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Magazine Chapter 4: Problem 27
Express all probabilities as fractions. One of the longest words in standard statistics terminology is "homoscedasticity." How many ways can the letters in that word be arranged?
Short Answer
Expert verified
1307674368000
Step by step solution
01
Identify the Total Number of Letters
Count the total number of letters in the word 'homoscedasticity'. There are 16 letters in total.
02
Identify Repeated Letters
Identify the letters that repeat and count their frequency. The word 'homoscedasticity' has the following repeated letters:- h: 1- o: 2- m: 1- s: 2- c: 2- e: 1- d: 1- a: 1- t: 2- i: 2- y: 1
03
Use Permutation Formula for Repeated Letters
To find the number of unique arrangements of the word, use the permutation formula for repeated items:different arrangements formula: \[ P = \frac{n!}{n_1! \times n_2! \times n_3! \times ... \times n_k!} \]where \( n \) is the total number of letters, and \( n_1, n_2, ... n_k \) are the frequencies of each repeated letter.For 'homoscedasticity': \[ P = \frac{16!}{1! \times 2! \times 1! \times 2! \times 2! \times 1! \times 1! \times 1! \times 2! \times 2! \times 1!} \]
04
Compute the Factorials
Compute the factorial values:\[ 16! = 20922789888000 \]\[ 2! = 2 \]The denominator will be:\[ 1! \times 2! \times 1! \times 2! \times 2! \times 1! \times 1! \times 1! \times 2! \times 2! \times 1! \]Calculate: \[ 2 \times 2 \times 2 \times 2 = 16 \]So the denominator is:\[ 1 \times 2 \times 1 \times 2 \times 2 \times 1 \times 1 \times 1 \times 2 \times 2 \times 1 = 16 \]
05
Calculate the Final Result
Apply the formula: \[ P = \frac{20922789888000}{16} = 1307674368000 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Permutations
Permutations are about arrangements and order. In simple terms, it's finding how many ways we can arrange items. For example, the word 'cat' has different permutations: 'cat', 'cta', 'act', 'atc', 'tca', and 'tac'.
When dealing with repeated letters, it becomes more complex. The formula changes to account for duplicates so we're not counting them more than once. The formula used in the exercise is:
\[ P = \frac{n!}{n_1! \times n_2! \times n_3! \times ... \times n_k!} \]Here, 'n' is the total number of items, and 'n_1, n_2,..., n_k' are the frequencies of the repeated items.
Understanding this makes it clear that 'homoscedasticity' will have fewer unique arrangements because several letters repeat.
When dealing with repeated letters, it becomes more complex. The formula changes to account for duplicates so we're not counting them more than once. The formula used in the exercise is:
\[ P = \frac{n!}{n_1! \times n_2! \times n_3! \times ... \times n_k!} \]Here, 'n' is the total number of items, and 'n_1, n_2,..., n_k' are the frequencies of the repeated items.
Understanding this makes it clear that 'homoscedasticity' will have fewer unique arrangements because several letters repeat.
Combinatorics
Combinatorics is the math of counting things. It helps us find out how to count combinations and permutations. In our exercise, we used combinatorics to find the number of ways to arrange 'homoscedasticity'.
The distinction between combinations and permutations is important: permutations consider order, while combinations do not. Calculating permutations with repeated letters is a specific combinatorics problem.
For example, if you have 3 apples and 2 oranges, combinatorics helps determine how many different ways you can order them. The core of the solution lies in these principles of 'counting' effectively and accurately.
The distinction between combinations and permutations is important: permutations consider order, while combinations do not. Calculating permutations with repeated letters is a specific combinatorics problem.
For example, if you have 3 apples and 2 oranges, combinatorics helps determine how many different ways you can order them. The core of the solution lies in these principles of 'counting' effectively and accurately.
Factorials
Factorials are mathematical functions that grow quickly. The factorial of a number (written as 'n!') is the product of all positive integers up to that number. For instance, 4! equals 4 脳 3 脳 2 脳 1 = 24.
In permutations with repeated letters, factorials help calculate the total possible arrangements. In our solution:\[ 16! = 20922789888000 \]It's a giant number! Factorials also calculate smaller groups (like the repeated ones), for example:\[ 2! = 2 \]So, to simplify large calculations and avoid counting duplicates, factorials are essential.
In permutations with repeated letters, factorials help calculate the total possible arrangements. In our solution:\[ 16! = 20922789888000 \]It's a giant number! Factorials also calculate smaller groups (like the repeated ones), for example:\[ 2! = 2 \]So, to simplify large calculations and avoid counting duplicates, factorials are essential.
Probability
Probability measures the chance of something happening. In the context of permutations with repeated letters, it tells us how likely a specific arrangement is. Though not directly used in our permutation calculation, understanding probability gives a deeper insight into why we don't double count.For example, if we consider all possible arrangements of letters as equally likely, the probability of any single arrangement happening is:\[ \text{Probability} = \frac{1}{\text{Number of Unique Permutations}} \]In the case of 'homoscedasticity', with 1,307,674,368,000 unique permutations, the probability of one specific arrangement is very small. This shows how combinatorics and probability often intersect in these types of problems.
