找出可以由单词 DAUGHTER 的字母组成的不同的 8 个字母排列，以便所有元音一起出现
在数学中，排列涉及将一个组的所有成员排序为某个系列或排列的函数。换句话说，如果该组已经被定向，则其组件的重定向称为置换过程。排列几乎在每个数学领域都以或多或少重要的方式发生。当在某些有限的地方观察到不同的命令时，它们经常出现。
排列
排列被称为按顺序组织组、主体或数字的过程，从集合中选择一个或多个数字，被称为组合，使得整数的序列不受影响。
置换公式
在排列中，从一组 n 个项目中收集 r 个项目，没有任何替换。在这个收集物质的序列中。
nPr = (n!)/(n – r)!
Here,
n = set dimensions, the total number of object in the set
r = subset dimensions, the number of objects to be choose from the set
组合
组合是从组中选择对象的一种方式，这样（与排列不同）选择的顺序无关紧要。在较小的情况下，总而言之，可以想象组合的数量。组合是指一次取k个对象，不重复地组合n个对象。提到允许重复的组合，经常使用表达式 k-selection 或 k-combination with repeat。
组合配方
结合起来，从一组 n 个对象中选择 r 个对象，其中选择的顺序无关紧要。
nCr =n!⁄((n – r)! r!)
Here,
n = Number of objects in group
r = Number of objects selected from the group
找出可以由单词 DAUGHTER 的字母组成的不同的 8 个字母排列，以便所有元音一起出现。
解决方案：
Total number of letters in DAUGHTER = 8
Vowels in DAUGHTER = A, U, E (vowels are a, e, i, o, u)
Arranging all vowels, since all vowels occur together, they can be AUE, UAE, EAU and so on.
Number of Permutation 3 vowels,
= 3P3
= 3!/(3 – 3)!
= 3!/0!
= 3 × 2 × 1 = 6 ways
Arranging 6 letters,
Number we need to arrange = 5 + 1 = 6
Number of permutations of 6 letters,
= 6P6
= 6!/(6 – 6)!
= 6!/0!
= 6 × 5 × 4 × 3 × 2 × 1 = 720
Thus, total number of arrangements = 720 × 6 = 4320
类似问题
问题 1：找出 FATHER 的字母组成的不同的 6 个字母排列的数量，以便所有元音一起出现？
解决方案：
Total number of letters in FATHER = 6
Vowels in FATHER = A, E (vowels are a, e, i, o, u)
Arranging all vowels, since all vowels occur together, they can be AE, EA and so on.
Number of Permutation 2 vowels,
= 2P2
= 2!/(2 – 2)!
= 2!/0!
= 2 × 1 = 2 ways
Arranging 5 letters,
Number needed to arrange = 4 + 1 = 5
Number of permutations of 5 letters,
= 5P5
= 5!/(5 – 5)!
= 5!/0!
= 5 × 4 × 3 × 2 × 1 = 120
Thus, total number of arrangements = 120 × 2 = 240
问题 2：找出可以由单词 EDUCATION 的字母组成的不同 8 字母排列的数量，以使所有元音不一起出现？
解决方案：
Total number of letters in EDUCATION = 8
Vowels in EDUCATION = E, U, A, I, O (vowels are a, e, i, o, u)
Arranging all vowels
First, calculate when all vowels occur together, they can be EUAIO, UAIOE, AIOUE, IOEUA, OEUAI and so on.
Number of Permutation 5 vowels,
= 5P5
= 5!/(5 – 5)!
= 5!/0!
= 5 × 4 × 3 × 2 × 1 = 120 ways
Arranging 4 letters,
Number we need to arrange = 3 + 1 = 4
Number of permutations of 4 letters,
= 4P4
= 4!/(4 – 4)!
= 4!/0!
= 4 × 3 × 2 × 1 = 24
Thus, total number of arrangements = 120 × 24 = 2,880
So, when all vowels do not occur together, total possible arrangements = 8! – 2880 = 40320 – 2880 = 37440.
问题3：使用HARYANA这个词的字符可以建立多少种不同的表达方式？
解决方案：
Total number of characters in HARYANA = 7
The character A repeats 3 times i.e, = 3!
Number of expression that can be formed
= 7!/3!
= 7 × 6 × 5 × 4 × 3!/3!
= 840 words
问题4：以辅音开头和结尾的不同表达有多少个，可以完成单词“EQUATION”？
解决方案：
Total number of characters in EQUATION = 8
8 characters i.e. 3 consonants 5 vowels.
The consonants are to settled 1st and last place and it can be done in 3P2 ways.
Now 5 vowels and 1 consonant are left i.e. 6 letters which can be organized in 6! ways. Hence the number of expression under given condition is
3P2 × 6! = 6 × 720 = 4320.
问题5：“ALLAHABAD”这个词的字母可以组成多少种不同的表达方式？
解决方案：
There are total 9 character in the word ‘ALLAHABAD’ in which 4 are ‘A’ s, 2 are ‘L’ and remaining all are definite.
So, the needed number of words
= 9!/4!2! = (9 × 8 × 7 × 6 × 5 × 4!)/4! × 2!
= 7560