5. Planet Beta has 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, the number of possible alliances of three countries that are all friendly with each other is $\qquad$.
【Answer】 7
【Explanation】
Use $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{1}$ to represent seven countries, and use dashed lines to connect them to indicate enemy relationships, and solid lines to connect them to indicate friendly relationships. Then each country has 2 dashed lines and 4 solid lines. There are a total of $7 \times 2 \div 2=7$ dashed lines, and the rest are solid lines.
First, it is necessary to explain that these 7 points must be connected by 7 dashed lines in a closed loop. $A_{2}$ must be connected to two points with dashed lines, which we can denote 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 relationships;
$A_{5}$ 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 $135,136,146,246,247,257,357$, these 7
ways. (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)