Look at this picture first, because 1, 2, 3 are adjacent, and 4 is adjacent to 3, so at least 3 colors are used, and at most 4 colors are used.
Method one
There are two situations.
1, first look at the use of three colors:
In the first step, because we have four colors to choose from, we choose three from these four colors, so there are four arrangements for a * * * (that is, we choose the combination number of 3 from 4, and the specific calculation is 4! /[(4-3)! 3! ])。
In the second step, we will paint the four areas of the picture with the selected three colors. But in this step, we have to draw 1, 2, 3 first. They must use different colors. But there is a problem with different arrangements. Three out of three colors is six (that is, three! Species)
[Of course, the first step and the second step can also be merged directly, that is, the arrangement of 3 of 4 can be directly calculated (4! /(4-3)! ), that is, 24 kinds, are the same as the numbers obtained in the first step and the second step. ]
Step three, now paint area 4. 4 can't be the same color as 3, so there are only two choices-the same as 1 or 2.
Multiply the numbers obtained in the three steps to get 48.
2. Look at the situation of painting four colors:
This is a direct calculation of the arrangement of 4 in 4, which is 4! =24
The two situations add up to 72, which is the final answer.
Method 2
We didn't separate the three colors from the four colors at first and painted them directly.
Of course, Zone 4 is special. Draw 1, 2, 3 first. That is, 24 is obtained by arranging 3 out of 4 (this step is the same as the first two steps of method 1 case 1). Next, let's draw area 4. As long as it is not the same color as area 3, it can choose three of the four colors to draw.
So the total is 24*3=72, and 72 is the final answer.
How did Supplement 4, Supplement 3, Supplement 3 and Supplement 2 come out?
Then we don't classify, and we don't distinguish region 4 from the beginning. We color 1, 2, 3 and 4 in turn. (Of course, it can also be in other order, and the idea is the same. )
We have four colors to choose from, so the first painted area has four color schemes (according to the order we set, it is the area 1).
Next, draw area 2. Because zone 2 and zone 1 are adjacent, and the colors of adjacent zones cannot be the same, we used a color in zone 1 before. So we only have three colors to choose from when we color area 2.
Next, paint area 3, which is adjacent to areas 1 and 2. 1 and 2 have each used one color, so there are only two colors left in area 3 to choose from.
Finally, let's draw area 4. (Note: The particularity of Area 4 is finally revealed at this moment. ) Zone 4 is only adjacent to Zone 3, so Zone 4 can ignore the existence of 1 and Zone 2, as long as it is not the same color as Zone 3. Among the four colors to choose from, one is for Zone 3, and there are three other colors to choose from.
To sum up, in this order, the optional colors for each step are 4, 3, 2 and 3, which can be multiplied.
(So strictly speaking, it cannot be said that the four areas of 1234 can be filled with several colors in isolation, but can be filled with several colors in a certain order [for example, in the order of 1234]. No matter which area it is, if we draw it in the first step, there will be four choices)
Question: What if you draw in the order of 4, 3, 2, 1?
Ok, just follow the above ideas step by step. You will find that in this order, the optional colors for each step are 4, 3, 3 and 2. Try it yourself ~