Why counting $C(n,r)$ needs to divide $k$?

Why do we divide by $k$ when counting $C(n,r)$?

For example, I have 4 different balls (A,B,C,D).

I want to pick 3 of them and not put back.

I know the formula and I got the answer is 4.

$$c_r^n = \frac{p_r^n}{r!}=4$$

1. ABC

2. ABD

3. ACD

4. BCD

But what's the idea of formula to divide 3 and also divide 2?

Thank you~

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Once you select the three balls, counting order (which is what you do with $P^n_r$), how many times will a particular choice of $3$ balls occur? There are $P^3_3$ ways of listing those three balls, so by taking order into account you are "counting" each selection of $3$ objects $P^3_3$ times. So you must divide by $P^3_3 = 3 = 3\times 2$ to get the right answer.
(Alternatively, you have $3$ ways of picking a "first" ball, $2$ ways of picking a "second" ball from the remaining ones, and $1$ way of picking the third ball from the remaining ones; this gives $3\times 2\times 1=3!$ repetitions).
The general formula is $$C^{n}{r} = \frac{P^n_r}{P^r_r}$$ because when counting with $P^n_r$, each selection of $r$ objects will be counted $P^r_r=r!$ times.