Why does choose in combinatorics work? Say, the question asks for the number of $8$ bit strings with exactly $2$ $1$'s, why is it $8\choose 2$? I don't know how the formula relates to all the ways the two bits can be arranged in the string. Also, I'm not completely sure why it says it's a factorial, when the total number of strings is $2^8$. I'm confused about the relation between all the things. How does it come to the answer?
Visualize: You have a muffin tin with eight holes in it. You can fill up $2$ holes with ones and leave to rest empty with zeros. The question boils down to nothing more or less than: how many ways are there to choose which $2$ of the $8$ holes to fill.

So that is why $ {8\choose 2}=$ number of ways to choose $2$ out of $8$ elements is the correct answer.

From the comments, you seem to understand why "number of ways to choose $k$ out of $n$ elements" $= \frac {n!}{k!(n-k)!}$ so I won't try to explain that.  (Unless you want me to.)

It's a factorial rather than $2^8$ because you do not have any choice in how many holes to fill. If you have a choice for all eight holes there are $2^8$ options. But once you choose the two that will be $1$ you will not have any more choices. If you manage to get to $7$ and only chose one $1$ you will not have any choice for the last bit; it must be $1$. If you get to 6 with all $0$s, you will have no choice for the last two; they must both be $1$.

If you want to think of this in terms of $2^8$ there are $2^8$ ways to filled $8$ bits.  But you must toss out all $11xxxxxx$ except for $11000000$ which you may keep.  SO that that leaves $2^8 - (2^6 - 1)$.

Now you must also toss out all $101xxxxx$ except for $10100000$ which you must keep and you must toss out all but one $1001xxxx$ and all but one $10001xxx$ etc.

So that leaves  $2^8 - (2^6 - 1) - (2^5 - 1) - (2^4 - 1) - (2^3 - 1) - (2^2 - 1) - (2^1 - 1)$

But now we mus toss out all the $011xxxxx$ except $01100000$ and all the $0101xxxx$ but one and so on.  so the answer is:

$2^8 - (2^6 - 1) - (2^5 - 1) - (2^4 - 1) - (2^3 - 1) - (2^2 - 1) - (2^1 - 1)$

$-(2^5-1) -(2^4-1)-(2^3-1) - (2^2-1) -(2^1-1)$

$-(2^4-1)-(2^3-1)-(2^2-1) -(2^1-1)$

$-(2^3-1) -(2^2 -1) -(2^1 -1)$

$-(2^2 -1) - (2^1 -1)$

$-(2^1-1)$

$-(2^5-1) -(2^4-1)-(2^3-1) - (2^2-1) -(2^1-1)$

$-(2^4-1)-(2^3-1)-(2^2-1) -(2^1-1)$

$-(2^3-1) -(2^2 -1) -(2^1 -1)$

$-(2^2 -1) - (2^1 -1)$

$-(2^1-1)$

$-(2^4-1)-(2^3-1)-(2^2-1) -(2^1-1)$

$-(2^3-1) -(2^2 -1) -(2^1 -1)$

$-(2^2 -1) - (2^1 -1)$

$-(2^1-1)$

$-(2^3-1) -(2^2 -1) -(2^1 -1)$

$-(2^2 -1) - (2^1 -1)$

$-(2^1-1)$

$-(2^2 -1) - (2^1 -1)$

$-(2^1-1)$

$-(2^1-1)=K$

And as it turns out  $K = \frac {8*7}{2}$.  (!!) I'll leave it to you to contemplate an algebraic reason that may be so.

But I recommend that you study up on "stars and bars" to get a sense of how these different methods of interpretation tactilely yield the same and correct answers.
Because C(8,2) is simpler notation for 8!/6!2!

This question looks pretty bad on you as it shows you didn't try to properly understand the concept of a combination before you decided to create an account only so you could ask this.