Question

Tyler 进入一条自助餐队伍，他要选择一种肉、两种不同的蔬菜和一种甜点。如果食物选择的顺序不重要，他可能选择多少种不同的餐食？

肉类：牛肉、鸡肉、猪肉

蔬菜：烤豆、玉米、土豆、番茄

甜点：布朗尼、巧克力蛋糕、巧克力布丁、冰淇淋

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

$4$

$24$

$72$

$80$

$144$

答案：C

知识点：组合乘法原理

难度评级：1270

解答：

他有 $3$ 种肉类选择和 $4$ 种甜点选择。

他必须从 $4$ 种蔬菜中选 $2$ 种。因为顺序不重要，蔬菜选择有 $\dfrac{4\cdot3}{2}=6$ 种。

因此可能的餐食数为 $3\cdot6\cdot4=72$ 。

所以正确答案是 C。

He has $3$ choices for the meat and $4$ choices for dessert.

He must choose $2$ of the $4$ vegetables. Since order does not matter, there are $\dfrac{4\cdot3}{2}=6$ vegetable choices.

This gives $3\cdot6\cdot4=72$ possible meals.

Thus, C is the correct answer.

2001 AMC 8 第 14 题

答案：C

难度评级：1270

解答：

其他年份的第 14 题