Question

Pat 想从充足供应的三种甜甜圈中买四个：糖霜、巧克力和糖粉。共有多少种不同选择？

Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?

$6$

$9$

$12$

$15$

$18$

答案：D

知识点：隔板法组合

难度评级：1340

解答：

不同选择的数量等于方程 $g+c+p=4$ 的非负整数解个数。由插板法，解的个数为 $\dbinom{4+2}{2}=\dbinom62=15$ 。

所以正确答案是 D。

The number of selections is the number of nonnegative integer solutions of $g+c+p=4.$ By stars and bars, this is $\dbinom{4+2}{2}=\dbinom62=15.$

Thus, the correct answer is D.

2001 AMC 10 第 19 题

答案：D

难度评级：1340

解答：

其他年份的第 19 题