4. (10 points) On Beta Planet, there are seven countries, each of which has exactly four friendly countries and two enemy countries. There are no three countries that are all enemies with each other. For such a planetary situation, a total of $\qquad$ three-country alliances, where all countries are friends with each other, can be formed.
【Analysis】Use $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}$ to represent seven countries, and use dashed lines to connect to indicate enemy relations, and solid lines to connect to indicate friendly relations. Then each country has 2 dashed lines and 4 solid lines. There are a total of $7 \times 2 \div 2=7$ (lines) dashed lines, and the rest are solid lines. First, it must be explained that these 7 points must be connected by 7 dashed lines to form a closed loop. $A_{2}$ must be connected to two points with dashed lines, let's denote them as $A_{1}, A_{3}$, and $A_{3}$ must be connected to another point with a dashed line, denoted as $A_{4} ; A_{4}$ is connected to $A_{5}$ with a dashed line, otherwise the remaining 3 points would have mutual enemy relations; $A_{5}$ is connected to $A_{6}$ with a dashed line, otherwise the remaining two points cannot be connected by 2 dashed lines; $A_{6}$ is connected to $A_{7}$ with a dashed line, and finally $A_{7}$ can only be connected to $A_{1}$ with a dashed line. The final connection diagram is as follows. As long as the three selected points do not have any two adjacent, the condition is met, there are $135,136,146,246,247,257,357$, a total of 7 kinds. (For clarity, we use $1,2,3,4,5,6,7$ to represent $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}$ respectively)【Solution】Solution: Use $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}$ to represent seven countries, and use dashed lines to connect to indicate enemy relations, and solid lines to connect to indicate friendly relations.

Selecting three points such that no two are adjacent meets the condition, there are $135,136,146,246,247,257,357$, a total of 7 kinds.
(For clarity, we use $1,2,3,4,5,6,7$ to represent $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}$ respectively), so for such a planetary situation, a total of 7 alliances of three countries that are all friendly can be formed. Therefore, the answer is: 7.