I'm not understanding combinations and counting The question:
An urn has 10 black balls numbered from 1 to 10, and 10 white balls numbered from 1 to 10. In how many ways can we choose 5 balls from the urn?
(There are more questions, which is why there's balls that are different colors and numbered. This is just one of the)
I did answer ${20 \choose 5}$ which is the correct answer. But I don't know why, other than the word choose being in the question. 
Intuitively, I don't understand why $20 \times 19 \times 18 \times 16 \times 17$ isn't an answer for this question. I know it's incorrect and that they're very different answers, but I don't know why. My thinking is that the first ball you choose, you pick out of 20 possible choices, then there's 19, then 18, then 17 choices to pick from. 
 A: You've started correctly, thinking 
 $$20×19×18×17×16 $$
picks the five balls one at a time.
Now realize that when you see five of the balls you might have chosen the same five in one of the $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ possible orders. That means your starting guess counted each possible choice $120$ times. So divide
$$\frac{20×19×18×17×16}{5!} . $$
That's exactly how you compute $\binom{20}{5}$.
A: 
I did answer ${20 \choose 5}$ which is the correct answer. But I don't know why, other than the word choose being in the question.
Intuitively, I don't understand why $20 \times 19 \times 18 \times 16 \times 17$ isn't an answer for this question.

The main reason is that, usually when problems are worded like this, order doesn't matter, i.e. it's a problem about combinations instead of permutations.
Say, if the balls were numbered, I were to pick up balls $\#1-\#5$, in that order. As far as the problem is concerned, this is no different than if I were to pick them up in reverse order. Or, really, any of the $5! = 120$ orderings of the balls. So long as I get those specific balls, the order in which I grab them is unimportant. (Granted the problem could have been worded better to make it clearer. Of course, "choose" and no specifications of order usually are a key sign that order doesn't matter.)
This generalizes for any other group of five balls in that, for any given combination of balls, there are $5!=120$ equivalent ways to have that combination (including the given pairing). Accordingly, your idea of $20 \times \cdots \times 16$ needs to account for this overcounting: it gives us $5!=120$ times more combinations than we want! Luckily, this is easily remedied, and that's why we divide by $5!=120$, and thus the correct answer is given by
$$\binom{20}{5} = \frac{20 \times 19 \times 18 \times 17 \times 16}{5!}$$
This idea leads to the more general formula for this: were there $n$ total balls, and $r$ were drawn, we can get them in
$$\binom n r =  \frac{n \times (n-1) \times (n-2) \times \cdots \times (n-r+2) \times ( n-r+1)}{r!}$$
ways. You might sometimes see alternate notations for the top, sometimes called a "falling factorial." It may be denoted $n^{\underline r}$, or in the iterated product notation by $\prod_{k=1}^r (n-r+k)$. Either way, you basically start at $n$, and then go back by one until you have a total of $r$ numbers, then multiply them all together and divide by $r!$.
Sometimes people are introduced to this briefer, if less intuitive, formula for $\binom n r$. It only takes a little algebra to convince yourself of the equivalence, but you should probably focus on the first for intuition:
$$\binom n r = \frac{n!}{r! \cdot (n-r)!}$$
A: You have to choose $5$ balls from $20$ balls; so the order of the balls does not matter. There are $20×19×18×17×16 $ number of ways contains only  $\frac{20×19×18×17×16 }{5!}$ different combinations of the balls. Although,if order of the drawn balls did matter,then yours would be the correct answer.
