two-variable generating function for all number partitions 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).
 A: This is an elaboration on Haiman's notes that you found.
By "weight," he means the monomial contribution of a partition in the expanded expression for the generating function.  For example, just using $x$ terms, the partitions of 4 in the expansion correspond to
$$ x^4 + x^3x^1 + x^2x^2 + x^2x^1x^1 + x^1x^1x^1x^1 = 5x^4.$$
As a colleague likes to say in his classes, "you can't stop me from" replacing the $(1-x^i)$ terms with $(1-yx^i)$.  What does this $y$ do to the partitions of 4?
\begin{gather}
yx^4 + (yx^3)(yx^1) + (yx^2)(yx^2) + (yx^2)(yx^1)(yx^1) + (yx^1)(yx^1)(yx^1)(yx^1) \\
= y^1x^4 + 2y^2x^4 + y^3x^4 + y^4x^4.
\end{gather}
The $y$ terms record the number of parts.  We see algebraically that there is 1 partition of 4 with 1 part, 2 partitions of 4 with 2 parts, 1 partition of 4 with 3 parts, and 1 partition of 4 with 4 parts.  Note that setting $y=1$ reduces this expression to $5x^4$.  (Euler came up with this approach.)
For the transformation from the product to the sum, think about the operation of partition conjugation.  If you draw the Ferrers diagram of a partition, where each part corresponds the the number of dots in a row, conjugation reflects along the diagonal and has the effect of swapping rows and columns.  Thus $\lambda \in P(n,k)$ with $k$ parts has first column of height $k$, and its conjugate has first row / part $k$. Therefore, $p(n,k)$ counts both partitions with $k$ parts and partitions with largest part $k$.
The infinite sum is over these conjugate partitions.  They are grouped by largest part $k$: the numerator insures that $k$ is a part and the $t^k$ indicates that the partition has $k$ columns, while the denominator allows for any number of parts 1 through $k$.
[Graphics idea beyond my MSE skills: In a Ferrers diagram of, say, $(4,3,2)$, put a $yx$ in the 3 boxes of the first column and an $x$ in the other 6 boxes.  Conjugating gives $(3,3,2,1)$ where the first row has $yx$ in each of its 3 boxes while the subsequent rows consist of boxes with $x$.  The products of the box content for both is $y^3x^9$.  In the first partition, that corresponds to a 3-part partition of 9.  In the second, it's a partition with largest part 3, the first 3 corresponding to $(yx)^3$.]
You're on the right track for (b).  The equivalent infinite sum expression is derived on pp4-5 of Haiman's notes.
