High school math coloring problems, high school sophomore math coloring problems

This question. No.

First of all, I don't know if the ring in the middle is painted.

Secondly, I have never heard of the problem of coloring more than 2 concentric circles, because then the position of the dividing rings will lead to many different situations.

Another: I know a formula about the coloring of rings, but I don't know if it's the answer you want.

I don't remember the specifics.,Now push it:

For example, if a ring is divided into n parts, there are k colors to color, and the adjacent parts of the color cannot be the same, so how many coloring methods there are.

The first air must have K painting methods.

The second air has a K-1 painting method.

There are k-1 coating methods from the third space to the nth space.

So, if the nth empty color is the same as the first one, it doesn't fit the topic.

However, when the nth empty color is the same as the first one, it can be seen as dividing the ring into n-1 colors and coloring it (this step needs to be thought through).

Therefore, the n-empty coloring method is to subtract the number of n-1 empty coloring methods by the n-1 power of k times (k-1).

The n-1 empty coloring method is to subtract the n-2 empty coloring method by the n-2 power of k times (k-1).

By analogy, the n-space coloring method is to subtract the n-2 power of k times (k-1) from the n-2 power of k times (k-1).

Subtract the n-3 power of k-times (k-1) and subtract the n-4 power of k-times (k-1). Subtract k

The above equation proposes k, and uses the equation of the equal ratio sequence to find the method of n empty coating k color.

k*<(k-1)^(n-1)-[1-(k-1)^(n-2)/(2-k)]>

It is curly braces that indicate power.

That's the rough way

Dizzy, as long as you use the permutations and combinations of the sophomore year, this kind of topic is just pediatrics, take a good look at the sophomore math book! I don't even have to look at the formula with the method of the sophomore year of high school.

I'm confused too. I didn't understand the formula. The answer should be fine.

You're a ring problem. It's not the same as normal lattice coloring. Because it is evenly divided, there is no difference between the No. 1 position and the No. 2 position.

The difference can only be seen when it is all painted at the end. So there are relatively few methods.

It is better to send the formula in more detail and then study it.

Make the topic clear, preferably with a diagram.

The answer is no problem, it is right.

To understand it this way, we call the four colors 1, 2, 3, and 4. You may wish to record the color of b as color 1,1) When b is the same color as e, then e is also 1, so d cannot be 1, we mark the color of d as 2, and f can be filled in with 2, 3 or 4. If f fills in 2, c has two options: 3 and 4; If f does not fill in 2 (if f fills in 3, then c can only fill in 4; If f is 4, then c can only be 3), which is also the case.

So there are a total of cases of filling in b and then f and then c in this way: 1x2+2x1 - this is a bit right with the answer.

2) When b is the same color as d, then d is also 1, we mark e as 2, and f can be filled in with 3 or 4.

If f is filled with 3, c has two options: 2 and 4; If f fills in 4, c has two options: 2 and 3;

Then there are a total of cases of filling in b and then f and then c: 1x2 + 2x1.

Add (1) and (2) above together, that is: 2*(1x2+2x1) kinds, so the result is consistent with the answer.

I'm throwing bricks and stones, and I believe there will be other better ways to understand.

There are four situations to choose the surface of the triangular pyramid P-ABC for ABP coloring, and then choose the surface BB1C1C for coloring, which is divided into two possibilities: the same color of the surface BB1C1C and the surface ABP, and the different colors of the surface BB1C1C and the surface ABP.

If the surface BB1C1C and the surface ABP are the same color, then the surface PBC has 3 cases, the surface APC has 2 cases, and then divided into two possibilities, the first one: the surface PBC and the surface ABB1A1 color is the same, then the surface AA1C1C has a case, the surface AA1B1C1 has a case, the total 4 * 3 * 2 * 1 * 1 * 1 * 1 = 24 (kinds) The color of the surface PBC is different from the surface AB1A1, and the color of the surface AB1A1 has two possibilities, one is that it may be the same as the surface PAC, It may also be different from the surface pac, if the face abb1a1 is the same as the surface pac, then the surface aa1c1c has two cases, and it is easy to know that a1b1c1 has a case, counting 4*1*3*2*1*2*1=48;If the noodles abb1a1 is different from the noodles pac, there is a case of the noodles abb1a1, and there is a situation between the noodles aa1c1c and the noodles a1b1c1, which is 4 1 3 2 1 1 1 24

In the same way, if the face BB1C1C and the face ABP are not the same color, there are 96 cases with a total of 192 types.

Solution: First apply the three sides of the triangular pyramid P-ABC, and then paint the three sides of the triangular prism, with a total of C31 C21 C11 C21=3 2 1 2=12 different coating methods

So the answer is: 12

Solution: The two endpoints of each line segment in the figure are painted with different colors, which can be classified according to the type of color painted, b, d, e, f use four colors, then there are a44 1 1 = 24 coloring methods;

B, D, E, F with three colors, then there are A43 2 2 + A43 2 1 2 = 192 coloring methods;

b, d, e, f with two colors, then there are a42 2 2 = 48 coloring methods;

According to the principle of differential counting, there are 24 + 192 + 48 = 264 different coloring methods.

