Problem 31 - Algebra for College Students

Question

How many different seven-letter permutations can be formed from the seven letters of the word ALGEBRA? 2520

Step-by-Step Solution

Answer

There are 2520 different permutations of the word ALGEBRA.

Step 1: Understanding the Problem

We need to find the number of different permutations of the seven letters in the word "ALGEBRA". A permutation refers to the arrangement of all items (letters, in this case) in all possible orders.

Step 2: Counting Unique Letters and Repetitions

The word "ALGEBRA" is composed of seven letters, with the letter 'A' appearing twice. The letters 'L', 'G', 'E', 'B', and 'R' each appear once.

Step 3: Using the Permutation Formula

For permutations involving repeated letters, we use the formula \[\frac{n!}{p_1! \times p_2! \times \, ... \, \times p_k!}\]where \(n\) is the total number of letters, and \(p_1, p_2, ..., p_k\) are the frequencies of the repeated letters. In this case, \(n = 7\), and there is one repetition: two 'A's, so \(p_1 = 2\).

Step 4: Calculating the Factorials

Calculate the factorials: - The factorial of 7 is \(7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040\).- The factorial of 2 for the two 'A's is \(2! = 2 \times 1 = 2\).

Step 5: Applying the Formula

Substitute the values into the permutation formula:\[\frac{7!}{2!} = \frac{5040}{2} = 2520\]This calculation gives us the total number of distinct permutations of the letters.

Key Concepts

Factorials

Factorials are foundational in understanding permutations and combinations. A factorial of a non-negative integer is represented by the notation \( n! \). It is the product of all positive integers less than or equal to \( n \). For example, the factorial of 7, denoted as \( 7! \), is calculated as:

• \( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \)

Factorials are crucial when determining permutations since they calculate how many ways you can arrange \( n \) distinct items. Notice how quickly factorials grow: even with only seven items, the total arrangements reach 5040, illustrating how factorial calculations expand possibilities rapidly.

Repeated Letters

When computing permutations with repeated letters, we must adjust our calculations to account for these repetitions. A repeated letter means that some arrangements will look identical, which reduces the actual number of unique permutations.

In the word 'ALGEBRA,' the letter 'A' appears twice. To find distinct permutations, we divide by the factorial of the number of repeats. This is expressed in the permutation formula as dividing by \( p_1! \), where \( p_1 \) is the count of the first repeated letter. For 'ALGEBRA', we calculate using \( 2! \) for the two 'A's, which equals 2.

• \( 2! = 2 \times 1 = 2 \)

By dividing by \( 2! \), we remove duplicates caused by repeating 'A's, ensuring only unique permutations are counted.

Arrangement of Letters

Arrangement of letters in permutations deals with ordering elements in all possible ways. In the context of the provided problem, we want to arrange the letters in 'ALGEBRA' such that each arrangement is unique and all letters are used.

• 'A', 'L', 'G', 'E', 'B', 'R', 'A'

This is crucial as permutations rely on positions being distinguishable. Even though two 'A's are not distinct, every permutation considers position. Thus, with different letters, arrangements would differ significantly in number. The order matters highly in permutations.

Permutation Formula

The permutation formula for handling repeated items is an adaptation of the basic permutation concept. For a string of letters where one or more letters repeat, the formula

• \( \frac{n!}{p_1! \times p_2! \times \ldots \times p_k!} \)

is used to calculate the unique arrangements. Here, \( n \) refers to the total number of items, and \( p_1, p_2, \ldots, p_k \) are the factorials of the counts of each repeating letter. In our case, 'ALGEBRA' uses:

• \( n = 7 \), since there are 7 letters
• \( p_1 = 2 \), related to the twice occurring 'A'

The calculation involved:

• \( \frac{7!}{2!} = \frac{5040}{2} = 2520 \)

This formula efficiently yields the correct number of distinct permutations, considering all repetitions.