Because when sampling—with or without replacement—all we are interested in is are the distinct items selected—not the order in which they were selected. All combinations really are is permutations where order doesn't matter. So you can start with permutations but then must address the lack of interest in the order by treating each "sampling-identical" set as the same. This is done by dividing the total number of permutations by the number of permutations each sample may have. For example, when choosing 3 items from a pool of 4, selecting $(A, B, C)$ is functionally equivalent for sampling purposes as selecting any of $(A, C, B), (B, A, C), (B, C, A), (C, A, B)$ or $(C, B, A)$. There are six such items because there are $k!$ ways to distinctly order $k$ items.
Let's enumerate the possibilities:
- A, B, C (1)
- A, B, D (2)
- A, C, D (3)
- A, C, B (1)
- A, D, B (2)
- A, D, C (3)
- B, A, C (1)
- B, A, D (2)
- B, C, D (4)
- B, C, A (1)
- B, D, A (2)
- B, D, C (4)
- C, A, B (1)
- C, A, D (3)
- C, B, A (1)
- C, B, D (4)
- C, D, A (3)
- C, D, B (4)
- D, A, B (2)
- D, A, C (3)
- D, B, A (2)
- D, B, C (4)
- D, C, A (3)
- D, C, B (4)
Note all of the above can be broken into four categories (listed by numerals) which correspond to using the same letters regardless of order. Which is what we expect ${4\choose 3} = \frac{4!}{3!\cdot(4 - 3)!} = 4$