Problem 47 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the number of distinguishable permutations of the group of letters. \(\mathrm{A}, \mathrm{A}, \mathrm{G}, \mathrm{E}, \mathrm{E}, \mathrm{E}, \mathrm{M}\)

Step-by-Step Solution

Answer

The number of distinguishable permutations is 420.

Step 1: Identify the elements and their frequencies

First, list the elements in the set and the frequency of each element. In this task, we have A appearing twice, G appearing once, E appearing three times, and M appearing once

Step 2: Calculate the total number of permutations

Next, calculate the total number of permutations of the seven letters by using the formula for the permutation of n objects taken as a whole, which is n!. This gives \(7!\)

Step 3: Adjust for indistinguishable elements

Since we cannot distinguish between the two As and three Es respectively, we divide the total number of permutations by the factorial of the frequency of As and Es. This gives \(\frac{7!}{2!3!}\)

Step 4: Calculate the final solution

Finally, compute the number by using the formula \(\frac{7!}{2!3!}\), which simplifies to 420 distinguishable permutations

Key Concepts

Combinatorics

Combinatorics is a branch of mathematics that deals with the counting, arrangement, and combination of objects. It is elemental when we aim to understand how many different ways items can be organized or selected. Think of it as the math version of solving a puzzle: finding all the possible solutions given a series of elements with certain conditions.

In our exercise involving the permutations of letters, combinatorics lets us calculate the different ways we can arrange the letters AAGEEE and M. This seemingly playful task has practical applications, such as arranging people in a line, creating passwords, or organizing tournaments. It's the backbone of probability theory, allowing us to predict the likelihood of certain events.

Factorial Notation

Factorial notation is a mathematical concept represented by an exclamation mark (!). It's used to describe the product of an integer and all the positive integers below it. For example, the factorial of 4, denoted as 4!, is calculated as 4 × 3 × 2 × 1, which equals 24.

Functionally, it is a shortcut to express the multiplication required when we are dealing with permutations or combinations. Factorials grow at an extremely fast rate with each increment of the number, playing a crucial role in combinatorics, particularly when we calculate the total possible permutations of a set, as seen in our exercise with the letters. When confronted with the task of finding permutations, our route typically starts with calculating factorials.

Permutations with Repetition

Permutations with repetition occur when we arrange items that include repeats of some elements, which is often the case in real-world scenarios. The fundamental principle here is that the permutation of the entire set of objects is divided by the permutation of identical objects, which avoids overcounting.

In case of our exercise, the repetition comes from the letters A and E. The presence of identical letters means some arrangements will look the same, which is why we need to adjust our counting. The adjusted formula to calculate permutations in this context is the total number of permutations divided by the product of factorials representing the repeated elements. It helps ensure each unique arrangement is only counted once, and this is why our final calculation is \(\frac{7!}{2!3!}\), which indeed reflects the correct number of distinguishable permutations, avoiding redundancy in arrangements.