@The-Blade-Dancer Hi again! The solution is counting these two separate cases: **Case 1.)** where the lower two second-from-bottom nodes are the \(\textcolor{red}{\text{same}}\) color, and **Case 2.)** where the lower two second-from-bottom nodes are \(\textcolor{red}{\text{different}}\) color.

For **Case 1**: There were \(48\) ways to color the top six nodes (with the two bottom nodes the same color)

For **Case 2**: There were \(96\) ways to color the top six nodes (with the two bottom nodes different color).

In the final formula, the \(48\) is multiplied by \(2\) because there are two options for the very bottom node color (e.g. \(\textcolor{green}{\text{green}}\) and \(\textcolor{orange}{\text{yellow}.})\) The \(96\) isn't multiplied by anything, because if you already have two colors for the second-from-bottom nodes, then there's only one choice for the very last bead's color (in this example, \(\textcolor{orange}{\text{yellow}}).\)

M3W3D12-y-part-3-solution.png

$$ 48 \times 2 + 96 = \boxed{192}$$

That's how the answer was found. 🙂