Combinatorics problems involve distinguishable and indistinguishable objects. In this lesson, we will go over distinguishable combinatorics.
I have 3 distinguishable balls and 2 distinguishable bins. How many ways can I put the ball into the bins? All of the balls have to be in the bins
For this problem, we have to recognize that whatever doesn’t go in the first bin must go in the second one. So, if we can figure out the possible orderings for the first bin, we can solve the problem.
For each ball, there are two options: it can either go in the first bin or not go in the first bin (it will go in the second bin). Since there are three balls, the number of arrangements for the first bin will be 2 x 2 x 2, which is 8. This is our final answer.
Mary goes to the gym three times a week. In a week, she works out her arms, chest, and legs. She works on one of these each time she goes to the gym. How many ways are there for her to schedule her gym sessions this week? (She only works out once a week).
This is actually a distinguishable combinatorics problem in disguise. You can think of it as the equivalent of putting 3 distinguishable balls (workout sessions) into 7 distinguishable bins (days of the week).
There are seven ways to select where the first ball goes. Since there are 6 bins left, there are now 6 possibilities for the second ball. Finally, there are 5 possible bins to put the final ball. Thus, the final answer is 7 x 6 x 5, which equals 210.
8 students enter an art competition. How many ways are there for the first and second places to be chosen?
Like the previous problem, there are 8 students that can win first place. Then, there are seven students that can be second. So, the final answer is 56, which is 8 x 7.
I have 3 distinguishable balls and 4 distinguishable bins. How many ways can I put the ball into the bins? Not all of the balls have to be in a bin.
The balls can either be in one of the 4 bins or outside all of the bins. To simplify the problem, we can consider the balls that are outside the bins to be in a 5th bin. So, now there are a total of 5 bins.
The first ball can go in any of the 5 bins. So can every other ball. So, the answer is 5 x 5 x 5, which is 125.