Permutations and combinations, simple, discernment, thank you

LZ: This is a typical mistake of yours, this calculation must be repeated, and to understand it this way: now there are athletes with numbers of , (the first 6 are male athletes, and the last 4 are female athletes) Think like you think about it, if you choose a female athlete with a number 7 for the first time, and choose four from the remaining 9 for the second time, if the selected four include an athlete with a number 8, this situation is the same as the first time the female athlete is numbered 8, and the second time you choose a number 7, so it is repeated So it can't be counted like that, it can only be classified like the answer:

1 female, 4 male, 2 female, 3 male, 3 female, 2 male, 4 female, 1 male, or according to the difference method

Hope it can help get you!

c c, your idea is good, you are very smart.

If you set them to be 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 and then connect them separately or select Yes, and you can also have this idea.

There are some minor problems with the thinking.

It should be sorted first, then step-by-step, or found using the indirect method. According to your train of thought, it seems to be right, but because you directly use the step-by-step solution, it is true that all the situations are considered, but some selection methods are repeated and superimposed in your calculations, so they are wrong. The reason why it is classified first and then divided into steps is to not add the same selection method without repeating it.

In fact, it doesn't matter whether there are 24 numbers or 30 numbers, the problem is to take any repeatable 4 or 5 numbers in 1 6.

Let's take the relatively simple AAAB type as an example:

First, choose 2 numbers from 1 6 as a and b respectively. There are 6 5 = 30 species.

Secondly, 3 A's and 1 B are arranged arbitrarily. There are 4 types.

Therefore, there are a total of 30 4 = 120 types of AAAB type.

The above steps are generic steps. Let's explain how steps 1 and 2 are solved

Step 1: Decide which of the 6 numbers to decide.

This step seems relatively simple, just take an orderly arrangement, but in fact there is a pitfall.

For example, AABC is taken from 3 out of 6 numbers, i.e. there are a(6,3)=120 species. However, there is only one b and only one c, and since the order is not considered in the end, it is the same whether a number is b or c. For example, 1231 can be regarded as b=2, c=3, or b=3, c=2.

Therefore, a(2,2) must be eliminated in the end, and the result of step 1 is a(6,3) a(2,2)=60 species.

For example, AABCD, taking 4 out of 6 numbers is A(6,4), and the order of any arrangement of BCDs is A(3,3), and the result is A(6,4) A(3,3)=60 species.

Step 2: Decide on the sort order.

The arrangement of AABC is A(4,4) A(2,2)=12. where a(4,4) represents the arbitrary arrangement of the four letters, and a(2,2) represents the arbitrary arrangement of the two a's.

Similarly, AABCD has an arrangement of a(5,5) a(2,2) = 60 species.

From the above two steps, we can find: AABC has 60 12=720 species. AABCD has 60 60 = 3600 species.

In permutations, a22 represents the number of cases where 22 elements are taken out of n elements and arranged, i.e., a22 = n!/(n-22)!

The reason for dividing by a22 is to translate the problem into a calculation of the number of combinations rather than permutations. The number of combinations is the number of m elements taken out of n elements, regardless of the order of the elements, whereas the number of permutations is the order in which the elements are considered.

When you need to calculate that 22 elements are taken out of n elements and arranged in an order, if you directly use the number of permutations formula, the result will include all the different permutations, i.e., n!/(n-22)!。But sometimes we only care about the choice of elements, without considering their order, and then we need to convert the result into a combination number.

Since 22 elements are taken out of n elements and arranged in a permutation, the number can be expressed as c22 = a22 22!where c22 is the number of combinations. Therefore, when calculating the problem, the answer needs to be divided by a22, that is, the final result obtained by Jianyou is c22.

Hope the above answers are helpful to you! If you have any questions, please feel free to ask.

Assuming that three people are A, B, and C have 3 sets of schemes A, B, or B, AC or C, AB, but there are 2 villages, 3 2 = 6 arrangements can be used.

Or consider the village selection, there is a plan of 1 person in village A, 2 people in village B, 2 people in village A, and 1 person in village B.

1 person in village A and 2 people in village B, there are c(3,1)=3 kinds of anmian.

2 people in village A and 1 person in village B, there are c(3,2)=3 arrangements.

So there are 3+3=6 arrangements.

Step 1: Divide into two groups, one group of two and one group of one.

Step 2: Two groups to two villages are all arranged.

Finally multiply. You are welcome to ask.

Choose 2 out of 5 people to go to the office with 2 people and the remaining 3 people to go to the office with 3 people missing.

c<5, 2> = 10;

Choose 2 out of 7 people to go to the office with 2 people and 3 out of the remaining 5 people to go to the office with 3 people out of the gap.

C<7, 2>C<5, 3> =21 10 = 210.

The subject can directly perform simple calculations on this, and the subject practices more example problems and uses formulas to perform calculations, as follows.

5532;Seed arrangement. The second method - interpolation method: n different elements are arranged in a row, and m elements are not adjacent to each other, then the rest (n-m) can be arranged first...

There are three scenarios for the election of 3 delegates, at least one female student:

1 female student and 2 male students.

2 female students and 1 male student.

3 female classmates.