Permutation of n Things Taking r at a Time, Each May Be Repeated Any Number of Times (n^r)

Permutations and combinations are fundamental concepts in combinatorial mathematics. They help us count the number of ways in which a set of objects can be arranged or selected. When it comes to permutations, we are interested in the arrangement of objects where the order matters.

Understanding Permutations with Repetition

When we talk about the permutation of n things taken r at a time with repetition, we mean that we are selecting r objects from a set of n distinct objects, and each object can be chosen more than once. This is different from permutations without repetition, where each object can only be selected once.

Formula

The formula for the number of permutations of n things taken r at a time with repetition is given by:

$$ P(n, r) = n^r $$

This formula is derived from the fact that for each of the r positions, there are n choices since repetition is allowed. Therefore, for r positions, there are $n \times n \times \ldots \times n$ (r times) ways to fill the positions, which is $n^r$.

Table of Differences

Aspect	Permutations without Repetition	Permutations with Repetition
Formula	$P(n, r) = \frac{n!}{(n-r)!}$	$P(n, r) = n^r$
Repetition	Not allowed	Allowed
Order	Matters	Matters
Example	Arranging 3 out of 5 books on a shelf	Choosing 3 ice cream scoops from 5 flavors, allowing duplicates

Examples

Example 1: Ice Cream Scoops

Suppose an ice cream shop offers 5 different flavors of ice cream, and you want to order a cone with 3 scoops. You can choose the same flavor more than once. How many different combinations can you have?

Using the formula for permutations with repetition:

$$ P(5, 3) = 5^3 = 125 $$

There are 125 different ways to arrange your ice cream scoops.

Example 2: Lock Combination

Imagine a lock that uses a 4-digit code, where each digit can range from 0 to 9. How many different combinations can the lock have?

Here, n = 10 (digits from 0 to 9) and r = 4 (four positions for the digits).

$$ P(10, 4) = 10^4 = 10,000 $$

The lock can have 10,000 different combinations.

In-Depth Explanation

To understand the concept of permutations with repetition more deeply, let's consider a scenario where we have a set of n distinct objects, and we want to arrange r of them in a sequence where repetition is allowed.

For the first position, we have n choices since we can pick any of the n objects. For the second position, we still have n choices because we are allowed to repeat the objects. This pattern continues for all r positions.

If we were to visualize this, it would look like a tree where each branch represents a choice of an object, and each level of the tree represents a position in the sequence. The total number of ways to arrange the objects is the number of unique paths from the root of the tree to the leaves, which is $n^r$.

Practice Problems

1. A password consists of 6 characters, each of which can be a lowercase letter (26 possibilities) or a digit (10 possibilities). How many different passwords are possible if characters can be repeated?

2. A car license plate consists of 3 letters followed by 3 digits. How many different license plates can be made if letters and digits can be repeated?

Solutions to Practice Problems

1. For the password, each character can be one of 36 possibilities (26 letters + 10 digits). Since there are 6 characters and repetition is allowed:

$$ P(36, 6) = 36^6 $$

1. For the license plate, the first 3 characters (letters) have 26 possibilities each, and the last 3 characters (digits) have 10 possibilities each. Since repetition is allowed:

$$ P(26, 3) \times P(10, 3) = 26^3 \times 10^3 $$

Understanding permutations with repetition is crucial for solving many real-world problems where the order of selection matters, and items can be chosen multiple times. Remember that the key difference between permutations with and without repetition is whether the same item can be selected more than once for a given arrangement.