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How can I deduce which operation removes redundacies?

Please don't answer by working backwards from the answer, or by appealing to arithmetic. Act as if you're learning this for the first time.
1. How can I deduce which operation ought fill in the red blank beneath?
2. Why can't it be subtraction?
### [I shortened the original explanation](https://betterexplained.com/articles/easy-permutations-and-combinations/):
>**Quandary:** How many ways can I give $3$ gifts to $8$ people, if nobody receives $>1$ gift?
>
>**Explanation:** For combinations, order doesn’t matter. This raises an interesting detail of redundancies.
>
>For a moment, let's figure out how many ways we can rearrange 3 people. We've 3 choices for the first person, 2 for the second, and only 1 for the last. So we have $3 · 2 · 1$ ways to re-arrange 3 people. But this is a permutation! If you want to know the number of arrangements for $N$ people, it’s just $N!$.
>
>So, for giving 3 gifts, there are $3! \; (= 6)$ variations for every choice we pick. To calculate how many combinations, just create all the permutations and $\color{red}{\text{_____ by all the redundancies}}$. [...]

Please don't answer by working backwards from the answer, or by appealing to arithmetic. Act as if you're learning this for the first time.
1. How can I deduce which operation ought fill in the red blank beneath?
2. Why can't it be subtraction?
### [I shortened the original explanation](https://betterexplained.com/articles/easy-permutations-and-combinations/):
>**Quandary:** How many ways can I give $3$ gifts to $8$ people, if nobody receives $>1$ gift?
>
>**Explanation:** For combinations, order doesn’t matter. This raises an interesting detail of redundancies.
>
>For a moment, let's figure out how many ways we can rearrange 3 people. We've 3 choices for the first person, 2 for the second, and only 1 for the last. So we have $3 · 2 · 1$ ways to re-arrange 3 people. But this is a permutation! If you want to know the number of arrangements for $N$ people, it’s just $N!$.
>
>So, for giving 3 gifts, there are $3! \; (= 6)$ variations for every choice we pick. To calculate how many combinations, just create all the permutations and $\color{red}\text{_____}$ by all the redundancies. [...]

probability