Subcategories

• Day 1

  Ordered Arrangements

  Module 3 Day 1 Challenge Part 1 Module 3 Day 1 Challenge Part 2 Module 3 Day 1 Challenge Part 3 Module 3 Day 1 Challenge Part 4 Module 3 Day 1 Challenge Part 5 Module 3 Day 1 Your Turn Part 1 Module 3 Day 1 Your Turn Part 2
  9
  Topics
  47
  Posts

  E

  The way I did it, and I thought he would too, was by considering the possible placements of B for any place of A. Since after we have defined 2 of the 4 elements, the other two can only be arranged in two ways. Then, you can add up all the ways to put A before B with considering the other two elements as one singular one, and then multiply by 2.
  So, with A in front 3 placements work, second, 2 and so on, so that we get 3+2+1=6. Then, we multiply by 2 to get 12.
  this also works on the your turn, except that there is actually only 1 arrangement of C and D for every placement of b.

• Day 2

  Venn Diagram

  Module 3 Day 2 Challenge Part 1 Module 3 Day 2 Challenge Part 2 Module 3 Day 2 Challenge Part 3 Module 3 Day 2 Challenge Part 4 Module 3 Day 2 Your Turn Part 1 Module 3 Day 2 Your Turn Part 2
  6
  Topics
  10
  Posts

  T

  $Hey everyone! Just wondering shouldn't the answer for this problem be something times 2^5 instead of something times 2^4? Mainly because that was the formula Professor Loh gave us!?$

• Day 3

  Overlapping Worlds

  Module 3 Day 3 Challenge Part 1 Module 3 Day 3 Challenge Part 2 Module 3 Day 3 Challenge Part 3 Module 3 Day 3 Challenge Part 4 Module 3 Day 3 Your Turn Part 1 Module 3 Day 3 Your Turn Part 2
  9
  Topics
  23
  Posts

  S

  I believe the choice A is also correct. Because these involve operations among sets, not numbers. So it doesn't matter if you subtract a certain part twice.

• Day 4

  Careful Casework

  Module 3 Day 4 Challenge Part 1 Module 3 Day 4 Challenge Part 2 Module 3 Day 4 Challenge Part 3 Module 3 Day 4 Your Turn Part 1 Module 3 Day 4 Your Turn Part 2 Module 3 Day 4 Your Turn Part 3 Module 3 Day 4 Your Turn Part 4 Module 3 Day 4 Your Turn Part 5
  13
  Topics
  39
  Posts

  P

  Why is it that (ii) is the best answer here. Should I be thinking that because this is a subcase of a given box dimension 1x2 , I need to keep the dimension fixed and consider the subcases ? Because to me at least (i) seems correct as well.

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