Permutations or other related questions, questions about permutations

These are very simple permutations.

1 question. It can be counted directly as a permutation problem. a3 4 = 24 species.

Or it can be seen as a combination first. 3 out of 4. c3/4..

Then put 3 different boxes in each A3 3...That is, c3 4 * a3 3 = 24 species.

2 questions. If the balls are all different, the same is true. Can be directly a8 12 = 19958400 species.

You can also draw 8 balls from 12 in combination. c8/12...Then continue the arrangement in 8 different boxes.

a8/8。Total: c8 12*a8 8=19958400.

But if it's 12 balls, it's all the same. You can directly arrange a8 8 = 40320....

Formula: an m=(m!)/(m-n)!

cn/m=(m!)/(m-n)!n!

m!=m*(m-1)*(m-2)*(m-3)..m-(m-1)]n!

n*(n-1)*(n-2)*(n-3)..n-(n-1)] should be understood. Pure hand-hitting.

I don't understand pm. or Q378252287

These two questions are not simple permutations and combinations:1Put 4 different balls into 3 different boxes, but there is no limit to how many each box can be packed, the installation method A3 4 is definitely not right, and it should be classified and calculated.

2.Different from question 1, because the 12 balls are the same, each box has 12 ways to pack, and the results of the empty 1, 2, and 3 boxes are different, so they should also be classified.

I suggest: study the simple ones first, such as the after-class exercises in Chapter 11 of the second year of high school in the textbook, and don't delve into the difficult problems first.

Formula: (n is the subscript, m is the superscript) an m = (n!)/(n-m)!

cn/m=(n!)/(n-m)!m!

n!=n*(n-1)*(n-2)*(n-3)..n-(n-1)]

For example: a5 3=(5!)/(5-3)! =(5*4*3*2*1)/(2*1)

c5/3=5!)/[(5-3)!3!]=(5*4*3*2*1)/[(2*1)*(3*2*1)]

I only remember how the numbers were calculated, and I forgot the formula.

The two problems are similar, for the first, each ball has 3 boxes to choose from, there are three methods, the total method uses the principle of multiplication to have 3*3*3*3=81 kinds, and the second one is similar.

I think so. I look like a simple permutation, but I'm not sure, because math isn't very strong.

anm(n below) = n(n-1)(n-2).n-m）

cnm=n（n-1）（n-2）..n-m）/m（m-1）（m-2）..1

The partition method can be used.

Seven balls are arranged in a row, three planks are inserted into the six gaps formed by the seven balls, and the balls contained in the four parts are divided into four different boxes, and there are c(6,3)=20 ways to distribute them.

c(6,3) is the number of combinations.

In fact, this problem can also be enumerated, because the base is small, it is easier to enumerate.

Solution: They sit next to each other, and if you look at it this way, this is equivalent to letting five mainland players sit first, and then inserting five Hong Kong and Macao players into the air around and in the middle of the mainland players.

5 of the 6 empty selections are put in, there are c(6,5) species, and they are arranged by 5 people, and there are a(5,5) species.

According to the principle of step-by-step multiplication counting, there are a total of c(6,5)*a(5,5)=720 species.

This topic does not seem to be so simple, first arrange 5 mainland players to sit apart, their position is arbitrary, there are a (5, 5) ways to sit, and then arrange five Hong Kong and Macao players, is to eat together should only be 5 empty, 5 people sit 5 empty, you can also sit at will, there are also a (5, 5) ways to sit, and then multiply.

But that's not the end of it, because it includes repetition, but the repetition is very extreme, "the player next to the left of each player is the same", and in each scheme, everyone moves one seat to the left or right, two seats, three seats ......It's the same as the original, which means that nine out of every ten are duplicates. So to remove the duplicates, divide by 10

So it should be a(5,5)*a(5,5) 10

5 mainlanders, sit down first, all the questions are arranged, 5! =5*4*3*2*1=120

Then all 5 Hong Kong and Macao people lined up, and it was also 5!= 120 kinds, respectively inserted in the middle of 5 mainlanders, there are 2 insertion methods, one in front of the mainlanders, the beginning of insertion, and the other is insertion behind the mainlanders. So total: 120*120*2=28800

It is not difficult to find that mainland players sit next to each other, and Hong Kong and Macao players are also like this, so there are 5! *5！species, but there are 20 ways for each scheme (counterclockwise, rotation coincides), so another 5! *5！20 = 720 species.

Q: 023456 these 6 numbers, 3 numbers are a group, which can form several groups of numbers.

Method: Special elements are preferred, solution: special attention can not be put in the first place. There are two cases: 0 and 0.

When you get to 0, let 0 choose one of the ten digits c(2,1) and then choose two a(5,2) from the remaining five digits

At this point, c(2,1) a(5,2)=40

0a(5,3)=60 was not selected

From 60 + 40 = 100

Therefore, 3 numbers are a group, which can form 100 groups of numbers.

I have already told you the method of "3 numbers as a group, you can form several groups of numbers", please find the number of groups of four, five, and six numbers yourself. You can't be lazy when you learn math well....

There are 6 kinds of a group of one number, 5*5=25 kinds of two numbers in a group, 5*5*4=100 kinds of three numbers, 5*5*4*3=300 kinds of a group of four numbers, 5*5*4*3*2=600 kinds of five numbers in a group, and 5*5*4*3*2*1=600 kinds of six numbers in a group. So there are 1,631 species in total. There are 100 types of three numbers in groups.

Because there are at least 3 of the same numbers in any two combinations of 7 numbers out of 11 numbers.

The two groups of combinations are required not to contain more than five identical numbers, which are divided into two situations

1) There are 3 identical numbers, and the different ways to take them are c(11,3)*c(8,4)*c(4,4) 2=5775;

2) There are 4 identical numbers, and the different ways to take them are c(11,4)*c(7,3)*c(4,3) 2=23100

There are 5775 + 23100 = 28875 different ways to take it.

The first group of combinations can be arbitrarily selected first, and the total sample points of the first group of combinations are: c[11,7], and each sample point may appear.

For each set of the first group of combinations, then take the second group, and the second group of combinations does not appear to contain more than five of the same numbers of the first group, then from.

1) 3 of the 7 numbers determined by the first group, and all the 4 numbers that are not taken by the remaining first group; A total of C[7,3]*C[4,4] methods;

2) or 4 out of the 7 numbers set by the first group, and 3 out of the remaining 4 numbers; A total of C[7,3]*C[4,4] methods;

There are c[11,7]*(c[7,3]*c[4,4]+c[7,3]*c[4,4]) combinations, and each of them may appear;

Idea: It happens to be taken out on the fifth time, which means that 2 of the first 4 times in the source state are defective, and the 5th time must be defective. The products are all blind, so you can use a combination (of course, you can use a permutation......）

p=(4c2) hail this source (10c3) = 1 20