位置9的问题 The number of random samples (without replacement) of size 3 that can be drawn from a population of size 5 is 题库 1/1[input].

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BlackTom AI · Accepted Answer

Question restatement: The problem asks for the number of random samples (without replacement) of size 3 that can be drawn from a population of size 5.

Key concept: When drawing a sample where the order of selection does not matter and there is no replacement, the count is given by combinations, denoted as C(n, k) or nCk.

Step 1: Identify the values: n = 5 (population size), k = 3 (sample size).
Step 2: Apply the combination formula: C(n, k) = n! / (k!(n - k)!). Here, C(5, 3) = 5! / (3! * 2!).
Step 3: Compute factorials: 5! = 120, 3! = 6, 2! = 2. So C(5, 3) = 120 / (6 * 2) = 120 / 12 = 10.
Step 4: Interpret the result: There are 10 distinct 3-element subsets that can be drawn from a 5-element population when order does not matter and without replacement.

Note on alternatives: If order mattered (i.e., sequences of 3 draws), the count would be P(5, 3) = 5 × 4 × 3 = 60, which is larger because permutations consider different orders as different samples. If one mistakenly used with-replacement counts, the result would differ even more, since replacements allow repeated elements.

位置9的问题 The number of random samples (without replacement) of size 3 that can be drawn from a population of size 5 is 题库 1/1[input].

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