On the last page, we mentioned that both permutations and combinations can be represented as a set of decisions that have certain choices available to them. The Fundamental Principle of Counting applies to situations such as these.
Fundamental Principle of Counting
The number of possible outcomes of an event is equivalent to the product of the the number of choices available to each decision made during that event.
So, for permutations, the event is arranging the components in a particular way, and the decisions are choosing which component to place into each position. Remember, in a permutation, the order matters. If Tosin makes a sandwich that has lettuce, cheese, and tomato (from the bottom up), that sandwich is a different permutation than the sandwich that has lettuce, tomato, and cheese.
There are two types of permutations: those that allow repetition and those that don't. Using the Fundamental Principle of Counting, can you figure out how many permutations Tosin can make if she plans on making a sandwich containing 3 ingredients and has 4 ingredients available to her? Repetition is allowed.
Tosin has 4 ingredients to choose from, and she can use any of them for any of three positions (bottom, middle, top) in her sandwich. Therefore, the answer is 4 × 4 × 4 = 43 = 64 possible permutations.
In fact, when repetition is allowed, the number of potential permutations can be represented as nr, where n is the number of choices available and r is the number of decisions, or positions, that need to be made/filled.
Permutations with Repetition
P(n, r) = nr
The above equation is true only when repetition is allowed. Can you recall the equation for calculating the number of possible permutations when repetition isn't allowed? See if you can write it down before you click the button below. Remember to use the terms n and r.
Permutations without Repetition
P(n, r) = \(\mathsf{ \frac{n!}{(n-r)!} }\)
We use the factorial because, as decisions are made, the number of potential choices available to later decisions decreases. The denominator of the above equation accounts for situations in which there are fewer decisions being made than there are choices. Even when the number of decisions is equal to the number of choices, n - r = 0 and 0! = 1, so the equation still holds true.
Question
How many possible permutations of a sandwich can Tosin make if she plans to use 3 ingredients in the sandwich and has 4 ingredients to choose from, assuming repetition isn't allowed?
So, the answer becomes \(\mathsf{ \frac{4!}{(4-3)!} = \frac{4!}{1!} = 4! = 24 }\). Although the answer was 4! in this case, the answer will NOT always be n!.
