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标题: [求助] math jijing [打印本页]

作者: olive 时间: 2002-8-23 10:49 标题: [求助] math jijing

某6个字母的sequence,由1a,2b,3c组成，问不同可能 [em17]

作者: noheart55 时间: 2002-8-23 11:46

尝试一下，
1。 a,bb,ccc 排列 p33=6
2. a,bb,cc,c p44-2*p33 = 12
3. a,bb,c,c,c =2
4. a,b,b,ccc p44-p33*2 =12
5. a,b,b,cc,c p55-(2)-(4)*2+(1) = p55-12-12*2+6=90
6. a,b,b,c,c,c 简单排列 =6

所以共128.

(5) 项好像不是很确信，请高人指正

作者: yulei69 时间: 2002-8-27 02:32 标题: My thought

Actual, it is put 1a,2b,3C into 6 slot, but the same word can not be counted more than twice. In fact, after 1a and 2b are in the slot, no matter how you put remaining 3c, it will be the same word. So the problem become how many different way you can put 1a and 2b into 6slot. Thus, we put 1a first, there are 6 possibilitys. After a is positoned, there left 5 slot. Among these 5, if you put 1b at the position 1, there are 4 way to put second b. Then put 1b in the 2nd postion, there are 3 way to put second b. So on and on. there are total 10 way to put there 2b into remaining 5 positions. Therefore, the total number of posibility would be 6*10=60.Correct or not.

作者: noheart55 时间: 2002-8-27 13:43

Agree yulei

作者: verse666 时间: 2002-8-29 14:16

吴强的书上有个差不多,是? 和*的排列.
此题其实很简单了 C 6 取1 乘 C 5 取 2

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