Hello, shay!
pka is absolutely correct . . . you need a system.
There are two possible "values": \(\displaystyle \,\{B,G\}\)
If there are \(\displaystyle n\) children, there are \(\displaystyle 2^n\) possible outcomes.
So with 3 children, there are \(\displaystyle 2^3\,=\,8\) outcomes.
We will make a 3-column table.
In column-1, write B's and G's four-at-a-time: \(\displaystyle \;\begin{array}{cccccccc}B\\B\\B\\B\\G\\G\\G\\G\end{array}\)
In column-2, write B's and G's two-at-a-time: \(\displaystyle \;\begin{array}{cccccccc}B\\B\\G\\G\\B\\B\\G\\G\end{array}\)
In column-3, write B's and G's one-at-a-time: \(\displaystyle \;\begin{array}{cccccccc}B\\G\\B\\G\\B\\G\\B\\G\end{array}\)
Reading across the table, we have all eight possible outcomes:
\(\displaystyle \;\;\;\begin{array}{cccccccc}1&B&B&B\\2&B&B&G\\3&B&G&B\\4&B&G&G\\5&G&B&B\\6&G&B&G\\7&G&G&B\\8&G&G&G\end{array}\)