Subcategories

• Module 0 Day 3 Challenge Part 1

  Counting Numbers With Repeated Digits

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  You asked a really interesting question here! The Day 3: Challenge (Part 1) question explanation jumps a little bit ahead to show you what tricks you will see later on. To answer your question in a nutshell, both methods work equally well for simple problems, but maybe not for more complicated ones. Professor Loh likes to choose problems that are doable by what we call "brute force," so everyone has a chance to attack them. However, if you asked a different question that was on a larger scale, like how many 10-digit numbers you can make using the digits 1, 2 and 3, there would be so many ways to list that you couldn't list them all easily. In this case, alternative counting tricks are much more efficient to use!

  This relates to math in general: there are often many different ways of doing a problem. Once you have solved a problem with one technique, there is a huge pleasure that comes from seeing that there was another, perhaps very different, way to solve the same problem. Almost all of the problems in this course have this characteristic. It's not just a fun way to learn math, it stretches your mind in a good way!

  Happy Learning!

  The Daily Challenge Team

• Module 0 Day 3 Challenge Part 2

  Tree Diagram, Counting With No Restriction

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  That's a really great question, and this principle is one of the reasons why counting and combinatorics can be so tricky. It's wonderful that you're trying to delve into the meaning of why we multiply the number of choices together. The basic idea is that multiplication represents groups, and the number of ways to combine different choices can also be represented using groups.

  There is another question from Professor Loh's original Mock Mathcounts Sprint Round relating to this concept. There is a free video explanation of one of the questions, which happens to relate to the multiplication principle. In the video, there are three mini-questions, one of which asks to find the number of ways to choose 2 people out of 8 to stand side-by-side for a photo:

  https://daily.poshenloh.com/courses/take/2020-mathcounts-mock/lessons/10033528-welcome

  I'll copy the explanation again here:

  The important idea here is to use multiplication, not addition, to find the ways to arrange the two people. Multiplication can be used to express the number of things that are arranged into groups.

  Let's first draw 8 people and name them Person 1, Person 2, Person 3, Person 4, Person 5, Person 6, Person 7 and Person 8.

  30e6fec3-3557-4edd-bbbe-e454e2be2149-image.png

  The ways to pick two of them to stand side-by-side for a picture can be arranged into eight groups, each of which has seven ways in it:

  6cbe068d-b19f-496b-86d8-14e413bf46fa-image.png

  In the picture above, a ways described as "48" would mean that Person 4 stands on the left, and Person 8 stands on the right. Order matters here, because the photo would look different if Person 8 were on the left and Person 4 were on the right, which is described as "84."

  There are eight groups because there are eight people who can stand on the left, and given that choice of person on the left, there are seven choices for the person on the right.

  Please ask if you have any more questions!

  Happy Learning!

  The Daily Challenge Team

• Module 0 Day 3 Challenge Part 3

  Multiplication Principle

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  [Originally posted in Discussions]

  Module 0 Week 1 Day 3 Something to Think About

  Yes, you could! The reason Dr. Loh doesn't focus on this way is because you might have thought of that already, and he wants to show you a neat, different way of doing it.

  The casework method would be as follows:

   

  Case 1: All different digits

  The digits have to be 1, 2, and 3, and there are 3! = 6 ways to mix them up to create a three-digit number.

   

  Case 2: Two same digits, the third different

  There are three possibilities for the double-digit: 1, 2 or 3. After we have chosen the double-digit, we have two choices for the other, different digit. This gives \(3 \times 2 = 6\) ways. However, we can mix them up. For example, 122 can be mixed up in three ways \((122, 212, \text{ and } 221).\) So we have \( 6 \times 3 = 18\) ways for this case.

   

  Total:

  Ways from Case 1 + Ways from Case 2 =
  $$ 6 + 18 = 24 \text{ total ways } $$

  There's an urge in math-aficionados and math-philes to always search for new, clever or surprising ways of solving problems. It's the same urge that makes a computer programmer spend hours writing a program to do a mundane calculation that might take him the same amount of time if he did it by hand. And it's the same urge that makes you want to tell a little kid to count his legos by making piles of 10, rather than counting one by one.

  If you still have any more questions, please don't hesitate to ask!

  Happy Learning,

  The Daily Challenge Team

• Module 0 Day 3 Your Turn Part 1

  Complementary Counting

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  Hey @小馄饨chris, sorry for the late response, not sure if you still need it 😅
  @Legendaryboy991 is right: basically, 901, 109, 890, aren't numbers that count consecutively up or down because 0 isn't after 9. It's just to remove confusion in case some people were thinking that way. Hope this helps!