排列組合問題在MATLAB中的實(shí)現(xiàn)方法大全

1、排列組合問題在 MATLAB 中的實(shí)現(xiàn)方法大全 matlab 做排列組合：比如要 ABCD 的全排列，可以用 perms 函數(shù) perms(ABCD) 運(yùn)行結(jié)果DCBADCABDBCADBACDABCDACBCDBACDABCBDACBADCABDCADBBCDABCADBDCABDACBADCBACDACBDACDBABCDABDCADBCADCB以下是幾個(gè)常用的排列、組合與階乘等函數(shù)。1、combntns(x,m)列舉出從 n 個(gè)元素中取出 m 個(gè)元素的組合。其中， x 是含有 n 個(gè)元素的向量。2、perms(x) 給出向量 x 的所有排列。3、nchoosek(n,m)從 n 各元素中

2、取 m 個(gè)元素的所有組合數(shù)。 nchoosek(x,m) 從向量 x 中取 m 個(gè)元素的組合4、factorial(n) 求 n 的階乘。% 求(2n-1)!或(2n)!5、 prod(n:m)% 求排列數(shù)：m*(m-1)*(m-2)* *(n+1)*n prod(1:2:2n-1)或 prod(2:2:2n)6、cumprod(n:m)輸出一個(gè)向量 n n*(n+1) n(n+1)(n+2) n(n+1)(n+2) (m-1)m7、gamma(n)求 n!8、v=n!;vpa(v)更詳細(xì)資料如下nchoosekBinomial coefficient or all combinationsS

3、yntax:C = nchoosek(n,k)函數(shù)描述： 從n個(gè)元素中一次選k個(gè)元素的所有組合數(shù) C (注意，C是一個(gè)數(shù)值)。C = n!/(n - k)! k!);C = nchoosek(v,k)函數(shù)描述：從向量v中一次選其中k個(gè)元素的所有組合C (注意：C是一個(gè)矩陣，列數(shù) 為k )DescriptionC = nchoosek(n,k)where n and k are nonnegative integers,returns n!/(n - k)! k!).This is the number of combinations of n things taken k at a time.

4、C = nchoosek(v,k),where v is a row vector of length n,creates a matrix whose rows consist of all possible combinations of the n elements of v taken k at a time.Matrix C con tai ns n!/(n - k)! k!) rows and k colum ns.Inputs n, k, and v support classes of float double and float single.Examples：The com

5、mand nchoosek(2：2：10,4)returns the even numbers from two to ten, taken four at a time：246824610248102681046810combntnsAll possible combinations of set of values從給定集合 set中列出所有可能的subset個(gè)元素的組合Syntaxcombos = combntns(set,subset)combos = combntns(set,subset) returns a matrix whose rows are the various co

6、mbinations that can be taken of the elements of the vector set of length subset.Many combinatorial applications can make use of a vector 1：n for the input set to return generalized, indexed combination subsets.DescriptionThe combntns function provides the combinatorial subsets of a set of numbers.It

7、 is similar to the mathematical expression a choose b, except that instead of the number of such combinations,the actual combinations are returned. In combinatorial counting, the ordering of the values is not significant.The numerical value of the mathematical statement a choose b is size(combo s,1).ExamplesHow can the numbers 1 to 5 be taken in sets of three (that is, whatis 5 choose 3)?combos = combntns(1：5,3)combos =1 2 31 21 31 31 42 32 32 43 4 size(combos,1) ans =54554555% 5 choose 310( 注意事項(xiàng) )： Note that if a val