Tanya prepared 4 different letters to be sent to 4 different address .For each letter ,she prepared an envelope with its correct address.if the 4 letters are to be put into 4 envelope at random.what is the possibility that only 1 letter will be put into the envelope with its correct address?

a.1/24,b.1/8,c.1/4,d.1/3,e.3/8

answer:d

求牛牛及斑竹告诉思路,谢谢

所有的组合是排列组合－24－分母

只有1各对应上的组合是：C41×C21×1＝8 第一步：选一个对应上信封的信 C41；第二步：第二封信只能从余下的3各信封中的另外两个抽取 C21；第三步：只能有一个选择：余下的两封信只能装入确定的信封。

我的概率一向比较糟糕，这道题我想了比较久；而且每次都是想错了，看了答案才恍然大误。希望能与楼主共勉。