如果不允许数字重复，使用数字 1 到 9 可以组成多少个 4 位数字？
排列被称为按顺序组织组、主体或数字的过程，从集合中选择主体或数字，被称为组合，其中数字的顺序无关紧要。
在数学中，排列也被称为组织一个群的过程，其中一个群的所有成员都被排列成某种顺序或顺序。如果组已经排列，则置换过程称为对其组件的重新定位。排列发生在几乎所有数学领域。它们大多出现在考虑某些有限集合上的不同命令时。
置换公式
在排列中，从一组 n 个事物中挑选出 r 个事物，没有任何替换。在这个挑选的顺序。
nPr = (n!)/(n – r)!
Here,
n = group size, the total number of things in the group
r = subset size, the number of things to be selected from the group
组合
组合是从集合中选择数字的函数，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。这种组合被称为一次合并n个事物而不重复。组合起来，顺序无关紧要，您可以按任何顺序选择项目。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 个东西是从一组 n 个东西中挑选出来的，挑选的顺序无关紧要。
nCr =n!⁄((n-r)! r!)
Here,
n = Number of items in set
r = Number of things picked from the group
用数字 1 到 9 可以组成多少个 4 位数字。如果数字不允许重复？
回答：
Repetition of the digit is not allowed. So, for the first digit we have 9 option for second digit we have 8 option for third digit we have 7 option and for fourth digit we have 6 option
There are total 9 digit from which we have to select 4, repetition is not allowed
Total no. of ways = 9P4
= 9!/(9-4)!
= 9!/5!
= 3024
类似问题
问题1：用数字1到9可以组成多少个5位数字。如果数字不允许重复？
回答：
Repetition of the digit is not allowed. So, for the first digit we have 9 option for second digit we have 8 option for third digit we have 7 option for fourth digit we have 6 option and for fifth digit we have 5 option
There are total 9 digit from which we have to select 5, repetition is not allowed
Total no. of ways = 9P5
= 9!/(9-5)!
= 9!/4!
= 15,120
问题2：用数字0、1、2、3可以组成多少个3位数。如果允许重复数字？
回答：
Repetition of digit is allowed. So, for the ones place we have 4 option i.e., 0,1,2,3 similarly for tens place we have again 4 option i.e., 0,1,2,3 and for the hundredth place we have 3 option i.e., 1,2,3 we can’t take 0 at hundredth place because if 0 will be filled at hundredth place it will not become 3 digit number it will be taken as two digit number.
Total no. of three digit number = 3 × 4 × 4
= 48
问题3：用数字0、1、2、3、4可以组成多少个5位数字。如果允许重复数字？
回答：
Repetition of digit is allowed. So, for the ones place we have 5 option i.e., 0,1,2,3,4 similarly for tens place we have again 5 option i.e., 0,1,2,3,4 for the hundredth place we have 5 option i.e., 0,1,2,3,4for the thousandth place we have 5 option i.e., 0,1,2,3,4 and for the ten thousandth place we have 4 option i.e., 1,2,3,4 we can’t take 0 at ten thousandth place because if 0 will be filled at ten thousandth place it will not become 5 digit number it will be taken as 4 digit number.
Total no. of five digit number = 4 × 5 × 5 × 5 × 5
= 2500
问题 4：如果不允许数字重复，可以用数字 (3,5,7,9,1,0) 组成多少个 4 位偶数？
回答：
For even number unit digit must be 0, Now the remaining digits are 5 i.e., 3,5,7,9,1 now for the thousand place we have 5 option for the hundredth place we have 4 option and for the tens place we have 3 option
Total no. of 4 digits even number can be formed = 5 × 4 × 3
= 60