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You have 8 people, and you have to pick 3 of them to get identical gifts. Even though the gifts are identical, let's give them labels: A, B, and C. Let's also number the people 1 to 8.

The first step is to figure out how many ways there are to hand out gifts in order. You have 8 choices for gift A, then 7 choices for gift B because you don't want to give to the same person who got gift A, and likewise 6 choices for gift C.

And now we get to the part about redundancies. Let's say for a moment that you gave gift A to person 3, gift B to person 5, and gift C to person 2. Let's call that outcome (3, 5, 2).

From the first step, you know you can make a list of outcomes like the above that is $8\cdot 7 \cdot 6 = 336$ items long. And since that list is all of the ways to give gifts in order to three people, you know that if (3, 5, 2) is on that list, then so is every rearrangement of those three numbers—(5, 3, 2), (2, 5, 3), etc. Because the gifts are identical, you can treat rearrangements as redundant.

So you can imagine organizing your list into groups, where two outcomes belong to the same group if, like (5, 3, 2) and (3, 5, 2), they are rearrangements of each other. (No outcome can belong to more than one group, because rearranging an outcome doesn't change the set of numbers that are in that outcome.) The answer you seek is the number of *groups* in the list, because each group corresponds to a single not-redundant outcome when all you care about is who gets a gift, not who gets which gift.

Is every group the same size? Yes, because no outcome can use the same number twice (that would mean someone got two gifts), so the numbers must all be different. And there are always $n!$ ways to rearrange $n$ different numbers.

 So you have a list of 336 outcomes, organized into groups of size $3! = 6$. How do you determine how many groups there are in the list? Hopefully the answer is now clearer.