Permutations
A permutation represents the number of ways \(r\) objects can be selected from a group of \(n\) total objects, without repetition, when the order that the objects are selected is important. So we often say this is an ordered arrangement of objects. Before we introduce the formula for permutations, we need a few more definitions, otherwise the formula won’t make sense.
Factorial: The product of every decreasing number from \(n\) to \(1\), represented as \(n!\)
\(n! = n(n - 1)(n - 2)(n - 3)....(3)(2)(1)\)
For example: \(5! = (5)(4)(3)(2)(1) = 120\)
In addition, \(0! = 1\) and \(1! = 1\)
Now that we have defined factorial, we can introduce the formula for permutations.
A permutation represents the number of ways \(r\) objects can be selected from a group of \(n\) total objects, without repetition, when the order that the objects are selected is important. So we often say this is an ordered arrangement of objects. Before we introduce the formula for permutations, we need a few more definitions, otherwise the formula won’t make sense.
Factorial: The product of every decreasing number from \(n\) to \(1\), represented as \(n!\)
\(n! = n(n - 1)(n - 2)(n - 3)....(3)(2)(1)\)
For example: \(5! = (5)(4)(3)(2)(1) = 120\)
In addition, \(0! = 1\) and \(1! = 1\)
Now that we have defined factorial, we can introduce the formula for permutations.
Permutation Formula: \(\large_nP_r=\frac{n!}{\left(n-r\right)!}; n\ge r\), where \(n\) is the total number of objects to choose from and \(r\) is the number of objects being selected and arranged.
What happens when \(n = r\)? Calculate using the formula above. What does it simplify to? Watch the video below for the explanation.
If \(n = r\), then the formula simplifies to just \(n!\)
This means that \(n!\) can be used to find the number of ordered arrangements if all of the objects are selected.
Example 1: Evaluate \(_7P_3\).
\(\large_7P_3=\frac{7!}{\left(7-3\right)!}=\frac{7!}{4!}=\frac{7\cdot6\cdot5\cdot4!}{4!}=7\cdot6\cdot5=210\)
Watch the video for an explanation to the work and how to do this on the calculator.
Let’s revisit example 2 from the previous page (Fundamental Counting Principle), using permutations.
One configuration of license plates in Illinois is \(3\) letters followed by \(3\) numbers. How many license plates are possible if letters and numbers cannot be repeated? (We can only use permutations in situations where repetition is not allowed).
\(\large_{26}P_3\cdot_{10}P_3=\frac{26!}{\left(26-3\right)!}\cdot\frac{10!}{\left(10-3\right)!}\)
\(\large_{26}P_3\cdot_{10}P_3=\frac{26!}{23!}\cdot\frac{10!}{7!}\)
\(\large_{26}P_3\cdot_{10}P_3=\frac{26\cdot25\cdot24\cdot23!}{23!}\cdot\frac{10\cdot9\cdot8\cdot7!}{7!}\)
\(\large_{26}P_3\cdot_{10}P_3=\normalsize26\cdot25\cdot24\cdot10\cdot9\cdot8\)
\(\large_{26}P_3\cdot_{10}P_3=\normalsize11,232,000\)
If you look back at Example 2 on the previous page, the problems simplify to the same value.
Each problem is different, so read carefully to determine if order matters and if repetition is allowed or not allowed. Key words or situations indicating order matters include:
- Order
- Arrange
- Line up
- Sequence
- Awards (first, second, third place)
- Election results (president, vice president, secretary, treasurer)
- Any other situation where the order of selection matters—these could include passwords, locker combinations, phone numbers, zip codes, license plates, etc.—but in these situations it would need to be specified that items cannot be repeated. If items can repeat in these situations, just use the fundamental counting principle.
Example 2: In the final grouping of skaters in the Olympic Figure Skating competition, \(6\) skaters take the ice. How many ways can the medals be awarded?
Example 3: How many ways can a family of \(4\) line up for a picture?
One other type of permutation problem involves finding the number of distinguishable permutations of \(n\) objects that exist. A distinguishable permutation is an ordered arrangement of letters that have distinct arrangements—we aren’t reading the same “word” more than once.
The formula for the number of distinguishable permutations is: \(\large\frac{n!}{\left(p!\cdot q!\cdot r!...\right)}\), where \(n!\) represents the number of ordered arrangements of all the letters in the word. The variables \(p\), \(q\), \(r\), etc. represent the number of times each letter in the original word appear, and \(p!\), \(q!\), \(r!\), etc. represent the number of ordered arrangements of each of those letters. By dividing by \(p!\), \(q!\), \(r!\), etc., we are eliminating the ordered arrangement of specific letters, thereby eliminating duplicate "words".
Example 4: Find the number of distinguishable permutations of the letters is the word REFEREE.