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    3. M3 Combinatorics Tools
    4. Week 3
    5. Day 12
    6. Module 3 Day 12 Your Turn Part 3
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    • P

      Complementary counting solution for mini-question
      • professionalbronco

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      quacker88

      @professionalbronco Really smart use of complementary counting! It's always super cool when a totally different solution is found. Great job!

    • victorioussheep

      Tree Diagram
      • victorioussheep

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      victorioussheep

      @quacker88 Oh, that is what I thought too...But I thought it might be easier, but maybe apparently not...

    • The Blade Dancer

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

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      debbie

      @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. 🙂

    • S

      Why can't you just do 3 × 2⁶ in the mini-question at the end?
      • spaceblastxy1428

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      debbie

      @spaceblastxy1428 Yes, you are right again! 🙂 This is a very fast way of getting the answer. Thanks for sharing!

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