Chapter 7: Problem 11
How many permutations are there for the 26 letters of the alphabet if the five vowels occur together?
Short Answer
Expert verified
The number of permutations is \(23! \times 5!\).
Step by step solution
01
Treat the vowels as a single unit
The problem states that the five vowels (A, E, I, O, U) must occur together. Treat these five vowels as a single unit or block. This means we are essentially arranging 22 consonants (non-vowel letters) and 1 block of vowels.
02
Calculate permutations of the consonants and vowel block
Since the consonants and the block of vowels need to be arranged, we have 22 consonants and 1 block. Thus, there are a total of 23 units to arrange. The number of permutations of 23 units is given by the factorial of 23, which is calculated as: \(23!\).
03
Calculate permutations within the vowel block
Within the vowels block, the five vowels (A, E, I, O, U) can be arranged among themselves. The number of permutations of these 5 vowels is \(5!\).
04
Calculate the total number of permutations
To get the total number of permutations, multiply the permutations of the 23 units (consonants and a vowel block) by the permutations within the vowel block: \(23! \times 5!\).
05
Compute the factorial values
Calculate \(23!\) and \(5!\). Specifically:- \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).- Calculate \(23!\), which may require a calculator as it is a large value.
06
Multiply to find the final answer
Finally, multiply the values: \(23! \times 5!\), using the computed result from Step 5.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factorial Calculation
Factorial calculation is a key concept in solving permutations problems. A factorial, represented by an exclamation mark (!), is the product of all positive integers up to a certain number. For example, for a number \( n \), \(!\) signifies the factorial of \( n \), which is \( n \times (n-1) \times (n-2) \times \cdots \times 1 \).
This operation is fundamental in determining permutations as it accounts for arranging \( n \) distinct items in a sequence. In the problem with the 26 letters, we use factorials to calculate how to arrange both the consonants plus the block of vowels \((23!)\), and then how to arrange the vowels themselves \((5!)\). These results give us the total possible arrangements when multiplied together: \(23! \times 5!\).
When calculating factorials, remember that the values grow rapidly:
This operation is fundamental in determining permutations as it accounts for arranging \( n \) distinct items in a sequence. In the problem with the 26 letters, we use factorials to calculate how to arrange both the consonants plus the block of vowels \((23!)\), and then how to arrange the vowels themselves \((5!)\). These results give us the total possible arrangements when multiplied together: \(23! \times 5!\).
When calculating factorials, remember that the values grow rapidly:
- \(5! = 120\)
- \(23!\) is a very large number, often requiring a calculator for exact calculation
Block Method
The Block Method is a clever trick to simplify permutation problems with restrictions. It involves treating a group of elements that must remain together as a single "block."
In our exercise, the vowels \(A, E, I, O, U\) are regarded as one block because they need to occur together. Instead of treating each vowel individually, we consider them as a collective unit. This reduces the number of separate units to arrange, simplifying the problem.
After defining the vowel block, the task is to arrange 22 consonants and this one vowel block. This results in arranging 23 distinct units. The permutations of these units are given by the number \(23!\), demonstrating how the Block Method streamlines calculations in permutation tasks with grouping restrictions.
In our exercise, the vowels \(A, E, I, O, U\) are regarded as one block because they need to occur together. Instead of treating each vowel individually, we consider them as a collective unit. This reduces the number of separate units to arrange, simplifying the problem.
After defining the vowel block, the task is to arrange 22 consonants and this one vowel block. This results in arranging 23 distinct units. The permutations of these units are given by the number \(23!\), demonstrating how the Block Method streamlines calculations in permutation tasks with grouping restrictions.
Permutations of Vowels
Permutations become even more interesting when they focus on a specific group, like vowels. In permutations, we can arrange elements in every possible sequence.
For the set of five vowels \(A, E, I, O, U\), we want to count how these can be rearranged among themselves. This is calculated using their factorial, \(5!\). This operation considers every position each vowel could occupy, creating multiple sequence possibilities.
Calculating \(5!\) gives us 120 ways to arrange the vowels. It emphasizes how the positioning of vowels affects overall arrangements when they must be kept together in the main problem. Thus, understanding permutations within a block is crucial for solving complex permutations problems.
For the set of five vowels \(A, E, I, O, U\), we want to count how these can be rearranged among themselves. This is calculated using their factorial, \(5!\). This operation considers every position each vowel could occupy, creating multiple sequence possibilities.
Calculating \(5!\) gives us 120 ways to arrange the vowels. It emphasizes how the positioning of vowels affects overall arrangements when they must be kept together in the main problem. Thus, understanding permutations within a block is crucial for solving complex permutations problems.
Combinatorics
Combinatorics is a branch of mathematics that studies counting, arranging, and exploring the structure of sets. It provides the foundation for understanding how we can select and organize items in different ways.
This exercise highlights using combinatorics to solve permutation problems with restrictions. The first task is to manage specific requirements, such as grouping the vowels. Combinatorics helps identify various permutations, ensuring the vowels stay together while conforming to other problem constraints.
By utilizing combinatorics principles, such as the Block Method and factorial calculation, we can systematically address complex restrictive conditions. It allows us to break down the problem into manageable segments, like arranging consonants separately from a collective vowel block, to efficiently find the total number of unique sequence arrangements possible.
This exercise highlights using combinatorics to solve permutation problems with restrictions. The first task is to manage specific requirements, such as grouping the vowels. Combinatorics helps identify various permutations, ensuring the vowels stay together while conforming to other problem constraints.
By utilizing combinatorics principles, such as the Block Method and factorial calculation, we can systematically address complex restrictive conditions. It allows us to break down the problem into manageable segments, like arranging consonants separately from a collective vowel block, to efficiently find the total number of unique sequence arrangements possible.
