Problem 41 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

In how many ways can the letters of the Web site MySpace be arranged if all of the letters are used and the vowels \(a\) and \(e\) must always stay in the order \(a e\) ?

Step-by-Step Solution

Answer

The letters of the website "MySpace" can be arranged in 180 different ways if all of the letters are used and the vowels "a" and "e" stay in the order "a e".

Step 1: Identify the number of distinct letters

Observe that the given word, "MySpace," has a total of 7 letters. In these 7 letters, "S" and "M" are repeated while the others (besides the vowels "a" and "e") are distinct. Now, we will treat the vowels "a" and "e" as a single unit, creating a total of 6 distinct elements to arrange: "MySP_ce"

Step 2: Find the number of ways to arrange the elements, including the combined vowels

With 6 distinct elements ("M", "y", "S", "P", "_", and "c"), we can use the permutation formula to find the number of ways to arrange them. The number of ways to arrange n distinct objects is given by n! (the factorial of n). For 6 elements, the number of ways will be 6! which can be calculated as: 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720 However, there is one more thing to consider here. Since "M" and "S" are repeated twice, we have to divide the total arrangements by the factorial of repetitions (2! for each, since there are two of each). So, the number of ways to arrange the elements = \(\frac{6!}{(2!)^2}\)

Step 3: Calculate the number of ways to arrange the elements

Now, we will calculate the number of ways to arrange the elements by plugging in the values into the formula: \(\frac{6!}{(2!)^2} = \frac{720}{(2 * 2)} = \frac{720}{4} = 180\) Hence, there are 180 ways to arrange the elements (with vowels "a" and "e" as a combined unit).

Step 4: Find the total number of ways to arrange the given word, including the vowels in order

Since the "a" and "e" vowels can only be arranged in one particular way as a combined unit (\((1 * 1)\)), the total number of ways to arrange the provided word with all the letters used and the required order of vowels is the same as the number of ways found in Step 3: Total number of ways = 180 In conclusion, the letters of the website "MySpace" can be arranged in 180 different ways if all of the letters are used and the vowels "a" and "e" must stay in the order "a e".

Key Concepts

Factorial

Factorial is a fundamental concept in permutations and combinations. It's used to find the number of ways to arrange a set of objects. The factorial of a number, denoted as \( n! \), is the product of all positive integers from 1 to \( n \). For example:

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

Factorials help calculate the total arrangements of distinct objects. If you have 6 distinct objects, they can be arranged in \( 6! = 720 \) ways.
Understanding factorials allows you to solve complex arrangement problems effectively. Remember, the larger the number, the more arrangements there will be!

Repetition in Permutations

Repetition in permutations occurs when some objects in a set are identical, making some arrangements look the same. To account for this, you have to divide the total number of arrangements by the factorial of each repetition.
Consider arranging the word "MySpace". Since both "M" and "S" appear twice, applying repetition helps: Given 6 items (M, y, S, P, ae, c), we first calculate \( 6! \). Then, to address repetition, divide by \((2!)^2\):

• \( \frac{6!}{(2!)^2} = \frac{720}{4} = 180 \)

Repetition ensures we don't overcount configurations that are actually the same.

Vowel Arrangement

In permutation problems, sometimes certain elements must stay together or in a specific order. This is common with vowels. For the exercise related to "MySpace," vowels "a" and "e" must always be in order "a e".
Treat these vowels as a single unit or "block" when organizing arrangements. This reduces the complexity by considering them combined: Think of "ae" as one element, alongside others like "M", "y", "S", "P", and "c". So, the arrangement becomes a problem of 6 elements, where "ae" is fixed in its order.

• This technique simplifies your calculation without disrupting the original constraint.

Such strategies are useful for preserving required orders among elements in problems.