一个人可以用加拿大这个词的字母生成多少个六个字母的词？
在数学中，排列被称为排列一个集合的过程，其中一个集合的所有成员被排列成一些系列或顺序。如果集合已经排列，则排列的过程称为对其组件的重新排列。几乎所有数学领域都以或多或少的重要方式发生排列。当考虑某些有限集上的不同命令时，它们经常出现。
置换公式
在排列中，从 n 个事物的集合中选择 r 个事物，没有任何替换。在这个选择的顺序。
nPr = (n!)/(n – r)!
where
n = set size, the total number of items in the set
r = subset size, the number of items to be selected from the set
组合
组合是从组中选择项目的行为，这样（不像排列）选择的顺序无关紧要。在较小的情况下，可以计算组合的数量。组合是指一次取k个不重复的n个事物的并集。组合可以以任意顺序选择项目。对于那些允许重复出现的组合，经常使用术语 k-selection 或 k-combination with replication。
组合配方
组合 r 个事物是从一组 n 个事物中选择的，其中选择的顺序无关紧要
nCr = n!⁄((n-r)! r!)
Here,
n = Number of items in set
r = Number of items selected from the set
一个人可以用加拿大这个词的字母生成多少个六个字母的词？
解决方案：
Case 1: Where one A can’t be right before or after another A (like AA or AAA)
There are six positions, numbered 1 through 6, to be assigned to the six letters.
The positions assigned to the 3 A’s may be (1,3,5) i.e., A*A*A, (1,3,6) i.e., A*A**A,
(1,4,6) i.e., A**A*A or (2,4,6) i.e., *A*A*A the asterisks must be substituted with
letters C, N, D in order to abide by the rules. So there are only 4 ways to assign
positions to the 3 A’s.
The remaining 3 letters are distinct, so they can be placed in 3! = 6 different ways.
Therefore, the number of words you can make using 6 letters from “CANADA” only once
where one A can’t be right before or after another A (like AA or AAA) is
4*6 = 24.
Case 2: Where one A can be right before or after another A (like AA or AAA)
The word ‘CANADA’ contains 3 A’s, 1 C, 1 N, and 1 D.
Number of permutations of the letters of the given word = 6!/3! = 120.
类似问题
问题1：一个人可以用印度这个词的字母生成多少个五个字母的词？
解决方案：
Case 1:- where one I can’t be right before or after another I (like II)
There are five positions, numbered 1 through 5, to be assigned to the five letters.
The positions assigned to the 2 I’s may be (1,3) i.e., I*I**, (3,5) i.e., **I*I or
(2,4,) i.e., *I*I* the asterisks must be substituted with letters N,D,A in order to
abide by the rules. So there are only 3 ways to assign positions to the 2 I’s.
The remaining 3 letters are distinct, so they can be placed in 3! = 6 different ways.
Therefore, the number of words you can make using 5 letters from “INDIA” only once
where one I can’t be right before or after another I (like II) is
3*6 = 18.
Case 2 :- where one I can be right before or after another I (like II)
There are 60 different ways to arrange the 5 letters in “INDIA”.
Explanation:
The word ‘INDIA’ contains 2 I’s, 1 A, 1 N and 1 D.
Number of permutations of the letters of the given word =5!⁄2!=60.
问题 2：一个人可以用单词 America 的字母生成多少个七个字母的单词？
解决方案：
Case 1:- where one A can’t be right before or after another A (like AA)
There are Seven positions, numbered 1 through 7, to be assigned to the seven letters.
The positions assigned to the 2 A’s may be (1,3) i.e., A*A**, (3,5) i.e., **A*A,
(2,4,) i.e., *I*I* (4,6) i.e., ***A*A* (5,7) i.e., ****A*A the asterisks must be
substituted with letters M,E,R,I,C in order to abide by the rules. So there are only
5 ways to assign positions to the 2 A’s.
The remaining 5 letters are distinct, so they can be placed in 5! = 120different ways.
Therefore, the number of words you can make using 7 letters from “AMERICA” only once
where one A can’t be right before or after another A (like AA) is
5×120 = 600.
Case 2 :- where one A can be right before or after another A (like AA)
There are 2520 different ways to arrange the 7 letters in “AMERICA”.
Explanation:
The word ‘AMERICA’ contains 2 A’s, 1 M, 1 E, 1R, 1I and 1C.
Number of permutations of the letters of the given word =7!⁄2!=2520.