Alphabet 可以组成多少种不同的 3 个字母组合？
顾名思义，数字系统是一种数学系统，用于使用各种符号和变量来表示数字。在数制下，可以绘制在数轴上的数字，通常称为实数，由一组值或数量表示。根据它们的各种特征，不同种类的数字被分为不同的集合或组。例如，有理数是可以以 p/q 形式表示的任何整数，其中 q 是非零整数。十进制、二进制、八进制和十六进制是不同类型系统的示例。
组合
它被定义为从给定序列中选择一个、两个或几个元素的过程，无论它们出现的顺序如何。如果您从只有两个元素开始的系列中选择两个组件，则这些元素的顺序无关紧要。
组合配方
当从序列中的 n 个元素中选择 r 项时，组合的数量为
nCr = n! / r! (n – r)!
例如，让 n = 7 和 r = 3，然后从 7 = 7 C 3 = 7!/3!(7 – 3)! 中选择 3 个元素的方法数！ = 35。
Alphabet 可以组成多少种不同的 3 个字母组合？
解决方案：
If the repetition of letters is allowed, then each alphabet can be chosen from the given 26 alphabets.
Hence total number of combinations = 26C1 × 26C1 × 26C1
= 26 × 26 × 26
= 17576
However, if repetition is not allowed, then the number of combinations = 26C1 × 25C1 × 24C1
= 26 × 25 × 24
= 15600
类似问题
问题 1. 给定一个存钱罐，里面有 20 个硬币，请确定它持有的五分硬币、一角硬币和四分之一硬币的排列数。
解决方案：
In this case, the order of the coins clearly does not matter. There’s also no mention of whether or not recurrence is permitted.
Following the formula, we have: n+r−1Cr = n+r−1Cn−1 , in the case of r number of combinations from a sequence with n number of elements where elements can be repeated:
The number of possible pairings = 20+3-1C20 = 22C20
= 22! / (22 – 20)! 20!
= 11(21)
= 231
问题 2. 如果我们只有一间三人间和两间双人间，请告诉我有多少种不同的方法可以让 7 名学生参加一次大学旅行。
解决方案：
This problem might be understood as having to divide the seven pupils into three groups of three, two, and two students.
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
The number of possible ways to choose two students from the remaining two students is equal to one.
Total number of arrangements = 35 × 6 × 1 = 210.
During a conference, 7 students can be assigned to 1 triple and 2 double hotel rooms in 210 different ways.
问题 3. 确定由 7 男 6 女组成的小组，其中至少有 3 名男子组成的五人委员会的数量。
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.
Number of arrangements when there are 3 men and 2 women on the committee = (7C3 x 6C2) = 525
Number of arrangements when there are 4 men and 1 woman on the committee= (7C4 x 6C1) = 210
Number of arrangements when there are all 5 men on the committee = (7C5) = 21
Total arrangements = 525 + 210 + 21
= 756
问题 4. 找出单词“LEADING”中的字母排列方式的数量，以使元音总是一起出现。
解决方案：
If the vowels are to appear together, they would form a separate letter in the word. Hence we are left with 4 + 1 = 5 letters, which can be arranged in 5! = 120 ways.
Furthermore, there are 3! = 6 ways to arrange the vowels together.
Total number of ways of arranging the letters = 120 x 6 = 720.
问题 5. 找出可以由八个辅音和五个元音组成的具有四个辅音和三个元音的单词的数量。
解决方案：
Number of ways of selecting 4 consonants out of 8 and 3 vowels out of 5 = 8C4 x 5C3
= \frac{8 ×7 ×6 ×5 ×4!}{4! × 4!} × \frac{5 ×4 ×3!}{3! × 2!}
= 70 × 10 = 700
Number of ways of arranging the 7 letters among themselves = 7! = 5040
Number of words that can be formed = 5040 × 700 = 3528000.