Problem 32 - College Algebra with Corequisite Support

Question

For the following exercises, find the distinct number of arrangements. The letters in the word "academia" that begin and end in "a"

Step-by-Step Solution

Answer

720 distinct arrangements.

Step 1: Identifying Constraints

The exercise requires the arrangements of the letters in "academia" such that they both start and end with 'a'. First, identify that the word "academia" has 8 letters, including 3 'a's.

Step 2: Fixing Positions of 'a'

Place the letter 'a' at the beginning and at the end of the sequence. This leaves 6 positions in between: _ _ _ _ _ _.

Step 3: Arranging Remaining Letters

Identify the remaining letters to arrange: 'c', 'a', 'd', 'e', 'm', and 'i', which includes one more 'a'. We need to find the distinct arrangements of these 6 letters.

Step 4: Calculating Total Arrangements of Inside Letters

Calculate the total arrangements of the letters 'c', 'a', 'd', 'e', 'm', and 'i'. The formula is \( \frac{6!}{1!} \) to account for the repetition of the letter 'a'. This results in \( \frac{720}{1} = 720 \).

Step 5: Conclusion

The number of distinct arrangements of the letters in "academia" that begin and end with 'a' is 720.

Key Concepts

Combinatorics

Combinatorics is a fascinating branch of mathematics that deals with counting, arranging, and grouping objects. It helps us find the number of possible arrangements or selections for a given set of items. When we dive into problems involving arrangements like the word "academia," we rely on combinatorial principles. Here, we want to know how many ways we can arrange letters under certain constraints, such as fixing the first and last letters as 'a'.

To solve such problems, we employ different combinatorial techniques:

• **Counting Principles**: These principles, such as the Addition and Multiplication Principles, allow us to determine the total number of ways to perform tasks.
• **Permutations**: Used when arranging items in a specific order, critical when each item has to occupy a unique position.
• **Combinations**: Deal with choosing items from a larger set where order does not matter.

Understanding these principles helps us tackle constraints effectively, as seen in the problem of arranging letters in "academia" with specific starting and ending conditions.

Factorials

Factorials are a key concept used in combinatorics, especially when working with permutations. A factorial, denoted by an exclamation mark (!), represents the product of all positive integers from 1 up to a given number. For example, the factorial of 6 is denoted as 6! and is calculated as:

• \[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \]

Factorials help us determine the number of ways to arrange a set of objects. In our exercise, we have 6 letters to arrange after fixing the positions of the first and last 'a', so we calculate 6! to find the total arrangements.

Factorials are essential when items need to be arranged in a line, and each item occupies a distinct position. They provide a straightforward method to calculate permutations without repetition, and adjustments can be made to account for repeated items as seen with an extra 'a' in our problem.

Permutations with Repetition

While permutations typically consider arrangements where each item appears only once, real-world problems often involve repetition. This is where permutations with repetition come into play. They allow us to account for repeated elements in a set of items, adjusting the total number of distinct permutations possible.

In our word "academia," there are multiple 'a's. If we were to arrange the letters without considering repetition, we would overcount identical arrangements. Hence, we use the formula for permutations with repetition, which is:

• \[ \frac{n!}{p_1! \times p_2! \times \ldots \times p_k!} \]

Here, \( n \) is the total number of items, and \( p_1, p_2, \ldots, p_k \) represent the frequencies of repeated items. For our set 'c', 'a', 'd', 'e', 'm', 'i', we have an extra 'a', leading us to calculate:

• \( \frac{6!}{1!} \)

This calculation ensures the additional 'a' is correctly accounted for, providing an accurate final count of arrangements as 720 distinct ways.