How many permutations are there from one to five?

If the 6 numbers are different from each other, then there are a(6,6)=720 permutations.

But there are 3 groups of two identical, so you need to divide by a(2,2)a(2,2)a(2,2)a(2,2)=8

So at the end there are 720 8 = 90 permutations.

From n different elements, any element m (m n, m and n are natural numbers, the same below) elements are arranged in a certain order, which is called taking out an arrangement of m elements from n different elements; The number of all permutations of m (m n) elements from n different elements is called the number of permutations of m elements from n different elements, which is represented by the symbol a(n, m).

Calculation formula: >< also specifies 0!=1(n!Represents n(n-1)(n-2).1, that's 6! =6x5x4x3x2x1

Whether there are duplicate numbers in the 6 numbers or not, the arrangement scheme is different. In addition, if you simply talk about permutation, 0 can also be put in the first place.

A full arrangement of 6 different numbers has a 6!= 720 kinds, if there are duplicate numbers in the middle, for example, there are n numbers in the middle, for example, the number a has n repeating arrangements, and the repeated arrangement of these n numbers should be subtracted, because these n same numbers are reversed, or the same arrangement. The method of subtraction is that the result of the full permutation is the number of all permutations of n, which is 6!

n!。Under the above premise, it is necessary to discuss the different situations according to the six figures. There are a total of 10 scenarios:

abcdef --6!= 720 species;

abcdee --6!/2!= 360 species;

abcddd --6!/3!= 120 species;

aabbcc --6!/2!/2!= 180 species;

abcccc --6!/4!= 30 species;

abbccc --6!/2!/3!= 60 species;

abbbbb --6!/5!= 6 species;

aabbcc --6!/2!/2!/2!= 90 species;

aabbbb --6!/2!/4!= 15 species;

aaabbb --6!/3!/3!= 20 kinds.

123456 has 6!arrangement, i.e. 720 kinds of Qi keys. This is because the first digit can be filled in any of the six digits (there are 6 choices), the second digit can choose one of the remaining five digits (there are 5 choices), and so on until the last digit is left with only one optional number.

Therefore, the total arrangement is 6 5 4 3 2 1 = 720.

123456 can repeat the 6 to the 6th power, not 6x5x4x3x2x1=720 permutations.

There are a total of 6 numbers to choose from, and each guess is used once, so each time there is one less choice for each digit spike or . There are 6, 5, 4, 3, 2, 1 choices in turn, and there are a total of 6 5 4 3 2 1 = 720 permutations.

If there is no 0, the result is a full permutation of 5:

If there is 0, 0 cannot be put in the first place, and the result is:

For example, a three-digit type consisting of 5*5*4=100 (species) consisting of a two-digit category: 5*5=25 (species).

There are a total of 125 kinds, if the product of the combination formula is the largest, then it must be the largest of three or two numbers, in this case, the product of these six combinations of three digits is the largest, all equal to 60, and the product of 45 and 54 is the largest, both equal to 20.

If there is no 0, the result is a full permutation of 5:

If there is 0, 0 cannot be put in the first place, and the result is:

1) Permutation: In general, take out m (m n) elements from n different elements and arrange them in a certain order, which is called a sequence, arrangement, permutatio...

Method 1: Make 6 2s and 5 1s all 1 with 11 10 9 ....2 1 methods, and 6 2 2 arrangement 6 5 4 3 2 1 kinds, 5 1 1 5 4 3 2 1 arrangement, the desired arrangement method has a total of traces:

11×10×9×…2 1) 6 5 4 3 2 1) 5 4 3 2 1) = 462 species.

Method 2: The number of methods to select 6 positions from 11 positions to arrange 6 cong letter 2 is c(11,6), and there is only 1 way to arrange the remaining 5 1s, and the total number of permutations sought is c(11,6)=462 kinds.

If there is no 0, the result is a full arrangement of Changsui 5:

If there is 0, then 0 can not be placed in the first place of the resistance to dismantling, and the result is:

c54=c51=5

Or. c54=(5*4*3*2) (4*3*2*1)=5 There are 5 different elements, divided into 4 groups:

c54=(comma 5*4*3*2) 4!=5<>

Definition of permutation:

From n different elements, any different elements of m (m n, m and n are natural numbers, the same below) different elements are arranged in a certain order, which is called taking out an arrangement of m elements from n different elements; The number of all permutations of m (m n) elements from n different elements is called the number of permutations of m elements from n different elements.