It's normally very difficult to count all the triangles yourself. However, Professor Loh showed a very neat shortcut.

Instead of counting every possible triangle, he counted every possible way to pick 3 of the lines in the picture. This is because when you draw 3 lines, you get a triangle (Unless 2 of them are parallel. Fortunately, none of those 6 lines are parallel to each other!).

Here are some pictures that show how this work:

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You can see how no matter which three lines I choose, they always make one of the triangles in the picture! So if I just figure out how many ways there are to choose 3 lines, then that's just the number of triangles in total.

How do we choose 3 of the 6 lines? Let's call the 6 lines A, B, C, D, E, and F. One way to do it is to just write down every possible combination of 3 lines. So, I could do A B C, A B D, A B E, A C D, A C E, A D E, B C D, ... You can see that this can take a long time! But, one important thing to see is that if I choose the three lines B C D, that's the same as choosing the three lines C D B. This is because the order in which we draw the 3 lines doesn't matter! No matter how I draw those 3 lines, I will get the same triangle.

Instead of writing them all down, though, we can use another shortcut! You can think of it as drawing three blank spots: _ _ _ Then, we choose which of the 6 letters from A to F we write in each spot. We can't use the same letter twice!

There are 6 possible letters to write in the first spot. Then, there are only 5 possible letters left, so there are 5 letters to write in the second spot. Finally, we have 4 letters left, so there are 4 possible letters we can write in the third spot. Altogether, there would be 6 x 5 x 4 possible combinations of letters, right?

However, remember that the order doesn't matter! We actually counted too many possible combinations, because we actually counted every triangle 6 times. How did this happen? Well, we actually counted the triangle from the lines A, B, and C every time we wrote down: ABC, ACB, BAC, BCA, CAB, or CBA. So we're counting everything 6 times more than we should: We only want the ABC, not any of the rest of the orders. So we have to divide by 6: 6 x 5 x 4 / 6 = 20 triangles

Notice that 6 = 3 x 2 x 1, the number of ways to order 3 letters! That's another way to see why we have to divide by 6.

]]>means "the number of ways to choose two things out of six things."

It's easy to mistakenly think this should be 6 × 5, because we learned before that the ways to line up six things in a row is equal to 6! = 6 × 5 × 4 × 3 × 2 × 1, and this looks similar, except you are not taking all of the six things, only two of them.

But.... it's not equal to 6 × 5! This is because "choose two things out of six things" doesn't imagine you putting them in a line. It just imagines them all mixed up. The 6 × 5 ways double-counts every way. For example, if you were choosing A and B out of six things, the ways AB and BA are both counted in 6 × 5.

This is why the number of ways to choose two things out of six things is really 6 × 5 × ½.

If you can be very careful and keep track of when order matters (i.e. AB and BA are different ways), and when order doesn't matter (i.e. when AB and BA count as the same way), then you're well on your way to understanding a lot of great combinatorial problems!

Happy Learning,

The Daily Challenge Team

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and we divided by 3!, which equals 6, because there are 6 copies of each way to make a triangle. These are the six ways to make a triangle with the red line, dark green line, and yellow line:

The number of copies per way depends on the question.

I hope this helps! Good luck on the rest of the course, and please ask if you have any more questions!

Happy Learning,

The Daily Challenge Team

]]>Prof. Loh solved two separate questions: 1.) How many intersection points are there? and 2.) How many triangles can you make?

For the first question, let's color the lines like this:

The 6 × 5 comes from the fact that there are 6 groups of ways, with 5 ways in each group. I'll illustrate two of the groups right here:

Group 1 (Ways start with Red)

Group 2: (Ways start with Green)

Group 3: (Ways start with Yellow) (not pictured)

Group 4: (Ways start with Blue) (not pictured)

Group 5: (Ways start with Black) (not pictured)

Group 6: (Ways start with Dark Green) (not pictured)

This is an easy way to count the ways, because without drawing the ways out, we can see that the 6 comes from the number of lines, and the 5 came from the number of other lines left over after we chose the first one! However, the downside to this easy way is that we have to consider the double-counted ways. In the Red Group and Green Group, there is a way that occurs in both: the intersection of the red line and the green line.

Actually, every way has a copy, and if you drew out all the lines for all the groups, you would see this!

So we divide 2 to the number of ways that we drew out, which gives

as the number of different intersection points. (Note that this was just a warm-up question; it was the answer to the hint. This is to show you the idea of how to choose x things from y things, and isn't needed for finding the number of triangles.)

The second question is to find the number of triangles.

I like to think of the top as meaning that there are 6 large groups, each with 5 small groups, each with 4 ways in them.

I'll only draw one of the largest groups, the one with first line red:

Yes, all those ways are in the first group (ways starting with red): 5 × 4 ways! The common characteristic between all these groups is that they have a red line. The picture above contains five sub-groups; you'll notice that in each group, there is always one other color that is present (for example, the last group includes ways with red and blue lines).

There are five other large groups (a group for ways which all contain the yellow line, a group for ways which all contain the black line, a group for ways which all contain the blue line, a group for ways which all contain the light green line, and a group for ways which all contain the dark green line). Each of those other five large groups also each contains five subgroups, each of which contain four ways.

However, we have counted some ways more than once. Each of the ways gets counted six times, as shown in the picture below. The number "1" means that this color is the large group color, the number "2" means that this color is the subgroup color, etc.

By symmetry, since each way is counted six times, we can divide the total ways we got earlier by 6 (which is 3!). So the number of triangles is

I hope this helps! We love helping our students through their struggles, so please don't hesitate to ask more questions. I am more than happy to answer them!

Happy Learning,

The Daily Challenge Team

]]>The term "6 choose 2" means "the number of ways to choose two things out of six different things." This came up in the lesson when Prof Loh was finding the answer to the question, "How many intersection points are there?"

To find out the value of 6 choose 2, it might help to color-code the six lines like this:

The 6 × 5 comes from the fact that there are 6 groups of ways, with 5 ways in each group. I'll illustrate two of the groups right here:

Group 1 (Ways start with Red)

Group 2: (Ways start with Green)

Group 3: (Ways start with Yellow) (not pictured)

Group 4: (Ways start with Blue) (not pictured)

Group 5: (Ways start with Black) (not pictured)

Group 6: (Ways start with Dark Green) (not pictured)

This is an easy way to count the ways, because without drawing the ways out, we can see that the 6 comes from the number of lines, and the 5 came from the number of other lines left over after we chose the first one! However, the downside to this easy way is that we have to consider the double-counted ways. In the Red Group and Green Group, there is a way that occurs in both: the intersection of the red line and the green line.

Actually, every way has a copy, and if you drew out all the lines for all the groups, you would see this!

So we divide 2 to the number of ways that we drew out, which gives

as the number of different intersection points. (Note that this was just a warm-up question; it was the answer to the hint. This is to show you the idea of how to choose x things from y things, and isn't needed for finding the number of triangles.)

I hope this helps! Let me know in the Discussions.

Happy Learning!

The Daily Challenge Team

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