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《组合数学》课程复习

Many counting problems are solved by establishing a bijection between the set to be counted and some easy-to-count set. This kind of proofs are usually called (non-rigorously) combinatorial proofs.

 

The number of k-compositions of n is equal to the number of solutions to x_1+x_2+\cdots+x_k=n in positive integers. 

count the number of solutions to x_1+x_2+\cdots+x_k=n in nonnegative integers  We call such a solution a weak k-composition of n.

 

 Formally, a multiset M on a set S is a function m:S\rightarrow \mathbb{N}. For any element x\in S, the integer m(x)\ge 0 is the number of repetitions of x in M, called the multiplicity of x. The sum of multiplicities \sum_{x\in S}m(x) is called the cardinality of M and is denoted as | M | . 

 we have already evaluated the number \left({n\choose k}\right). If S=\{x_1,x_2,\ldots,x_n\}, let zi = m(xi), then \left({n\choose k}\right) is the number of solutions to z_1+z_2+\cdots+z_n=k in nonnegative integers, which is the number of weak n-compositions of k, which we have seen is {n+k-1\choose n-1}={n+k-1\choose k}.

 

We can think of it as that n labeled balls are assigned to m labeled bins, and {n\choose a_1,a_2,\ldots,a_m} is the number of assignments such that the i-th bin has ai balls in it.

 

posted on 2014-08-31 15:52  Vingstar  阅读(216)  评论(0)  编辑  收藏  举报