0}^\infty \left(\frac{3}{4} \right)^k = 4,
$$
hence $C = \frac 14$.
