drawing n balls randomly, *without* replacement from a bag containing b black balls and w white balls ---Part1

Question

A bag has $b$ black balls and $w$ white balls. balls of the same color are indistinguishable.

1. find, $z_k=$ the number of ways in which $n$ balls can be drawn randomly, without replacement from this bag containing $b$ black balls and $w$ white balls, when $k$ of the $n$ balls drawn are white

2. find, $x=$ the number of ways in which $n$ balls can be drawn randomly, without replacement from this bag containing $b$ black balls and $w$ white balls

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What I have done

1. There is just one way of drawing $k$ white balls and $n-k$ black balls without replacement from this bag containing $b$ black balls and $w$ white balls, iff $\space k\leq w$ and $n-k \leq b$. And there are $\frac{n!}{k!(n-k)!}$ ways of ordering the $k$ white balls and $n-k$ black balls. So, the number of ways of ordering $n$ balls from a bag of $(w+b)$ balls, when $k$ of the $n$ balls are white is $$z_k = \frac{n!}{k!(n-k)!}\mathbb{I}\{k \leq w\}\mathbb{I}\{n-k \leq b\}$$

2. obviously, $$x = \sum_{k=0}^nz_k = \sum_{k=0}^n \frac{n!}{k!(n-k)!}\mathbb{I}\{k<=w\}\mathbb{I}\{n-k<=b\}$$, where $\mathbb{I}(x) =\begin{cases} 1 & x=True \\ 0 & x=False \end{cases} $.

now, assuming $n \leq b,n \leq w$, we have,

$$z_k = \frac{n!}{k!(n-k)!}$$ and $$x = \sum_{k=0}^nz_k = \sum_{k=0}^n \frac{n!}{k!(n-k)!} = \sum_{k=0}^n \binom{n}{k} = 2^n$$

my question is: are the calculations done above correct?

note:- the current question and this question are very similar. I know. the question I gave a link to is just too long to answer, I guess. So, I am trying to break down my question into small pieces so that they can be answered easily. I will change/delete the question(that I gave a link to) based on whether there is a mistake in the calculations done in this question.

edit:- the follow-up question to this question is here

Answer 1

This looks good to me. A quick way to check your work would be to solve the problem another way. For example, provided there are enough balls of each color, each of the $n$ balls you draw can be either of two options (black or white), thus there are $2^n$ many ways to draw them, agreeing with your answer.

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I hope this helps ^_^