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From numbers (1) to (30), (3) numbers that are multiples of (3) and (2) numbers that are not multiples of (3) are to be selected. How many ways are there?

Answer and explanation

Correct answer: (22800)

The direct answer is option A, 22800. From 1 to 30, the multiples of 3 are 3, 6, ..., 30, so there are \(30/3=10\). The remaining \(30-10=20\) numbers are not multiples of 3. We independently choose 3 of the first group and 2 of the second group, so the multiplication principle gives \(\binom{10}{3}\binom{20}{2}=120\times190=22800\). Option A is exactly this value. B, 11400, is half the correct result and misses valid selections. C, 19000, is not the product of the required independent choices. D, 45600, doubles the correct count and therefore overcounts. The groups are disjoint, so combinations are multiplied, not added.

Related tags

CombinationsClass11Hard

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