#### Objective

Here you will learn how to calculate the number of permutations possible when repeated uses of set items are allowed.

#### Concept

If a bank account consists of eight digits between zero and nine, and repeated digits are allowed, how many possible account numbers are there?

This is not the same question as it would be if repeats were not allowed. At the end of this lesson, we’ll return to this question and review the difference.

#### Watch This

http://youtu.be/L_v4e4Yaypw Ten Marks – Permutations with Repetition

#### Guidance

When we consider permutations, we often specify that repeated values are not allowed, simply because many real-world situations would not support the concept. For instance, if you are calculating the number of possible seating arrangements for six people, it would be pretty silly to include any possibilities with the same person in multiple seats (unless you have a cloning machine in the closet!).

That said, there are times when you do need to include duplicates, such as the bank account question in the concept section above. The number of possible permutations in a set can be much greater when repeated values are allowed, but the calculation is actually simpler in concept.

To calculate the number of possible permutations of \begin{align*}r\end{align*} items from *\begin{align*}n\end{align*}* available items, simply raise \begin{align*}n\end{align*} to the power of \begin{align*}r\end{align*}:

\begin{align*}n^r: (number \ available)^{number \ chosen}\end{align*}

**Example A**

The ice cream shop on the corner carries 27 flavors of ice cream, how many different 4-scoop cones can be created there?

**Solution:**

There are 27 flavors, so \begin{align*}n=27\end{align*}. We are creating 4-scoop cones, so \begin{align*}r=4\end{align*}.

\begin{align*}Number \ of \ unique \ cones=27^4=531,441\end{align*}

**Example B**

Lockers in your school are each three digits 0-9. How many different combinations are possible?

**Solution:**

There are ten digits, so \begin{align*}n=10\end{align*}. We are looking for arrangements of three digits each, so \begin{align*}r=3\end{align*}.

\begin{align*}Number \ of \ locker \ combinations=10^3=1000\end{align*}

**Example C**

Keith’s trivia challenge team competes in trivia competitions all over the U.S. During a competition, the members are numbered 1-6, and a die is cast before each of the twenty questions in the challenge to decide which team member must answer the question. How many possible ways are there for Keith’s team to step up to the podium and answer trivia questions in a single meet?

**Solution:**

Keith’s team has six members, and there are twenty questions. There are six possible choices for the first question, and six possible \begin{align*}2^{nd}\end{align*} choices for each of them, resulting in \begin{align*}6^2=36\end{align*} possibilities for the first two questions. Since there are twenty questions, the total number of possible lineups is \begin{align*}6^{20} \approx 3.656 \times 10^{15}\end{align*}.

#### Concept Problem Revisited

*If a bank account consists of eight digits between zero and nine, and repeated digits are allowed, how many possible account numbers are there?*

Each digit has ten possibilities, and there are eight digits: \begin{align*}10^8=100,000,000\end{align*} possible bank accounts.

#### Vocabulary

** Permutations** are unique arrangements of items.

** Combinations** are unique groups of items.

#### Guided Practice

- How many different ice cream cones can be made with four scoops of ice cream, if there are 12 flavors to choose from?
- How many permutations are possible with seven units composed of the digits 0-9, duplication allowed?
- How many four-letter permutations are possible using the letters of the alphabet?

**Solutions:**

1. Since duplication is allowed, the number of possible permutations can be calculated with the formula \begin{align*}n^r\end{align*}:

\begin{align*}12^4=20, 736\end{align*}

2. There are ten digits to choose from, and we are making permutations of seven digits each:

\begin{align*}10^7 =10, 000, 000\end{align*}

3. There are twenty-six letters, and we are building permutations of four letters each:

\begin{align*}26^4=456,976\end{align*}

#### Practice

For questions 1-12, calculate the number of possible permutations, duplicate values are allowed.

- Using five decks of cards, permutations of five cards each.
- Using the letters A-G and numbers 1-5, arrangements of eight units each.
- Using the letters in the word “combine”.
- Seven digit arrangements of the numbers 0-9.
- Ice cream cones with three scoops chosen from 19 flavors.
- 1 cent, 5 cent, 10 cent, and 25 cent coins, in arrangements of five coins at a time.
- Letters A-F, in arrangements of six letters each.
- Roll a 10-sided die seven times.
- Roll a standard die five times.
- How many locker combinations are possible using three digits on a ten-digit dial?
- How many unique passwords can be made from the letters of the word “remix”?
- How many unique passwords can be made from the letters in “portable”?

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 8.3.