AMC10 2007 A

AMC10 2007 A · Q12

AMC10 2007 A · Q12. It mainly tests Basic counting (rules of product/sum), Combinations.

Two tour guides are leading six tourists. The guides decide to split up. Each tourist must choose one of the guides, but with the stipulation that each guide must take at least one tourist. How many different groupings of guides and tourists are possible?

有两个导游带领六名游客。导游决定分开。每个游客必须选择一个导游，但规定每个导游至少带一个游客。有多少种不同的导游和游客分组方式？

(A) 56 56

(B) 58 58

(D) 62 62

(E) 64 64

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): The first guide can take any combination of tourists except all the tourists or none of the tourists. Therefore the number of possibilities is \[ \binom{6}{1}+\binom{6}{2}+\binom{6}{3}+\binom{6}{4}+\binom{6}{5} =6+15+20+15+6=62. \]

答案（D）：第一位导游可以带游客的任意组合，但不能带全部游客或一个游客都不带。因此可能的数量为 \[ \binom{6}{1}+\binom{6}{2}+\binom{6}{3}+\binom{6}{4}+\binom{6}{5} =6+15+20+15+6=62. \]

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