5.3. Permutations
Permutations
In a previous example, we were asked to find the word sequences formed by using the letters {A,B,C} if no letter is to be repeated. The tree diagram gave us the following six arrangements:
ABC, ACB, BAC, BCA, CAB, and CBA
Arrangements like these, where order is important and no element is repeated, are called permutations.
Example 5.3.1
There are four choices for the first letter of our word, three choices for the second letter, and two choices for the third.
| 4 | 3 | 2 |
Applying the multiplication axiom, we get 4 · 3 · 2 = 24 different arrangements.
Example 5.3.2
| 4 | 3 |
Since there are no more restrictions, we can go ahead and make the choices for the rest of the positions. So far we have used up 2 letters, therefore, five remain. So for the next position there are five choices, for the position after that there are four choices, and so on. We get:
| 4 | 5 | 4 | 3 | 2 | 1 | 3 |
So the total permutations are 4 · 5 · 4 · 3 · 2 · 1 · 3 = 1440.
Example 5.3.3
a. The number of four-letter word sequences is 5P4 = 120.
b. The number of three-letter word sequences is 5P3 = 60.
c. The number of two-letter word sequences is 5P2 = 20.
Before we give a formula for nPr, we’d like to introduce a symbol that we will use a great deal in this as well as in the next chapter.
Factorial: n! = n(n − 1)(n − 2)(n − 3)··· 3 · 2 · 1.
Where n is a natural number.
0! = 1
Now we define nPr.
The Number of Permutations of n Objects Taken r at a Time:
nPr = n(n − 1)(n − 2)(n − 3)···(n − r +1), or
nPr =
Where n and r are natural numbers.
The reader should become familiar with both formulas and should feel comfortable in applying either.
Example 5.3.4
Next we consider some more permutation problems to get further insight into these concepts.
Example 5.3.5
Example 5.3.6
| 4 | 3 | 2 | 5 | 4 |
Clearly, this makes sense. For every permutation of three math books placed in the first three slots, there are 5P2 permutations of history books that can be placed in the last two slots. Hence the multiplication axiom applies, and we have the answer (4P3) (5P2). We summarize.
- Permutations: A permutation of a set of elements is an ordered arrangement where each element is used once.
- Factorial: n! = n(n − 1)(n − 2)(n − 3)···3 · 2 · 1. Where n is a natural number. 0! = 1
- Permutations of n Objects Taken r at a Time: nPr = n(n − 1)(n − 2)(n − 3)···(n − r + 1), or nPr = . Where n and r are natural numbers.
Circular Permutations and Permutations with Similar Elements
In this section we will address the following two problems.
- In how many different ways can five people be seated in a circle?
- In how many different ways can the letters of the word MISSISSIPPI be arranged?
The first problem comes under the category of Circular Permutations, and the second under Permutations with Similar Elements.
Circular Permutations
Suppose we have three people named A, B, and C. We have already determined that they can be seated in a straight line in 3! or 6 ways. Our ne