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    • F

      M0 Week 4 Challenge Q20

      Week 4
      • • • fantasticcrow
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    • T

      Ad?

      Funny
      • • • Todymaster
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      @enlivenedporcupine-0 Yeah that's CSS web development add just skip it sir.

    • S

      Can anyone give me feedback on how can I improve my app{It's a game}

      Programming
      • • • studiouszebra
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      @Desolate_101 I made that using Python and In python I used Tkinter. If you can contribute to this project you or wana tell me something you can contact me a this email : harshsc2007@gmail.com.
      Thank you for checking out my project. After 1 year some one has sent some reply. And now I have stopped development on my project cause of studies. if you want to develop you can contact me at above email and if you want we can partnership in coding. on projects

      Thank You
      Harsh

    • G

      A little confusing

      Module 0 Day 15 Your Turn Part 1
      • • • genuineopossum
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      yeah, i think it would better if they mentioned something. but usually questions like that are a follow-up of the your turn problem or else the problem would be meaningless. 😆

    • G

      Something interesting...

      Module 0 Day 16 Challenge Part 3
      • • • genuineopossum
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      F

      @genuineopossum to clarify, 9 is supposed to be (10-x) for 1 and 8is supposed to be (10-x) for 2 and so on.

    • F

      Module 0 Week 3 Challenge Q20

      Comments & Feedback
      • • • fantasticcrow
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      @audrey said in Module 0 Week 3 Challenge Q20:

      @fantasticcrow Yay, I'm glad that helped!! And as for counting versus probability, it all depends on what the question asked for. To give a simple example:

      How many ways are there to make a three digit number where all the digits are even? [This would be a counting problem, because you want to count how many 3-dig #'s there are where all the digits are even]

      Vs.

      If you choose any three digit number at random, what is the probability that all its digits are even? [This would be a probability problem, because you're asked what the chance of something happening is-- so in this case, you would take the answer from #1, and divide by how many three digit numbers there are!]

      In the context of this problem, we can't use simple counting because not all of the options are equally likely to be chosen. In other words, you should use counting when all the options are equally likely ("weighted")-- for example, choosing a three digit number at random. But in this case, they're not, so we have to actually multiply out probabilities, if that made sense.

      Very helpful. Thank you!

    • F

      Module 0 Week 3 Challenge Q19

      Comments & Feedback
      • • • fantasticcrow
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      @lucky_ducky1 thanks! you gave a great explanation.

    • R

      Type in as many math problems as you can find

      Math Problems
      • • • reliabledove
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      E

      -1x-2+3-4+(5x6-7^2)- -18=?

    • Bulba_BulbasaurB

      Wow

      Programming
      • • • Bulba_Bulbasaur
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      Desolate_101D

      @RZ923 wait... What???? Prof.Loh = Married ???

    • E

      I get the same questions on a test.

      Comments & Feedback
      • • • excitedarmadillo 0
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      Desolate_101D

      @excitedarmadillo-0 No, it is just the same test, it doesn't change

    • G

      There was a typo...

      Module 0 Day 14 Challenge Part 1
      • • • genuineopossum
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    • T

      Rubik's cube algorithms that make you faster (if you cube)!

      General Discussion
      • • • tranquilsnail
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      Hypixel Skyblock

      Hobbies and Recreation
      • • • ShadowStrikes
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      @Desolate_101 yes

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      Why do you multiply by 4 and 6?

      Module 5 Day 5 Challenge Part 2
      • • • consideratewallaby 0
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      MethM

      @daringhorse i think you meant 3^0+3^1 and 5^0+5^1 right, i just want to check

    • K

      This way works if you have graph paper

      Module 4 Day 14 Challenge Part 1
      • • • kindbobcat
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    • P

      What is that geometric series thing he was talking about?

      Convert to a Fraction
      • • • professionalbronco
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      Thank you

    • P

      Different way of solving this problem

      Module 3 Day 14 Challenge Part 4
      • • • professionalbronco
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      quacker88Q

      So sorry about that @professionalbronco !!

      But yes, your solution is absolutely right. Another way of looking at is the "algorithm" Professor Loh mentioned in the video: by sorting alphabetically, you'll get that there are 7 options for person A to make a pair, 5 options for the second pair, 3 for the third, and 1 for the last. $$7\cdot5\cdot3\cdot1=\boxed{105}$$ as well.

      Also from the lesson, using the expression you had but expanding out the 4!:
      $$\frac{\binom{8}{2}\cdot\binom{6}{2}\cdot\binom{4}{2}\cdot\binom{2}{2}}{4\cdot3\cdot2\cdot1}$$
      and then expanding the binomial coefficients and cancelling:
      $$\frac{\frac{8\cdot7}{2}\cdot\frac{6\cdot5}{2}\cdot\frac{4\cdot3}{2}\cdot\frac{2\cdot1}{2}}{4\cdot3\cdot2\cdot1} \implies \frac{(\cancel{4}\cdot7)\cdot(\cancel{3}\cdot5)\cdot(\cancel{2}\cdot3)\cdot(\cancel{1}\cdot1)}{\cancel{4}\cdot\cancel{3}\cdot\cancel{2}\cdot\cancel{1}}$$

      which leaves us the same expression for \( 7!! \) 🙂

      also !!
      did you know that the exclamation in the factorial is kind of like the "index" of how much you're shifting each number (double factorial shifts by 2, triple by 3, and so on). Soooo... technically we could get to triple factorials (or higher) maybe:
      \(8!!!=8\cdot5\cdot2=80\)
      \(11!!!!=11\cdot7\cdot3=231\)

    • E

      Nuuuuuuu

      M3 Combinatorics Tools
      • • • energizedpanda
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      Desolate_101D

      @RZ923 agreeed

    • T

      Specs of computer that you use?

      Funny
      • • • Todymaster
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      Π

      @Todymaster 2017 macbook air 13' 8gb ram 1600mhz ddr3 1.8 chz dual core intel i5 with intel hd graphics 6000 1536mb

    • M

      I do not understand where Po shen loh got the 3√3. How did he get that?

      Module 2 Day 3 Challenge Part 3
      • • • mirthfulostrich
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      Π

      @debbie he's done 1,000 exams? That's a sheesh moment

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