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    Should same-angles but opposite order combinations be allowed?

    Module 2 Day 8 Challenge Part 3
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    • The Blade DancerT
      The Blade Dancer M0★ M1★ M2★ M3★ M4 M5
      last edited by debbie

      Module 2 Week 2 Day 8 Challenge Part 3

      Whether repeat (like same angles but opposite order) combinations are allowed should be specified in mini-question

      The Blade Dancer
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      T debbieD 2 Replies Last reply Reply Quote 3
      • T
        Tylenol M2 M3★ M4 M5 @The Blade Dancer
        last edited by Tylenol

        @The-Rogue-Blade I don't completely get what you're asking. (assuming you are talking about the opposite order of lines)I think that repeats are not allowed because that would be the same thing. I considered each angle being surrounded by a pair of lines and never thought of the order.
        🙂
        🙂
        🙂

        1 Reply Last reply Reply Quote 3
        • debbieD
          debbie ADMIN M0★ M1 M5 @The Blade Dancer
          last edited by debbie

          @The-Rogue-Blade I see what you're asking! You're saying that perhaps this pair of angles should be counted twice, depending on which is the first angle and which is the second angle:

          M2W2D8-ch-part-3-question-pairs-lines-need-specify-order.png

          The double-counting occurs if order matters. For example, in questions like, "How many ways are there to choose two out of five people to stand in a line?" 👩‍👩‍👧 👨‍👨‍👧 or "How many ways are there to make an ice cream cone with two scoops if you can choose from vanilla, chocolate, mocha, mint chip, or raspberry?" 🍦 🍦 etc., then order matters.

          If we say "How many ways are there to choose a pair of students out of 5 to sing in the performance?" 🎵 then, does order matter? No, it doesn't.

          We have a convention of counting "the ways to choose two things out of \(n\) things," and it's called a "binomial coefficient," which looks like \( \binom{n}{2}\), or it's called more casually a "choose." You will see more of this in Module 3: Combinatorics. 🙂 We also saw chooses in Module 0 with the Day 8 lesson about counting the number of triangles out of a bunch of criss-crossing lines. In that lesson, we saw

          $$ \text{ number of ways to choose } 2 \text{ lines out of } 6 \text{ lines } = \frac{6 \times 5 }{2} $$

          We divided by \(2\) because order didn't matter; whether we chose a line first or second didn't affect the answer.

          This mini-question is asking for the number of pairs in the same way, so that's why we don't count the highlighted \(40^{\circ}\) and \(20^{\circ}\) angles twice as in the diagram above.

          🙂

          RZ923R 1 Reply Last reply Reply Quote 4
          • RZ923R
            RZ923 M0★ M2★ M3★ M4★ M5 @debbie
            last edited by

            @debbie
            That’s getting into combinatorics in a geometry question explanation lol 🙂

            Very Interesting

            debbieD 1 Reply Last reply Reply Quote 3
            • debbieD
              debbie ADMIN M0★ M1 M5 @RZ923
              last edited by

              @RZ923 I wonder if it's as easy to put geometry into a combinatorics problem....... 🙂

              RZ923R 1 Reply Last reply Reply Quote 4
              • RZ923R
                RZ923 M0★ M2★ M3★ M4★ M5 @debbie
                last edited by

                @debbie
                Prof Loh does it in Module 3, Day 4 Challenge Question 🙂

                Very Interesting

                1 Reply Last reply Reply Quote 1

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