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

    Module 3 Day 12 Your Turn Part 3
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    • The Blade DancerT
      The Blade Dancer M0★ M1★ M2★ M3★ M4 M5
      last edited by debbie

      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?

      The Blade Dancer
      League of Legends, Valorant: Harlem Charades (#NA1)
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      debbieD 1 Reply Last reply Reply Quote 2
      • debbieD
        debbie ADMIN M0★ M1 M5 @The Blade Dancer
        last edited by

        @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. 🙂

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