Chapter 14: Problem 12
How many different-appearing arrangements can be created using all the letters AAABBC?
Short Answer
Expert verified
60 different-appearing arrangements.
Step by step solution
01
Calculate the total number of letters
Count the total number of letters in the word AAABBC. There are 6 letters in total.
02
Identify and count repetition of each distinct letter
There are three distinct letters: A, B, and C. Count how many times each letter repeats within the word. A appears 3 times, B appears 2 times, and C appears 1 time.
03
Apply the formula for permutations with repetition
Use the formula to calculate permutations with repetition: \(P = \frac{n!}{n_1! \cdot n_2! \cdot n_3! \cdot \ldots \cdot n_k!}\) where \(n\) is the total number of items to arrange, and \(n_1, n_2, \ldots, n_k\) are the frequencies of each distinct item. For the word AAABBC, this becomes \(P = \frac{6!}{3! \cdot 2! \cdot 1!} = \frac{720}{6 \cdot 2 \cdot 1} = \frac{720}{12} = 60\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factorial
To solve problems involving permutations, we first need to understand the concept of a factorial. The factorial of a number, represented as \( n! \), is the product of all positive integers up to that number.
For example, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \. \) Factorials are crucial in combinatorics because they help us calculate the total number of ways to arrange a set of items.
An intuitive way to think about it is imagining arranging books on a shelf. If you have 3 books, there are \( 3! = 6 \) different ways you can place them.
Factorials grow very fast, making them vital for solving more complex problems involving larger numbers.
For example, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \. \) Factorials are crucial in combinatorics because they help us calculate the total number of ways to arrange a set of items.
An intuitive way to think about it is imagining arranging books on a shelf. If you have 3 books, there are \( 3! = 6 \) different ways you can place them.
Factorials grow very fast, making them vital for solving more complex problems involving larger numbers.
Combinatorics
Combinatorics is the field of mathematics that studies counting, arranging, and combining items. It's like the science of arranging things!
A significant part of combinatorics involves finding the number of different ways to arrange a set of items.
We often use factorials in these calculations. Problems can vary from simple arrangements to complex ones involving repetitions and constraints.
For example: *Calculating the number of ways to arrange books on a shelf*
*Finding distinct ways to form teams from a group*
Combinatorics helps us solve real-world problems efficiently by providing tools to manage and calculate arrangements and combinations.
A significant part of combinatorics involves finding the number of different ways to arrange a set of items.
We often use factorials in these calculations. Problems can vary from simple arrangements to complex ones involving repetitions and constraints.
For example: *Calculating the number of ways to arrange books on a shelf*
*Finding distinct ways to form teams from a group*
Combinatorics helps us solve real-world problems efficiently by providing tools to manage and calculate arrangements and combinations.
Repetition in permutations
When items repeat in a set, the arrangement rules change. This is called permutations with repetition.
The formula for permutations with repetition considers repeated items by dividing by the factorial of the counts of each repeated item: \( P = \frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!} \).
Where \( n \) is the total number of items, and \( n_1, n_2 \ldots n_k \) are the counts of each repeated item.
Let's take the word 'AAABBC'. It has 6 letters, where A repeats 3 times, B repeats 2 times, and C appears once.
Using the formula: \( P = \frac{6!}{3! \cdot 2! \cdot 1!} = 60 \).
This adjusts for repeated items, giving us the number of distinct permutations.
The formula for permutations with repetition considers repeated items by dividing by the factorial of the counts of each repeated item: \( P = \frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!} \).
Where \( n \) is the total number of items, and \( n_1, n_2 \ldots n_k \) are the counts of each repeated item.
Let's take the word 'AAABBC'. It has 6 letters, where A repeats 3 times, B repeats 2 times, and C appears once.
Using the formula: \( P = \frac{6!}{3! \cdot 2! \cdot 1!} = 60 \).
This adjusts for repeated items, giving us the number of distinct permutations.
Distinct arrangements
Distinct arrangements focus on finding unique ways to order a set, considering any repetitions.
In simpler terms, we want to count the different ways items can be ordered so that each arrangement looks unique.
We achieve this by using the permutations with repetition formula to adjust for duplicate items. For example, in the word AAABBC, calculating distinct arrangements involves:
In simpler terms, we want to count the different ways items can be ordered so that each arrangement looks unique.
We achieve this by using the permutations with repetition formula to adjust for duplicate items. For example, in the word AAABBC, calculating distinct arrangements involves:
- Counting total letters (6)
- Identifying repetitions (A=3, B=2, C=1)
- Applying the formula: \( \frac{6!}{3! \cdot 2! \cdot 1!} = 60 \).