Problem 4 - College Algebra

Question

Answer the following questions. Describe how the permutation of \(n\) objects differs from the permutation of choosing \(r\) objects from a set of \(n\) objects. Include how each is calculated.

Step-by-Step Solution

Answer

Permuting all \( n \) objects uses \( n! \), while permuting \( r \) from \( n \) uses \( \frac{n!}{(n-r)!} \).

Step 1: Understanding Permutations of n Objects

The permutation of all the objects in a set of size \( n \) involves arranging all \( n \) items in sequential order. The number of permutations of \( n \) distinct objects is calculated as \( n! \) ("n factorial"), which is the product of all positive integers up to \( n \).

Step 2: Understanding Permutations of r Objects from n Objects

When selecting \( r \) objects from a set of \( n \) objects, and arranging them in sequential order, we use the formula for permutations of choosing \( r \) objects from \( n \), given by \( P(n, r) = \frac{n!}{(n-r)!} \). This accounts for the fact that only a subset of the full set is chosen and reflects every possible arrangement of that subset.

Step 3: Comparing and Contrasting the Two Cases

In the permutation of all \( n \) objects, every object is used in the arrangement, so the result is simply \( n! \). In contrast, the permutation of \( r \) objects from \( n \) objects involves only a subset of the total, so the permutation is not just \( n! \) but rather the arrangements possible when working with just \( r \) objects, calculated by \( \frac{n!}{(n-r)!} \).

Key Concepts

Factorials

The concept of a factorial is central to understanding permutations. A factorial, represented by an exclamation point (!), refers to the product of all positive integers up to a given number. For instance, if you have a number \( n \), then the factorial of \( n \) is denoted as \( n! \) and calculated as \( n \times (n-1) \times (n-2) \times \, ... \, \times 2 \times 1 \). It shows how many ways the objects can be arranged or ordered.

Factorials grow exceedingly fast as the number \( n \) increases. For example:

• \( 3! = 3 \times 2 \times 1 = 6 \)
• \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)

The calculation of permutations with factorials forms the basis of more complex combinatorial problems, like permutations of choosing \( r \) objects from \( n \). This makes the understanding of factorials fundamental in the realm of permutations.

Combinatorics

Combinatorics is the field of mathematics that deals with counting, arranging, and combination of objects. This is crucial when dealing with permutations. Its principles help us understand how to arrange objects effectively.

When it comes to permutations, combinatorics helps differentiate between the arrangements of all objects (with no selections made) and only a subset of those objects. These count how many different ways a set or a subset of objects can be organized when the order is important.

For example, in combinatorics, when we consider the number of ways to arrange all \( n \) different objects, it involves calculating \( n! \). On the other hand, when considering selecting \( r \) objects from \( n \) and allowing their order to vary, the formula \( P(n, r) = \frac{n!}{(n-r)!} \) is used. This formula takes into account all permutations of those chosen \( r \) objects. It highlights how thoroughly combinatorics covers both probability questions and arrangement possibilities.

Arrangement of Objects

The arrangement of objects underlies the calculation of permutations. In permutations, the focus is always on the order of arrangement. When you arrange a set of objects, each ordering is considered unique. This is why permutations differ from combinations, where order does not matter.

For a given number of objects \( n \), arranging all objects in order is a straightforward application of \( n! \), as each object plays a role in the arrangement.

However, when only \( r \) objects from \( n \) need to be arranged, the consideration becomes slightly more complex. Here, only a subset is selected, and it needs each possible way that subset can be arranged. The formula \( \frac{n!}{(n-r)!} \) is used, which first chooses which \( r \) objects will be arranged, and then calculates the arrangement of these \( r \) objects.

• Example: Choose 3 letters from ABCDE to arrange.
• Here, \( n=5 \) and \( r=3 \), so \( P(5, 3) = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{120}{2} = 60 \).

This calculation suggests that there are 60 different ways to arrange 3 letters chosen from a set of 5. This idea of arrangement based on selection is fundamental in understanding the permutations of different subsets from a whole.