The question asks: "How many different simple random samples of size 4 can be obtained from a population whose size is 43?"

This is a combinatorics problem, where we are asked to calculate the number of possible combinations of 4 items that can be selected from a total of 43 items. The formula for combinations is:

$\binom{n}{r} = \frac{n!}{r!(n - r)!}$

Where:

• $n$ is the total population size (43),
• $r$ is the sample size (4).

So, the formula becomes:

$\binom{43}{4} = \frac{43!}{4!(43-4)!} = \frac{43 \times 42 \times 41 \times 40}{4 \times 3 \times 2 \times 1}$

Let's compute this:

$\binom{43}{4} = \frac{43 \times 42 \times 41 \times 40}{24} = \frac{43 \times 42 \times 41 \times 40}{24} = 123410$

So, the number of simple random samples that can be obtained is 123,410.

Would you like further explanation on how combinations work, or any other details?

Here are 5 questions to consider:

1. What is the difference between combinations and permutations?
2. How do factorials work in combinatorics?
3. How does the sample size affect the number of possible combinations?
4. In what situations would you use combinations instead of permutations?
5. How does the formula change if we are sampling with replacement?

Tip: Remember that in combinations, the order of selection does not matter.