Question about combinatorics in case the arrangement is not important. So, I know that the number of combinations of $n$ objects taken $r$ at a time (where the order does not matter) is given by $\frac{n!}{(n - r)!}$ divided by $r!$. Based on my understanding, I would divide the product of given formula $\frac{n!}{(n-r)!}$ by just $r$. I don't understand why it should be divided by $r!$. Can someone explain? For example, if we were to select two managers from five people then the possible ways to do it would be $5\cdot4 \cdot \frac{1}{2} = 10$. Also, I only partly understand the reason you have to divide it by anything when the order of the selected objects is not important. How to better understand the case when the order is not important?
 A: To answer your question, let us look at a very simple example.
Imagine we were asked to find the number of different ways to arrange the letters in the word 'SHEEP' (the words formed do not have to be meaningful). How would this be done?
Since we were asked to arrange the letters in the word 'SHEEP', we will have to use permutations (since the order of the letters in the word matters).
Therefore, the number of arrangements, assuming that all the letters are distinct, is given by:

 $$^5P_5 = \frac {5!}{(5 - 5)!} = \frac {5!}{0!} = \frac {5!}{1} = 5! = 120$$

However, all the letters in the word 'SHEEP' are not distinct; the letter 'E' is repeated twice. Therefore, we will have to divide the number of arrangements by $2!$. This is because, in the above calculation, we have assumed that all the letters are distinct.
Note: We divide by $2!$ not $2$. This is more evident with more repeated letters, such as with the word 'DELETE'. If we notate each 'E' differently, like '$DE_1LE_2TE_3$', then the calculation above sees '$DE_1LE_2TE_3$' differently from '$DE_1LE_3TE_2$' and '$DE_3LE_2TE_1$'. This is clearly not the case, as the 'E' is the same no matter where it is written. The same concept can be applied to words like 'INTERNET' and 'BOOKSHELF', where more than one letter is repeated.
Therefore, the number of arrangements of the letters in the word 'SHEEP' is given by:

 $$\frac {^5P_5}{2!} = \frac{\frac {5!}{(5 - 5)!}}{2!} = \frac{\frac {5!}{0!}}{2!} = \frac{\frac {5!}{1}}{2!} = \frac {5!}{2!} = 5\cdot 4 \cdot 3 = 60$$

Let us now extend this concept with combinatorics where the arrangement does not matter.
Let us take your example - selecting two managers from five people.
Let us first look at arranging two people from a group of five i.e. the order does matter. This is given by:

 $$^5P_2 = \frac {5!}{(5 - 2)!} = \frac {5!}{3!} = 5\cdot 4 = 20$$

Let us now look at selecting two people from a group of five i.e. the order does not matter.
Let us first notate the people who are selected as '$S$' and the people who are not selected as '$N$'. Therefore, the group of people can be rewritten as:

 $$S \; S \; N \; N \; N$$

To solve this problem, all we have to do is find out the number of ways we can arrange the letters above. There are $5!$ total ways to arrange the letters above. However, the letter 'S' is repeated twice and the letter 'N' is repeated thrice. With that in mind, we have to look back at our problem with the word 'SHEEP'. When we were asked to arrange the letters in the word 'SHEEP', we noticed that the letter 'E' was repeated twice. We, therefore, divided the total number of arrangements by $2!$ i.e. the factorial of the number of times a letter was repeated.
Thus, we can apply the same concept here.
The number of ways to select 2 managers from 5 people is given by:

 $$\frac {5!}{2!\cdot 3!} = \frac {5\cdot4}{2\cdot 1} = 10$$

This explains why we have to divide $\frac {5!}{(5 - 2)!}$ by $2!$ i.e. divide $\frac {n!}{(n - r)!}$ by $r!$. It is because of that extra $r!$ group which we are taking note of (the second repeated group). This also explains why $^5C_2$ $=$ $^5C_3$, since selecting 2 from 5 is the same as selecting 3 from 5; the complementary group is left behind.

I hope this helps! I'm new to MSE, so I would really appreciate any and all feedback. Thank you!
