AMC 8 2001 Problem 14

AMC8 2001 年第 14 题

基本计数原理★★★☆☆

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

Tyler 进入了自助餐排队队列等待选餐，他可以选择一种肉，两种不同的蔬菜和一种甜点。如果食物的顺序不重要，这顿饭他能有多少种不同的组合？ ·肉：牛肉、鸡肉、猪肉 ·蔬菜：烤豆、玉米、土豆、西红柿 ·甜点：布朗尼，巧克力蛋糕，巧克力布丁，冰淇淋

A.4
B.24
C.72
D.80
E.144

正确答案 · Correct Answer

解析 · Solution

Solution 1

There are 3 possibilities for the meat and 4 possibilites for the dessert, for a total of 4×3=12 possibilities for the meat and the dessert. There are 4 possibilities for the first vegetable and 3 possibilities for the second, but order doesn't matter, so we overcounted by a factor of 2. For example, we counted 'baked beans and corn' and 'corn and baked beans' as 2 different possibilities, so the total possibilites for the two vegetables is

\displaystyle \frac{4×3}{2} = 6

, and the total number of possibilites is 12×6=72. To find the total ways, we multiply all the combinations of all 3 items: For meat, it's asking how many ways can you pick 1 item out of 3 or

\displaystyle \binom{3}{1} = \frac{3!}{(3-1)! \cdot 1!} = \frac{3 \cdot 2 \cdot 1}{2 \cdot 1 \cdot 1} = 3

(plain logic.) For vegetables, it's the amount of ways you can pick 2 items out of 4 or

\displaystyle \binom{4}{2} = \frac{4!}{(4-2)! \cdot 2!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{2 \cdot 1 \cdot 2 \cdot 1} = \frac{12}{2} = 6

For dessert, it's the amount of ways you can pick 1 item out of 4 or

\displaystyle \binom{4}{1} = \frac{4!}{(4-1)! \cdot 1!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1 \cdot 1} = 4

(also logic). Therefore, it's

3×6\times4=12×6=72.

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