Sampling without replacement from a population

Question

While counting number of ways to select a sample of size n, from a small(<=60 elements) population, we use combinations to count. For example, number of ways to select a sample of size 3 from a population of size 60, let's say is: 60C3 or 60 choose 3 i.e the usual combinations formula.

My question is:

Why the order does not matter while counting the number of ways we can sample without replacement from a population ? or why don't we use permutations to calculate number of ways to sample? If someone could elaborate with simple example would be nice.

Note: I am studying hypergeometric random variable.

Answer 1

Does not matter to whom and for what?

The number of possible ordered samples of $n$ items from a population of $N$ items is indeed $N!/n!$, and the number of possible unordered samples of $n$ items from a population of $N$ items is ${N \choose n}$. Both of these are legitimate counts of the number of possible outcomes of a certain aspect of the sample. As to whether or not the order matters, that begs the question: matters to whom and for what?

When we sample via simple-random-sampling without replacement, this means that samples containing the same items (in any order) are equally likely, which means that we will typically use procedures on the sample that are invariant to their order. If we use the sample in ways that are order invariant, then the order doesn't matter to the way in which we use the sample. In such cases, it is quite natural for us to ask the number of possible unordered samples we can get and to perform probability calculations on this basis. Contrarily, if we use the sample in ways that are not order invariant, then the order matters to the way in which we use the sample, and we would therefore have to consider order. Even in the latter case, we know that the probability of the unordered sample is $(N-n)!$ times the probability of any ordered sample of the same items, so the conversion is quite simple.

Answer 2

Hi: You use combinations when its without replacement but there's a multiplication involved also. Say you had numbers 1,2 and 3 and you wanted to calculate the number of ways you could pick 2 of these numbers without replacement. Then, you could pick 3 numbers on the first trial and 2 numbers on the second trial so the total ways of picking two numbers without replacement is 3 * 2 = 6.

But this is (3 choose 1) * ( 2 choose 1 ). So, you're using combinations through the choose function but you're multiplying two of them because, in the second trial, you only have 2 numbers left.

If it was with replacement, then it would be (3 choose 1) * (3 choose 1 ) = 9 ways because, since you're putting them back, you still have 3 numbers to pick from on the second attempt.

Does that make sense ?

Answer 3

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:

1. A, B, C (1)
2. A, B, D (2)
3. A, C, D (3)
4. A, C, B (1)
5. A, D, B (2)
6. A, D, C (3)
7. B, A, C (1)
8. B, A, D (2)
9. B, C, D (4)
10. B, C, A (1)
11. B, D, A (2)
12. B, D, C (4)
13. C, A, B (1)
14. C, A, D (3)
15. C, B, A (1)
16. C, B, D (4)
17. C, D, A (3)
18. C, D, B (4)
19. D, A, B (2)
20. D, A, C (3)
21. D, B, A (2)
22. D, B, C (4)
23. D, C, A (3)
24. 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 ${4\choose 3} = \frac{4!}{3!\cdot(4 - 3)!} = 4$