Tosin decides that using each ingredient only once is too limiting. She figures that she'll have more options if she lets herself use ingredients over again.
It turns out that deriving the calculation for finding combinations with repetition is a bit more complicated than what we've already derived. Watch the video below. When you're done, try answering the question at the bottom of the page. Feel free to watch the video as many times as you need to get a solid grasp of this formula.
As you watch this video, use the study guide to follow along if you'd like. Click the button below to download the study guide.
Let's say you are at a dinner buffet and the plates being used are those that have ridges that divide the plate into three sections. Now suppose you decide to fill one plate with just vegetables. How many ways could three vegetables be selected from the five choices: asparagus, broccoli, carrots, dill pickles, and eggplant. You might choose to fill your plate with three options like this with asparagus, broccoli, and carrots, or asparagus, broccoli, and dill pickles, or asparagus, broccoli and eggplant. If we were to list every combination of three vegetables it might look something like this… You may recall, that this is an example of a permutation calculation. This list of sixty options shows all of the different ways a plate can be filled with three different choices from the five vegetables, but technically there are a lot of repeated plate patterns. For example, each permutation in the first column contains the same three vegetables; asparagus, broccoli and carrots. But these vegetables are in different positions on the plate. Now since these arrangements are technically the same since we have one serving of each of these vegetables, we can eliminate the repeated patterns. And we can do the same for all unique combinations in this list… Now we are left with ten combinations of unique groups of vegetables.
This number of unique combinations can be found in a similar way that permutations are calculated. Using the combination formula, instead of the permutation formula, we have a method to calculate the number of unique ways that three DIFFERENT vegetables can be selected from five choices…
Now let's take this one step further. What if you wanted to fill two sections on the plate with the same vegetable, or perhaps all three sections with the same vegetable? This is known as a combination WITH REPETITION. The formula needed to find combinations with repetition is slightly different still. Trying to find the amount of combinations with repetition is like trying to calculate the number of combinations of actions performed on each potential component. So the new formula might look like this…
When used with our original example in the video we can show that there are 35 combinations of vegetable selections with repetition. How many combinations with repetition are available if we have five choices and a plate that had four divisions?...
Question
Given the equation you just learned, if there are 7 choices and 4 decisions to make, how many combinations can we have with repetition allowed?
