在会议期间,7 名学生可以分配到 1 个三人间和 2 个双人间酒店有多少种方式?
给定一定数量的元素,它们的不同排列,或者通过一个接一个地选择,或者其中一些,或者一次全部选择,称为排列。它们被视为将线性序列分配给给定序列的成员的过程。它也被称为对给定序列或系列的元素重新排序的过程。
换句话说,排列一个序列意味着列出该序列的所有可能排列。
例如,系列 {1, 2} 可以写成两种方式:{1, 2} 或 {2, 1}。
在给定序列中的n 个元素中排列r个元素时的排列数
n P r = n!/(nr)!
例如,让 n = 5 {A, B, C, D, E} 和 r = 2(所有排列都有两个元素)。答案是5 P 2 = 5! /(5-2)! = 20。这 20 个排列是:AB、AC、AD、AE、BA、BC、BD、BE、CA、CB、CD、CE、DA、DB、DC、DE、EA、EB、EC 和 ED。
组合
它被定义为从给定序列中选择一个或两个或一些元素的过程,而不管元素的顺序如何。比如说,要选择一个系列中只有两个元素的两个元素,首先,这些元素的顺序无关紧要。
当从给定序列中的 n 个元素中选择 r 个元素时的组合数,
n C r = n!/r!(nr)!
例如,设 n = 5 和 r = 2,然后从 5 = 5 C 2 = 5!/2! 中选择 2 个元素的方法数(5-2)! = 10。
在会议期间,7 名学生可以分配到 1 个三人间和 2 个双人间酒店有多少种方式?
解决方案:
This problem can be interpreted as having to put the 7 students into groups of 3, 2 and 2.
Number of ways to choose 3 students in the triple = 7C3 = 7! / 3!4! = 35
Number of ways to choose 2 out of the remaining 4 students = 4C2 = 4!/ 2!2! = 6
Number of ways to choose 2 students out of the remaining two students = 1
Total number of arrangements = 35 × 6 × 1 = 210.
Hence, 7 students can be assigned to 1 triple and 2 double hotel rooms during a conference in 210 ways.
类似问题
问题 1. 有多少种方法可以选出 5 人组成一个委员会,以使 7 名男性和 6 名女性组成的委员会中至少有 3 名男性?
解决方案:
At least three men on the committee means we can have either exactly three, four or all five men in the committee.
Required number of ways = (7C3 × 6C2) + (7C4 × 6C1) + (7C5)
= + +
= 525 + 210 + 21
= 756
问题 2. 'LEADING' 单词的字母有多少种排列方式可以使元音始终连在一起?
解决方案:
There are 3 vowels in the word. Number of ways to arrange these vowels among themselves = 3! = 6
Now as for the 4 letters, number of ways to arrange = 5! = 120
Total number of ways of arranging the letters = 120 × 6 = 720.
问题 3. 8 个辅音和 5 个元音,4 个辅音和 3 个元音可以组成多少个单词?
解决方案:
Number of ways of selecting 4 consonants out of 8 and 3 vowels out of 5 = 8C4 × 5C3
=
= 70 × 10 = 700
Number of ways of arranging the 7 letters among themselves = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
Number of words that can be formed = 5040 × 700 = 3528000.
问题 4. 如果不允许重复,“LOGARITHMS”这个词可以组成多少个四个字母的词?
解决方案:
Since there are 10 different letters in the word ‘logarithms’.
Required number of words = 10P4
= 10!/6!
= 5040