Solve the math arrangement problem schedule !

Categorical discussions. a) When Mathematics is in Lesson 1:

The class meeting can be in the afternoon period 1 or in the afternoon period 2 i.e. c(1 2) Because the first period has been stopped by the math class, the physical education class is arranged in the remaining 4 positions like other classes, i.e. a(4 4).

So the total method a(4 4)c(1 2) = 48 b) when math class is not ranked first.

Math class can only be in the morning, so it can be the 2nd, 3rd, 4th period in the morning, so c(1 3), the class will be as usual c(1 2), physical education class cannot be ranked in the 1st period, so there are only 3 places left, i.e. c(1 3), and the remaining three classes can be ranked a(3 3) casually

So c(1 3)c(1 2)c(1 3)a(3 3)=108 In summary, a total of 48+108=156 methods.

Class meeting: c(2,1)=2 types.

1.Mathematics is ranked in the first section in the morning, then all are arranged, a(4,4)=24 kinds, count: 2*24=48 kinds;

2.Mathematics is not ranked in the first section, then mathematics c(3,1)=3 kinds, mathematics, class meeting, and the first section can not be arranged, so physical education c(3,1)=3 kinds, and the rest of the full arrangement a(3,3)=6 kinds, count: 2*3*3*6=108 kinds;

Total: 48 + 108 = 156 species.

a(subscript 4) (superscript 4) *c (subscript 2) (superscript 1) *c (subscript 4) (superscript 1) -a (subscript 3) (superscript 3) *c (subscript 2) (superscript 1) *c (subscript 3) (superscript 1) = 156

There are several ways to do this.

Here's a list of one: first ignore the condition of "no sports in the first session of the morning". Mathematics must be arranged in the morning, put it up first, the class meeting must be scheduled in the afternoon, and it will also be put aside, and the remaining four subjects will be arranged, the first three in the arrangement will be the morning class, and the last one will be the afternoon class.

Then there is a vacant section in the morning, and there are four ways to arrange in the morning; There is also a vacant seat in the afternoon, and the class will be arranged in the afternoon, and there are two ways to arrange it. Counting and sorting method a (subscript 4) (superscript 4) * c (subscript 2) (superscript 1) * c (subscript 4) (superscript 1).

Then consider that the first class is a physical education and deduct it. In that case, the first class is a fixed physical education. The rest of the method is the same as above, but the conditions change:

Arrange 5 classes, 3 in the morning, 2 in the afternoon, math in the morning, class meeting in the afternoon". Counting and sorting method a (subscript 3) (superscript 3) * c (subscript 2) (superscript 1) * c (subscript 3) (superscript 1).

a (subscript 4) (superscript 4) * c (subscript 2) (superscript 1) * c (subscript 4) (superscript 1) minus a (subscript 3) (superscript 3) * c (subscript 2) (superscript 1) *c (subscript 3) (superscript 1).

List the following one by one, don't schedule the class meeting in the morning, the first period of physical education, there are 192 kinds in total, and then after excluding the math class in the afternoon, there will be 156 kinds left.

I didn't arrange it carefully, the method was right.

I wouldn't make a formula like that, and I hope you can understand it

A22 times A44 + C21 times C31 times C31 times A33

Summary: The standard arrangement is to ,.. the natural numbers 1 and 2Pick finger n in order from small to large, such as 12....

Permutation, in general, takes out m (m n) elements from n different elements and arranges them in a certain order, which is called taking out a permutation of m elements from n elements. In particular, when m=n, this arrangement is called all permutation.

Classification:

The key is how to pair up four people in groups (first group and then row):

In the first category, four people are divided into two groups, 3 people and 1 person, and then the two groups are divided into two groups to stand on one of the seven steps.

c(4,1) a(7,2) = 168 different stations;

In the second category, four people are divided into two groups, two people in each group, and then the two groups are divided into two groups to stand on one of the seven steps.

c(3,1) a(7,2) = 126 different stations;

The third group, four people are divided into three groups: 2 people, 1 person and 1 person, and then the three groups are divided into three groups to stand on one of the seven steps.

c(4,2) a(7,3) = 1260 different station methods;

In the fourth category, four people each stand on one of the seven steps, and there are a(7,4)=840 different ways of standing;

Therefore, there are 168 + 126 + 1260 + 840 = 2394 kinds of different station methods.

Divide people first, there are 1, 3 points, 2, 2 points, 2, 1, 1 points, 1, 1, 1 points, 3 points: 4*7*6=168

2,2 points: 4 * 3 2 * 7 * 6 = 252

2,1,1 points: 4*3 2*2*7*6*5=25201,1,1,1 points: 7*6*5*4=840

So there are a total of 168 + 252 + 2520 + 840 = 3780 species.

The simplest solution:

A, B, C, D can stand on any step, each person has 7 ways to stand, a total of 7*7*7*7, minus the situation where 4 people are standing on the same step (7 types), that is, what is sought, 7 4 - 7 = 2401 -7 = 2394.

The answer is very clear, there are two kinds, the first is 5 men choose 2 and 3 women choose 1, the second is 5 men choose 1 and 3 women choose 2, the sum of the two is that both men and women are asked to participate. And the answers you listed are exactly repeated, like Male A and Female 1, and then choose 1 male B from the remaining 6, so that it is exactly the same as Male B and Female 1, and then Male A, so you have twice as many answers.

You're doing it repetitively. For example, I took Male 1 and Female 1 first, then Male 2, and I took Male 2 and Female 1 first, and then Male 1, these two are the same.

There are duplicates in the results of 4*4*3, that is, 3 times are arranged '3 classes are in factory A', '3 classes are all in factory A', and 3 classes are arranged 2 times (3*3*2=18 times), so it is necessary to remove the extra (3-1)+18 2=11

Finally, 48-11 = 37 times.

There are duplicates.

For example, the first class goes to A, the second class goes to A, and the third class goes to B.

When the second shift goes to A, the first shift goes to A, and the third shift goes to B.

Just subtract these duplicates.

It is clearer to use the "repulsion" method, and there will be no weight

There are any kinds of classes: u(4,3)=4 3=64 kinds;

2. Factory A does not have a class to go, that is, 3 classes go to another 3 factories at will, there are u(3,3)=3 3=27 kinds;

So, in total: 64-37 = 27 species.

Take two groups as an example: 1, select 7 out of 12 people as a group, and the remaining 5 people in a group c12,7*c5,5.

2. 7 out of 12 people were selected as Group A, and the remaining 5 were selected as Group B C12,7*C5,5.

3. Select 7 from 12 people as a group of cracked dust, and the rest of the foundation digging a group of 5 people, and then arrange group A and B C12, 7 * C5, 5 * A2, 2.

4. 6 out of 12 people were selected as the A Fengzen group, and the remaining 6 were selected as the B group C12,7*C5,5.

5. Select 6 from 12 people as a group, and the remaining 6 people in a group c12,6*c6,6. The same goes for the three groups. Welcome!