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> 1. How can I deduce which operation ought fill in the red blank beneath? 

You can't. It's a hideous phrasing. The issue at question is not "redundancies" (which would carry the implication that they're merely unnecessary) but multiple counting: that is, counting the same assignment more than once.

One way to approach it would be to point out that the original permutation problem can be seen as awarding 1 gold medal, 1 silver medal, 1 bronze medal, and 5 participation certificates. Then we get $8!$ presentation orders, but we reduced to $\frac{8!}{5!}$ distinct ways of assigning the items presented because the participation certificates are equivalent to each other. This isn't a proof of correctness, but it provides a justification for why making the medals equivalent to each other would further reduce to $\frac{8!}{5!3!}$.

To be more formal, take it from

> Well, in this case, the order we pick people doesn’t matter. If I give a can to Alice, Bob and then Charlie, it’s the same as giving to Charlie, Alice and then Bob. Either way, they’re equally disappointed.

The number of ways of ordering Alice, Bob, and Charlie is $3!$ by the same argument as before. So this means that we can cluster the arrangements into groups of $3!$ equivalent arrangements in such a way that each arrangement is in exactly one of the groups. But what we want to count is the number of these groups, or *equivalence classes*. In this special case where the equivalence classes all have the same size, we can just divide by the size of one equivalence class to get $\frac{1}{3!} \times \frac{8!}{5!}$.

For advanced readers only: in combinatorics we frequently get situations where objects must be grouped in equivalence classes which aren't all the same size. In this more general situation we count by *weighting* each arrangement by (one divided by the size of its equivalence class).