Confusion between these two combinatorial problems These two problems both come from Hossein Pishro-Nik's Introduction to Probability.
The first problem (which I will explain the set up and the solution):
Suppose we have 20 white phones and 10 black phones. We choose 9 phones at random. What is the probability that we have exactly four black phones.
Solution: The probability of this will be the number of ways we can have 4 black phones over the number of ways we can have any 9 phones.
$$P(\text{exactly four phones}) = \frac{\binom {10} {4} \binom {20} {5}}{\binom {30} {9}} $$
The numerator is the number of ways to select 4 black phones from 10 black phones times the number of ways to select 5 white phones from 20 white phones. The numerator is the number of combinations of 9 we can have from the 30 total phones.
What I don't understand is why selecting the 4 black phones doesn't determine the entire combination. Since the phones have to be black or white, if we say only four of them are black, then the rest have to be black so there's no need to multiply by $\binom {20} {5}$. This reasoning works for the next problem.
Second problem: Suppose we have a 20 by 20 grid. We start at the point $(0,0)$ and make only upward or rightward steps to get to the point $(40, 20).$
Solution: Here choosing the sequence of 10 rightward steps or choosing the sequence of 10 upward steps completely determines the path since we can only move up or right.
As a result, the answer is $ \binom {20}{10}$.
I don't see why the reasoning for the second question wouldn't work for the first question since the choice is binary (literally black or white).
 A: If you choose nine telephones at random, the chance that you will have four black telephones is the same as the chance that you will have five white telephones -- you can't have one without the other.
Following your argument, this probability would be
$$ P(\text{$5$ white telephones}) \stackrel?= \frac{\binom{20}{5}}{\binom{30}{9}}. $$
But this comes out to a different result than
$$ P(\text{$4$ black telephones}) \stackrel?= \frac{\binom{10}{4}}{\binom{30}{9}}. $$
So there is something wrong in this way of computing the probability.

Your second example is not a probability question. It's a simple counting question.
One thing that makes the path on the grid simpler is that at each step you really have only two choices: up or right. Unlike the telephones, where each black telephone was a distinct object before your choice (and remains one afterward), and where it makes sense to ask, if the first telephone you chose is black, which black telephone was it, if your first step on the grid is to the right, there's no question of "which rightward step" it was. It can only be the rightward step from $(0,0)$ to $(1,0).$
Also, unlike the telephones, where choosing one telephone as your first telephone does not prevent any other telephone from being chosen as the second one,
making a rightward step from $(0,0)$ to $(1,0)$ rules out a lot of steps that otherwise might have been possible.
For example, after $(0,0)$ to $(1,0)$ you cannot have a rightward step from
$(0,1)$ to $(1,1)$.
You also cannot have an upward step from $(0,0)$ to $(0,1)$.
Yet you could have chosen an upward step from $(0,0)$ to $(0,1)$
and then a rightward step from $(0,1)$ to $(1,1)$
if you had done those two steps first.
So the analogy "the step is either rightward or upward; the telephone is either black or white" is a bad analogy. There are all kinds of constraints on the steps that have no analogous constraints for the telephones, and freedoms of choice of telephones that do not exist for the steps on the grid.
This is why you have to actually consider the mechanics of a problem and not just grab a few numbers from it and throw them into an arbitrary formula.
