AMC12 2011 A

AMC12 2011 A · Q9

AMC12 2011 A · Q9. It mainly tests Basic counting (rules of product/sum), Casework.

At a twins and triplets convention, there were $9$ sets of twins and $6$ sets of triplets, all from different families. Each twin shook hands with all the twins except his/her siblings and with half the triplets. Each triplet shook hands with all the triplets except his/her siblings and with half the twins. How many handshakes took place?

在一个双胞胎和三胞胎大会上，有$9$对双胞胎和$6$组（三人一组的）三胞胎，且都来自不同家庭。每位双胞胎与除其兄弟姐妹外的所有双胞胎握手，并与一半的三胞胎握手。每位三胞胎与除其兄弟姐妹外的所有三胞胎握手，并与一半的双胞胎握手。总共发生了多少次握手？

(A) 324 324

(B) 441 441

(D) 648 648

(E) 882 882

Answer

Correct choice: (B)

正确答案：（B）

Solution

There are $18$ total twins and $18$ total triplets. Each of the twins shakes hands with the $16$ twins not in their family and $9$ of the triplets, a total of $25$ people. Each of the triplets shakes hands with the $15$ triplets not in their family and $9$ of the twins, for a total of $24$ people. Dividing by two to accommodate the fact that each handshake was counted twice, we get a total of $\frac{1}{2}\times 18 \times (25+24) = 9 \times 49 = 441 \rightarrow \boxed{\textbf{B}}$

共有$18$位双胞胎和$18$位三胞胎。每位双胞胎与不在同一家庭的$16$位双胞胎以及$9$位三胞胎握手，共与$25$人握手。每位三胞胎与不在同一家庭的$15$位三胞胎以及$9$位双胞胎握手，共与$24$人握手。由于每次握手被计算了两次，故总握手次数为$\frac{1}{2}\times 18 \times (25+24) = 9 \times 49 = 441 \rightarrow \boxed{\textbf{B}}$

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