Apply the A area first, there are 5 types.

There are 4 types of recoating the B area.

Re-coating C area, there are also 4 kinds, divided into two categories: the same as A, different from A, if C is the same as A, then D has 4 kinds, and not the same as A, so that E has only 3 ways to paint If C is different from A, then D has 4 kinds, and it should also be divided into two categories, the same as A, different from A, if D is the same as A, then E has 4 kinds; If D is different from A, then E has 3 types of finalities in common:

5*4*[1*4*3+3*(1*4+3*3)]=1020 If 4 colors are used, the analysis process is the same as above.

The result is: 4*3*[1*3*2+2*(1*3+2*2)]=240 species.

What about the figure??? Don't know how the 5 regions are distributed and the relationship between them? How to calculate...

Apply a first, 5 kinds: (5).

b, 4 types: (5*4).

c, 4 kinds: (5*4*3+5*4*1), because one of them (5*4*1) is the same as a.

d, 4 types: (5*4*3*3+5*4*3*1+5*4*1*4), when c and a are different, d may be the same as a (5*4*3*1).

e, when d and a are different, there are 3 kinds: (5*4*3*3+5*4*1*4)*3, and when d and a are the same, there are 4 kinds (5*4*3*1)*4

There are a total of (5*4*3*3+5*4*1*4)*3+(5*4*3*1)*4=1020 kinds.

4 types are the same as above, (4*3*2*2+4*3*1*3)*2+(4*3*2*1)*3=240 types.

In this case, it is best to use the recursive method to do it;

Let the circle be divided into n regions, and the coloring method is an;

Then when dividing the circle into (n+1) regions, paint the first n regions first;

1. If the color of the first area is different from the penultimate area, the first n area is colored by an (you can combine the last area with the penultimate area), and the last (n+1) area is colored, in this case, a total of 3an.

2. If the color of the first region is the same as that of the penultimate region (at this time, 1, n, n+1 are combined into one piece), then the first n areas are painted with a(n-1), and the last (n+1) area is painted with 4 methods, in this case a total of 4a (n-1).

Therefore, we find the recursive relation: a(n+1)=3an+4a(n-1); n≧3) a2=20，a3=60；

In this way, we can know that a4 = 3x60 + 4x20 = 260, and the a5 = 3x260 + 4x60 = 1020.

When the color number k=4, we can also get: a(n+1)=2an+3a(n-1).

At this time, a2=12, a3=24, a5=240;

When painted with k colors, you can get a more general formula:

a(n+1)=（k-2）an+（k-1）a(n-1)； a2=k(k-1), a3=k(k-1)(k-2) ,n≧3

When n is large, the formula for the an general term can be obtained by the elimination method.

You need graphics to get the right answer!!

There are 5 types in Figure A and 4 types in Figure B

1. If C color is the same as A, D has 4 colors, as long as it is not the same as AC.

2. The color of C is different from A, then C has 3 kinds, and D also has 3 kinds (except for A and Potato Pie C color hand elimination of the remaining 3 kinds).

Therefore, there are (4+3*3) kinds of graphs for c and d.

The total number of figures is 5*4*(4+3*3)=260

There are 5 ways to paint the No. 5 area first.

In the Tu 1 and 4 areas, there are two situations.

1) No. 1 and No. 4 area with the same color, then there are 4 kinds of colors, the last paint 2, No. 3 area, each area has 3 colors to choose from (2) No. 1 and No. 4 area different colors, then these two areas of coloring method is 4 3 = 12 kinds of final painting 2, 3 area, each area has 2 colors to choose from, so the total number of schemes = 5 (4 3 3 + 12 2 2) = 5 84 = 420 kinds.

Apply 5 first, then 1 and 4, then 2 and 3

1) 1 and 4 of the same color, painting how to have.

5 (4 1) 3 3 = 180 (species).

2) 1 and 4 different colors, how to paint.

5 (4 3) 2 2 = 240 (species).

To sum up, there are a total of coating methods.

180 + 240 = 420 (species).

For this problem, according to the two solutions you gave, the first solution is correct and the second solution is wrong!

The reason for the error is that the second solution does not understand the meaning of the question correctly.

According to the title: adjacent positions cannot be painted with the same color, but there is no provision for non-adjacent positions, so and can be the same color or different colors. If and are the same color, then there are 4 ways to color instead of 3, so the second solution is wrong.

The first category: and the same color, then there are 6 5 1 4 = 120 kinds of the second category: and different colors, then there are a total of 6 5 4 3 = 360 kinds of combined two types of cases, a total of 6 5 1 4 + 6 5 4 3 = 120 + 360 = 480 species.