AMC8 2004

AMC8 2004 · Q17

AMC8 2004 · Q17. It mainly tests Basic counting (rules of product/sum), Combinations.

Three friends have a total of 6 identical pencils, and each one has at least one pencil. In how many ways can this happen?

三个朋友总共有6支相同的铅笔，每人至少有一支。这种情况有多少种分配方式？

(A) 1 1

(B) 3 3

(D) 10 10

(E) 12 12

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) The largest number of pencils that any friend can have is four. There are 3 ways that this can happen: (4, 1, 1), (1, 4, 1) and (1, 1, 4). There are 6 ways one person can have 3 pencils: (3, 2, 1), (3, 1, 2), (2, 3, 1), (2, 1, 3), (1, 2, 3) and (1, 3, 2). There is only one way all three can have two pencils each: (2, 2, 2). The total number of possibilities is $3 + 6 + 1 = 10$.

（D）任何一个朋友最多能有四支铅笔。出现这种情况有 3 种方式：（4, 1, 1）、（1, 4, 1）和（1, 1, 4）。有 6 种方式使得某一个人有 3 支铅笔：（3, 2, 1）、（3, 1, 2）、（2, 3, 1）、（2, 1, 3）、（1, 2, 3）和（1, 3, 2）。三个人各有两支铅笔只有一种方式：（2, 2, 2）。可能性的总数是 $3 + 6 + 1 = 10$。

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