??? That just completely ignores the possibility of the two lower nodes being different colors, which isn't right, no?

  • M0★ M1★ M2★ M3★ M4 M5

    Hold up-

    1f835bd4-7252-4b84-a603-b699cbfba560-image.png

    48df87d0-0de6-475b-a999-9019cd5162ff-image.png

    That just completely ignores the possibility of the two lower nodes being different colors, which isn't right, no?

  • ADMIN M0★ M1 M5

    @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. 🙂