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High school mathematics coloring problem?
The answer is no problem, it is right.

It can be understood that we call the four colors 1, 2, 3 and 4 respectively. Let's write the color of B as color 1

(1) When B is the same color as E, then E is also 1, so D can't be 1, so we mark the color of D as 2.

At this time, f can be filled in 2, 3 and 4. If F is filled with 2, C has two choices: 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 filled with 4, then C can only be filled with 3), which is also two cases.

Then there is always a situation where 1x2+2x 1 kind of b, f and c is filled in this way-this is a bit right with the answer.

(2) When B and D are of the same color, then D is also 1. We mark the color of E as 2, and F can be filled as 3 or 4.

If F is filled with 3, C has two choices: 2 and 4; If F is filled with 4, C has two choices: 2 and 3;

Then there are always * * cases of filling in B, filling in F and filling in C: 1x 2+2x 1.

Add the above (1)(2), that is, 2*( 1x2+2x 1), and the result is consistent with the answer.

I am throwing a brick to attract jade, and I believe there will be other better ways to understand it.