之七

steven.wei

五个字母 D I G I T ,可以组成一个string,如 DIGIT.两个II不相临,可以组成多少string?
我想的做法以是从总排列数中减少两个II相邻的情况，也就是p(5,5)-P(4,4),可斑竹确认的答案是108，真烦。

有十个球，一个绿，一个蓝，依次取三个球不放回，求第一个是绿，其它二个不是蓝的概率
答案是9/30，可我认为应该是1/10 *8/9 *7/8　＝7/90

kitten369

1) P(55)/P(22) - P(44)=60-24=36 . both answers are wrong
  5 D I G I T have two same elements so it is not P(55)

2)you are right

steven.wei

hi, Miss Kitten369,

thanks for your help.但我还有疑问。为什么不是减而一定要除？当时我在想此题的时候，在想是P（5,5)-P(4,4),还是用除法十分犹豫，不是特别明白。我记得在新东方狒狒曾经提到过减法和除法的使用区别，当时一带而过，现在也不是明白。最后再问一下，为什么还要减去P(44)，反正此题一头雾水，还请好人做到底。先谢了。

kitten369

sorry I can not input chinese in this computer. hope I could explain clearlly.

5 elements to arrange(pai lei),if every element is different, there will be P(55), but if two(or 3,4) elements are same, then we will divide it byP(22)(or p33,p44).
I have an example you will remember it easily. if 5 elements all are same, it definitly is 1 ,in other words it is P(55)/P(55).

because减少两个II相邻的情况 we treat two II as one elment, then for 两个II相邻的情况 have P(44).
so final answer should be P55/P22-P44

Mush

我觉得也不对啊。
这样呢
p（55）-2*p(44)=72
有个乘2是因为两个i的互换位置。

steven.wei

hi,kitten
i guess i almost understand your ideas.your explanation is very clear. i must to say that this possibility-typed problem solving question is quite classical and representative. thanks for your help.