Sample Problem 1. How many triangles are in the diagram below?
There are no fancy techniques for this problem but to tough it out and count the possibilities. However, this doesn't mean we have to just look for triangles, make tally marks for each one we find, and hope that we don't over count or miss anything. Instead, in this problem and every casework problem, the most important thing you can do is to have an organized system through which you can count.
For this problem, there are many ways to organize your counting, none more correct than another. However, in this solution, we will detail 1 specific method of organization.
First of all, there are 5 triangles that don't have any triangles inside of them. These triangles are ABF, BGF, BCG, DCF, and EFD. We will proceed by casework on which triangles contain each of these triangles. First of all, we can easily see that ABF is contained in 3 triangles (excluding itself). Next, in order to not over count, we can easily see that there are 2 triangles that contain BGF but not ABF (excluding BGF). There are 0 triangles that contain BCG but not ABF or BGF (excluding BCG). There are 0 triangles that contain DCF but not ABF, BGF, or BCG (excluding DCF). There are 0 triangles that contain FED but not ABF, BGF, BCG, or DCF (excluding FED). Thus, there are a total of 5+3+2=10 total triangles.
Notice how we made it easier to count the triangles, avoided over counting easily, and we are certain that we counted each triangle, because we were organized, neat, and we had a method. What method we use or how we organize our casework isn't that important, as long as we are organized, neat, have a method. Also, note that other concepts can be used in combination with casework, such as complementary counting and the Principle of Inclusion and Exclusion, in order to reduce the number of cases or to make the casework easier.
For this problem, there are many ways to organize your counting, none more correct than another. However, in this solution, we will detail 1 specific method of organization.
First of all, there are 5 triangles that don't have any triangles inside of them. These triangles are ABF, BGF, BCG, DCF, and EFD. We will proceed by casework on which triangles contain each of these triangles. First of all, we can easily see that ABF is contained in 3 triangles (excluding itself). Next, in order to not over count, we can easily see that there are 2 triangles that contain BGF but not ABF (excluding BGF). There are 0 triangles that contain BCG but not ABF or BGF (excluding BCG). There are 0 triangles that contain DCF but not ABF, BGF, or BCG (excluding DCF). There are 0 triangles that contain FED but not ABF, BGF, BCG, or DCF (excluding FED). Thus, there are a total of 5+3+2=10 total triangles.
Notice how we made it easier to count the triangles, avoided over counting easily, and we are certain that we counted each triangle, because we were organized, neat, and we had a method. What method we use or how we organize our casework isn't that important, as long as we are organized, neat, have a method. Also, note that other concepts can be used in combination with casework, such as complementary counting and the Principle of Inclusion and Exclusion, in order to reduce the number of cases or to make the casework easier.
Exercises
- What is the probability of flipping at least 2 heads if you flip 6 coins?
- How many ways can 12 be made out of 1s, 2s, and 4s?
Solution 1
We will use complementary counting to reduce the amount of casework that we have to do. If we don't have at least 2 heads come up, then that means that either 0 or 1 head comes up. The probability that exactly 0 heads come up is \(\frac{1}{64}\). The probability that exactly 1 head will come up is \(\frac{6}{64}\) because any of the coins could come up heads. Thus, the probability that at least 2 heads come up is 1- \(\frac{1}{64}\)-\(\frac{6}{64}\)=\(\frac{57}{64}\).
solution 2
We will do casework on the number of 4s. If we have 0 4s, we can have 0-6 2s and fill the rest with 1s for a total of 7 possibilities. Similarly, for 1, 2, and 3 4s we have 5, 3, and 1 possibilities respectively. Thus, there are a total of 1+3+5+7=16 total ways.