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Anonymous users2024-02-07Combination + permutation problem People are divided into 3 groups, and there must be a group of two people, that is, 4 choose 2, c (2 groups go to 3 houses, it is a permutation problem, that is, p(3
3)＝6；3. The total number of schemes: c(2
4）*p（3
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Anonymous users2024-02-06A is in second place, B is in third place, and the remaining three have p3 = 6 combinations.
A is in second place, B is in fourth place, and the remaining three have p3 = 6 combinations.
A is in third place, B is in second place, and the remaining three have p3 = 6 combinations.
A is in third place, B is in fourth place, and the remaining three have p3 = 6 combinations.
A is in fourth place, B is in second place, and the remaining three have p3 = 6 combinations.
A is in fourth place, B is in third place, and the remaining three have p3 = 6 combinations.
A is in fifth place, B is in second place, and the other three have p3 = 6 combinations.
A is in fifth place, B is in third place, and the remaining three have p3 = 6 combinations.
A is in the fifth place, B is in the fourth place, and the remaining three have p3 = 6 combinations.
There are 9*6=54 combinations.
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Anonymous users2024-02-05To answer this question, you can consider whether A is ranked last.
1.A is the last place.
Select 1 person from the 3 people in Bing Ding E to rank first A under 3 on 1, and then the remaining 3 people (including B) are all ranked A under 3 and 3 on 3, a total of 18 kinds.
2.A is not the last one.
2 people are selected from the 3 people in the first and last place, and the position is 3 under 3 and 2 under A, and then the remaining 3 people are all ranked in a 3 and 3 up, a total of 36 kinds.
In summary, there are 18 + 36 = 54 kinds.
Welcome, and the review!
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Anonymous users2024-02-04This is not very easy to play, if it is from 50 people to choose one, then I hit a50 (1), this 1 is equal to 50 above, I hope the landlord understands.
There are a50 (13) * a37 (13) * a24 (12) * a12 (12) * a4 (4).
50 people are divided into 2 13s, 2 12s, and then each row can be arranged again, so there are a50 (13) * a37 (13) * a24 (12) * a12 (12).
These 4 rows have A4 (4) arrangement.
A and B are just next to each other, A and B are regarded as a person, the total number of people is seen as 49, divided into a 13, 3 12, and the other algorithms are the same as above, but pay attention to multiplying by 2, because A and B have two standing methods.
The answer is calculated, but it shouldn't be difficult, and it can be divided.
The second question is, if there are 13 people in a row, then what is the middle, as long as there is one person who is the 7th?
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Anonymous users2024-02-03This question is very popular! Rolling socks tremble. Hello landlord!
This problem can be calculated as follows: to arrange these six numbers in a good column, there are a6,6 a2,2=360 kinds; First subtract 4 in the third place case a5,5 a2,2=60;Subtract the number of cases adjacent to 3 and 5 a2,2*a5,5 a2,2=120;Finally, add 4 to the situation where the third place is defeated and 3 and 5 are adjacent to each other, the number a2,2*a4,4 a2,2=24. In summary, the obtained result is 360-60-120+24=204.
I hope that the landlord can answer correctly, don't be misled, don't mislead other friends, and don't let the correct person be depressed. Welcome!
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Anonymous users2024-02-02A is in second place, B is in third place, and the remaining three have p3 = 6 combinations.
A is in second place, B is in fourth place, and the remaining three have p3 = 6 combinations.
A is in third place, B is in second place, and the remaining three have p3 = 6 combinations.
A is in third place, B is in fourth place, and the remaining three have p3 = 6 combinations.
A is in fourth place, B is in second place, and the remaining three have p3 = 6 combinations.
A is in fourth place, B is in third place, and the remaining three have p3 = 6 combinations.
A is in fifth place, B is in second place, and the other three have p3 = 6 combinations.
A is in fifth place, B is in third place, and the remaining three have p3 = 6 combinations.
A is in the fifth place, B is in the fourth place, and the remaining three have p3 = 6 combinations.
There are 9*6=54 combinations.
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Anonymous users2024-02-01Combination + permutation problem People are divided into 3 groups, and there must be a group of two people, that is, 4 choose 2, c (2 groups go to 3 houses, it is a permutation problem, that is, p(3
3)＝6；3. The total number of schemes: c(2
4）*p（3
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Anonymous users2024-01-31First make a split and then consider the arrangement, by the title 12
Split into 4 different positive integer sums, only in the following cases since each case involves a full permutation of 4 people.
then the total number of requested funding methods is:
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Anonymous users2024-01-30There is already the correct answer ahead. Here's a different way of thinking:
Divide 120,000 yuan into 12 sticks, there are 11 empty sticks between the sticks, and take 3 of the 11 empty spaces to get a division of 4 people's contributions.
So there are a total of: 11c3 = 165 ways to contribute.
And because the amount of capital contributed by each person is different, the same contribution must be subtracted.
4 people contribute the same: (3,3,3,3) 1 type of 3 people contribute the same: (1,1,1,9); 2,2,2,6) A total of 8 kinds of 2 people contributed the same:
1,1,2,8); 1,1,3,7); 1,1,4,6); 1,1,5,5); 2,2,1,7); 2,2,3,5); 2,2,4,4); 3,3,1,5); 3,3,2,4);4,4,1,3) 96+12 = 108 species.
165 - 1 - 8 - 108 = 48 (answers).
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Anonymous users2024-01-29First of all, the second question is divided into three piles, two books in each pile, two of the six books are selected to form the first pile, and then two books are drawn from the remaining four books to form the second pile, and the last two books naturally become one pile. However, since we don't consider the issue of order, we need to consider the case of duplication. For example, "1st Pile AB, 2nd Pile CD, 3rd Stack EF" and "1st Stack CD, 2nd Stack AB, 3rd Stack EF" are the same stacking method.
Therefore, we need to find out the same way of dividing the pile to cause different arrangements, i.e., 6 types, and divide them by the arrangement to ensure the completeness of the way. So the column formula and the result are shown in the first equation in the figure.
Let's look at the first question. This question requires that the books that have been divided into piles be given to A, B and C, which requires a question of order (since when the piles are divided, A gets the first pile of books and B gets the first pile of books are two different divisions). So multiply the first question by 6 and the arrangement of 3 people.
I took the college entrance examination this year, 2 volumes of science in the country, 142 in mathematics, and if you don't understand anything, you can ask me.
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