2004年AMC10竞赛A第6题 2004年AMC12竞赛A第4题
$$2004-AMC10A-6$$  Bertha has $$6$$ daughters and no sons. Some of her daughters have $$6$$ daughters, and the rest have none. Bertha has a total of $$30$$ daughters and granddaughters, and no great-granddaughters. How many of Bertha\textquotesingle s daughters and grand-daughters have no children?  伯莎有 $6$$ 个女儿，没有儿子。 她的一些女儿有 $6$$ 个女儿，其余的没有。 伯莎总共有 30个女儿和孙女，没有曾孙女。 伯莎的女儿和孙女有多少没有孩子？
课内体系->知识点->计数原理->排列与组合->排除法 美国AMC10/12->Knowledge Point->Combination->Reasoning->Simple Logical Reasoning
Since Bertha has $$6$$ daughters, she has $$30-6=24$$ granddaughters, of which none have daughters. Of Bertha\textquotesingle s daughters, $$\dfrac{24}{6}=4$$ have daughters, so $$6-4=2$$ do not have daughters. Therefore, of Bertha\textquotesingle s daughters and granddaughters, $$24+2=26$$ do not have daughters. $$\boxed{ (\text{E}) 26}$$.  ▪  ▪ Alcumus**  Bertha has $$30-6=24$$ granddaughters, none of whom have any daughters. The granddaughters are the children of $$\dfrac{24}{6}=4$$ of Bertha\textquotesingle s daughters, so the number of women having no daughters is $$30-4=\boxed {26}$$.  Draw a tree diagram and see that the answer can be found in the sum of $$6+6$$ granddaughters, $$5+5$$ daughters, and $$4$$ more daughters. Adding them together gives the answer of $$\boxed{ (\text{E}) 26}$$. 