I'm stuck to solve the problems below:
(a) Let $p(n, k)$ be the number of partitions of $n$ into exactly $k$ parts. Show that
$$ \sum_{n, k \geq 0} p(n, k) x^{n} t^{k}=\prod_{i \geq 1} \frac{1}{1-x^{i} t}=\sum_{k \geq 0} \frac{x^{k} t^{k}}{(1-x)\left(1-x^{2}\right) \cdots\left(1-x^{k}\right)} . $$
(b) Find a similar generating function for the number $p_{d}(n, k)$ of partitions of $n$ into exactly $k$ distinct parts
I know the generating function of the partition number: $$ \sum_np(n)x^n=\prod_{i \geq 1}\frac{1}{1-x^i} $$ and how to prove it, but I've never heard about a two-variable generating function.
I did a little research about it, and I got some information about it:
Another way to get a generating function for $p(n, k)$ is to use a two-variable generating function for all partitions, in which we count each partition $\lambda=\left(\lambda_1, \lambda_2, \ldots, \lambda_k\right) \vdash n$ with weight $y^k x^n$, where $n$ is the size and $k$ is the number of parts. The monomial giving the weight contribution for a single part of $i$ now becomes $y x^i$ instead of just $x^i$ and accordingly, we get the generating function $$ P(x, y)=\sum_{n, k} p(n, k) y^k x^n=\prod_{i=1}^{\infty} \frac{1}{1-y x^i}=\frac{1}{(1-y x)\left(1-y x^2\right)\left(1-y x^3\right) \cdots} . $$
(I got it from : https://math.berkeley.edu/~mhaiman/math172-spring10/partitions.pdf)
From the information I got, I cannot understand how I can just change $x^i$ into $yx^i$, and also what the 'weight $yx^i$' means. Also, I have no idea how to start on problem (b).
Please help me.
UPDATE: I thought a little bit more, and I think this is the required answer from (b): $$ \prod_{i \geq 1}(1+yx^i) $$ by the same fashion in the blockquote(if I assume that I can understand that fully). But I'm stuck again to transform this formula into another form like part (a).