How many types of 3 numbers are there in a combination of N numbers?

This means that there are three choices (1,2,3) for each number

Then, all combinations are 3 n

The specific meaning is:

Let's say it's a 3-digit number:

Determine the number on the number of digits: Three.

Determine the number on the ten-digit number: three.

Determine the number on the hundreds: three.

So all the possibilities are 3*3*3=3 3=27.

Extended to n digits, that is:

Determine the number on the number of digits: Three.

Determine the number on the ten-digit number: three.

Determine the number on the hundreds: three.

Determine the number on the n number: three kinds.

So all the possibilities are 3*3*3*. n three = 3 n

So the answer is 3 n

The reason why multiplication instead of addition is used is because:

Addition principle: to do one thing, to complete it can have n types of methods, in the first type of methods there are m1 different methods, in the second type of methods there are m2 different methods ,......There are mn different approaches in the nth category of approaches, so there are common ways to accomplish this.

n=m1+m2+…+mn different methods.

Multiplication principle: to do one thing, to complete it can have n steps, in the first step there are m1 different methods, in the second step there are m2 different ways ,......There are mn different ways to do this in the nth step, then there are common ways to accomplish this.

n=m1×m2×…×mn

In short, addition applies to a thing in multiple ways, each of which can accomplish the thing, and there is no order between the various methods. Multiplication, on the other hand, has only one solution for one thing, but this solution requires multiple steps to complete, and there is a sequence between the steps.

In the matter of composing a number, each round of selecting numbers appears as a step rather than a method, and there is a sequence between the steps, so multiplication is used.

If this question is changed to: There are 3 methods in the selection method A arrangement, there are 3 methods in the selection method B arrangement, and there are 3 methods in the selection method C arrangement, then how many methods are there in total? In this case, the principle of addition applies, and the answer is 3+3+3=9

Hope you're satisfied.

1. Permutations and combinations of three numbers, such as 000, 001, 002....There are 1000 species in 999.

2. Solve the problem and take the course for the New Year:

It can be done by permutations and combinations:

In the single digit, you can take a total of ten numbers from 0 to 9, which means ten possibilities; In the ten digits, Chong Que Lap can take a total of ten numbers from 0 to 9, which means ten possibilities; In the hundreds, you can get a total of ten numbers from 0 to 9, which means ten possibilities, so 10 10 10 = 1000 possibilities to slip.

It can be counted as a number of numbers:

001 to 999 can be regarded as 1 to 999 calculation, 999-1+1=999, plus the number 000, that is, 999+1=1000.

1000 kinds of liquid preparation.

Permutations and combinations of 3 numbers, e.g. 000,001,002....In this way, 999 can choose 0 to 9, 10 numbers can be selected by 10 digits, 0 to 9, 10 numbers can be selected by 10 digits, and 0 to 9, 10 numbers can be selected by single digits.

So the total variety is 10 10 10 10 for a total of 1000 imitation halls.

How many combinations of three numbers are there to divide the situation:

1. There are 6 combinations of different three numbers (except zero) (such as: 1, 2, 3, etc.).

2. There are 3 combinations of two identical and one different numbers (except zero) (such as 2, 2, 3).

3. There is one combination of three identical numbers (except zero) (e.g. 2, 2, 2).

So, the three numbers are combined in a different way.

The calculation formula for permutations and combinations is: the number of permutations, take m from n and arrange them down, there are n(n-1)(n-2).n-m+1), i.e. n (n-m).

The number of combinations, taking m from n, is equivalent to not arranging, that is, n [(n-m)m].

The meaning of the title is not very certain. Let's discuss it on a case-by-case basis.

1. If it is "33 numbers are a group of 6 for every group, how many groups can be arranged?" ”

That's 33 6=,Only 5 groups can be arranged.

2. If it is "6 out of 33 to arrange, how many schemes are there?" ”

Then there are 33 in total!/(33-6)!797448960 permutation schemes.

3. If it means that "33 numbers are grouped in groups of 6, how many grouping schemes can there be?" “

This one is more complicated, and there are two steps. Because, up to 30 can be divided into 5 groups.

Therefore, the first step, 30 out of 33, 33!/30!/(33-3)!= 5456 cases.

In the second step, 30 numbers are divided into groups of 6, divided into 5 groups, a total of 30!/(6!5 = 1370874167589326400 scenarios.

Two-step overlay,There are a total of 5456*1370874167589326400 = 7479489458367364838400 schemes.

Hope it helps!

33 numbers are grouped with six numbers, and six groups can be excluded, because 33 6 = 5 still has three left, then the remaining three must also be a group, therefore, 33 numbers can be arranged in groups of six in a group of six numbers.

Then there are 33 in total!/(33-6)!797448960 permutation schemes.

The calculation process according to the question is as follows: 6 out of 33

1107568 group.

Regardless of the order, it is a matter of composition. 40 choices = 40!/7!/(40-7)!18643560 different combinations.

If you want to think about the order, it is the question of choosing the order. 7 out of 40.

40!/(40-7)!93963542400 permutation schemes.