Counting - Permutation & Combination

Most of us are already introduced to permutation and combination in high school math classes. I understand clearly what they separately define. But I get confused easily between their equations because they never made sense to me. This post is my attempt to make sense of the following equations.
$$P_{n,r} = \frac{n!}{(n-r)!} \\ C_{n,r} = \frac{n!}{(n-r)!\ r!}$$

Permutation

I will attempt to figure out the number of ways to select 2 characters out of 3. The characters are {ABC}, $n=3, r=2$. How many ways can I select 2 out of 3 characters? The answer is given by the permutation equation. Let me break it down to options. During the first selection, I have 3 options {ABC}, which will leave me 2 options for the second and for the final selection only 1 option will be left. As I get 3 options for the first, and for each of those first options, I get 2 sub-options, I have a total of $3 \times 2 = 6$ options. However, it is not close to the $P_{n,r}$ equation, but we will reach there. Now let me select 3 out of 3. We can assume the pattern will hold, $3 \times 2 \times 1 = 6$ options again. Similarly, 3 options to select 1 out of 3. Notice that, we are only multiplying $r$ numbers (even though we don’t know what those numbers are) for selecting $r$ out of $n$. This is where the equation makes sense. $n!$ gives us all the ways to select $n$ out of $n$.

$x!$ is read as ‘x factorials’. A factorial of a number is defined as $x! = x \times (x-1) \times (x-2) \times ... \times 1$. Here is an example, $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

Let’s make the example set {ABCDE}. For selecting 5 out of 5, I will have $5! = 5 \times 4 \times 3 \times 2 \times 1$ options. To select 3 out of those 5, $5 \times 4 \times 3$ options are available, which do not have $2 \times 1$ in it, meaning $5!$ is getting divided by $2 \times 1 = 2! = (5-3)!$ . There is a simple explanation for this division. Look at the numerator, we are selecting 5 characters when we only wanted 3. And, I have seen in earlier examples with {ABC} that for each first options there are 2 more sub-options. In the numerator, while selecting each 3rd character, instead of stopping, it also has $2!$ more sub-options and selecting them. Consider an iteration for example, while selecting 3 out of 5 in {ABCDE}, one of the iteration of $5!$ will select B, select D, select C and then continue on to select E and A. And another iteration will select B, select D, select C and then select A and E. Those sub-options {AE,EA} are extra and I don’t want them in my counting. As those two characters were selected in $2!$ ways, to remove them, put the same in denominator.

Combnination

Permutation gave us ways to select $r$ out of $n$ characters. $P_{3,3}$ of {ABC} is {ABC, ACB, CAB, BAC, BCA, CBA}. Note that, these are samne characters only in different order. So, order matters in permutation, unlike combination. $C_\{3,3\}$ of {ABC} is 1, only {ABC}. How does combination achieve this? From, the equation we see that it first calculates $P_{n,r}$ and then divides by $r!$ . In previous section, we learned that to remove some subset of permutation, we divide. The only thing to understand here is which permutations are we removing? For $C_{3,3}$ , we see that result 1 comes from $\frac{3!}{3!\ 0!} = \frac{6}{6}$ . It is evident by the same numerator and denominator that what is essentially happening here is, $\frac{\{ABC, ACB, CAB, BAC, BCA, CBA\}}{\{ABC, ACB, CAB, BAC, BCA, CBA\}}$ . It seems the denominator is duplicates of the parent characters {ABC} (remember, order does not matter here, ABC, BCA is same thing in combination). And how many ways can 2 characters be duplicate, for example, {AB}? That will be $2!$ ways, {AB,BA}. That is why for $C_{3,2}$ , $P_{3,2}$ is divided by $2!$ . We first get all the ways to select 2 characters out of 3 (which is $P_{3,2}$ ), and then divide by how many ways 2 characters can be duplicated.