1978年波兰数学奥林匹克有一道组合题:对于n元集合M的任何两个子集A和B,求得A∩B的元素个数,求证所有求得的个数之和为n.4n-1.下面给出原证明.证明我们用Ac表示A在M中的补集,即Ac=M-A,于是有∑A,B M|A∩B|=∑A M∑B M|A∩B|=∑A M2-1∑B M(|A∩B|+|A∩Bc|)=∑A M2n-1|A|=2n-1.n In 1978, the Polish Mathematical Olympiad has a combinatorial question: For any two subsets A and B of an n-ary set M, find the number of elements of A∩B and verify that the sum of all the numbers obtained is n.4n-1. The following is the original proof. Prove that we use Ac to represent the complement of A in M, that is, Ac=MA, so there is ∑A,BM|A∩B|=∑AM∑BM|A∩B|=∑A M2- 1∑BM(|A∩B|+|A∩Bc|)=∑A M2n-1|A|=2n-1.n