2005年上海高中数学竞赛有这样一道题： M={1,2,3,4,5},a,b,c,d,e∈M(允许重复)。求使abcd+e为奇数的概率。由于a,b,c,d有顺序,于是共有5~5种情况。下计算使abcd+e为奇数的概率。当abcd为奇数,有3~4种,e只能取2,4有3~4×2种, 当abcd为偶数,有5~4-3~4种,e只能为1,3,5有3·5~4-3~5种,于是abcd+e为奇数的概率为1/5~5(3~4·2+5~4·3-3~5)=1794/3125。由于组合计数中排列和组合是相对概念,于是很自然联想到如下问题：设M={1,2,3,4,5),A={x|x=abcd+e,a,b,c,d,e∈M},求A中元素为奇数的概率。这是一道比原问题更为困难的题目,下面 The 2005 Shanghai High School Mathematics Contest had such a question: M={1,2,3,4,5}, a,b,c,d,e∈M (allowing repetition). Find the odds that abcd+e is odd. Since a, b, c, and d have a sequence, there are a total of 5 to 5 cases. Calculate the probability of making abcd+e odd. When abcd is odd, there are 3~4 kinds, e can only take 2, 4 has 3~4×2 kinds, when abcd is even, there are 5~4-3~4 kinds, e can only be 1, 3, 5 have 3·5~4-3~5 kinds, so the probability of abcd+e is odd is 1/5~5(3~4·2+5~4·3-3~5)=1794/3125. Since the permutations and combinations in combination counting are relative concepts, naturally the following problems are naturally associated: Let M={1,2,3,4,5), A={x|x=abcd+e,a,b,c ,d,e∈M}, finds the probability that the elements in A are odd. This is a more difficult topic than the original question, below