A teacher has $5$ books to distribute to some of $20$ children in her class. How many ways are there for her to distribute the books if the books are all the same and no child gets more than one?
This is a combination since there is no repetition, but logically I don't get it. Let's say I give $1$ book to a child, now there are $19$ more students and $4$ books. So I give one to $1$ of the $19$, and so on and so forth such that $20\cdot19\cdot18\cdot17\cdot16$ are the amount of ways I can give a book. Obviously this is wrong, but I cannot wrap my head around why. Please, someone explain this to me.
If I first give a book to Alice, and then I give a book to Bob etc... it is no different than the scenario that I first give a book to Bob and then to Alice. By approaching via a permutations argument without division by symmetry, we will have counted each of the scenarios (alice,bob,charlie,david,ellen), (bob,alice,charlie,david,ellen), (bob,charlie,alice,david,ellen)... seperately, even though they are not "different" to us.
In general, you can consider this via permutations, for a current total of $20\cdot 19\cdot 18\cdot 17\cdot 16$, however noticing that for each group of five children, I will have counted that scenario exactly $5!$ number of times, you can divide by the amount of symmetry to correct the overcounting, for a total of $\frac{20\cdot 19\cdot 18\cdot 17\cdot 16}{5!}$. (Here, the "amount of symmetry" is 5! since for each different scenario we counted it a total of 5! times, the number of ways the five students could have been ordered to get the books)
The alternate approach uses combinations instead, where you choose 5 of the 20 students to come forward (without regards to order in which they arrive at your desk) in order to get books. Thus, the total is "20 choose 5" = $\binom{20}{5}$.
Notice that these answers are the same. $\binom{20}{5} = \frac{20!}{15!5!}=\frac{20\cdot 19\cdot 18\cdot 17\cdot 16}{5!}$.