Problem 14 · 2001 AMC 8
Hard
Counting & Probability
careful-countingcombinations
Tyler has entered a buffet line in which he chooses one kind of meat, two different vegetables, and one dessert. If the order of food items is not important, how many different meals might he choose?
Meat: beef, chicken, pork.
Vegetables: baked beans, corn, potatoes, tomatoes.
Dessert: brownies, chocolate cake, chocolate pudding, ice cream.
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Answer: C — 72.
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Hint 1 of 2
Each food slot is chosen independently, so multiply the choices — but the vegetables need care because you pick TWO and order doesn't matter.
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Hint 2 of 2
Picking 2 of 4 vegetables where order is irrelevant is a combination, C(4,2) = 6 — not 4 × 3, which would double-count corn-then-beans vs beans-then-corn.
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Approach: multiply the independent choices
- Handle the tricky slot first. "Two different vegetables, order doesn't matter": there are 4 × 3 = 12 ordered ways, but each pair (like beans+corn) got counted twice, so divide by 2 → 6 pairs. That's exactly C(4,2) = 6.
- The other slots are single picks: meat 3 ways, dessert 4 ways, all independent.
- Multiply independent choices: 3 × 6 × 4 = 72 meals. The reusable move: when a selection ignores order, count it as ordered then divide by the number of rearrangements (here, 2).
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