Problem 55 - College Algebra

Question

In Exercises \(49-58,\) solve by the method of your choice. How many different four-letter passwords can be formed from the letters \(A, B, C, D, E, F,\) and \(G\) if no repetition of letters is allowed?

Step-by-Step Solution

Answer

There are 840 different four-letter passwords that can be formed from the letters A, B, C, D, E, F, and G without any repetitions allowed.

Step 1: Understanding the Permutations Concept

A permutation is an arrangement or sequence of objects, people, elements, and so on. The order in which these elements are arranged is important. Since in this context, the repetition of letters is not allowed while making the password, we need to use a formula for permutations: \( P(n, r) = n! / (n-r)! \) , where \(n\) is the total number of options (the letters A to G), and \(r\) is the number of selections (4-letter password).

Step 2: Apply the Permutations Formula

For our problem, \(n\) = 7 (the letters A to G) and \(r\) = 4 (four letter password). Therefore, using the permutations formula, we get: \( P(7, 4) = 7! / (7-4)! = 7! / 3! \)

Step 3: Solve the Factorials and the Division

Calculate the factorial of 7 (7! = 7*6*5*4*3*2*1 = 5040) and the factorial of 3 (3! = 3*2*1 = 6). Then, divide the factorial of 7 by the factorial of 3 (5040 / 6) to get the total number of permutations possible.

Key Concepts

Factorials

To grasp permutations, understanding factorials is crucial. A factorial, denoted by an exclamation point (!), is the product of all positive integers up to a given number. For example, the factorial of 5, written as 5!, equals 5 x 4 x 3 x 2 x 1, which is 120.
Factorials are fundamental in permutations because they help calculate the number of possible arrangements. In our problem, we need to compute 7! because there are seven letters to choose from, and each letter needs a position in the sequence when forming passwords. Here’s how factorials break down in steps:

• 7! calculates the total arrangements if all seven letters were used. It's the product of counting down from 7 to 1, resulting in 5040.
• 3!, which is calculated as 3 x 2 x 1, gives us 6. This helps in determining how to adjust our total arrangements to fit scenarios where less than seven letters are used.

Remember, factorials grow rapidly as numbers increase, making them powerful tools in permutation and combination calculations.

Combinatorics

Combinatorics is a branch of mathematics focused on counting, arrangement, and combination of elements within a set. It is what underlies all permutation and combination problems. In this context, it helps us understand how different arrangements can form distinct outcomes.
When solving problems like forming a four-letter password from seven letters, combinatorics principles dictate counting how many different ways we can arrange our selections. Importantly, combinatorics deals with conditions imposed on arrangements, such as no repetition and the significance of order.
To apply these principles:

• Consider all possible elements – here, the letters A through G.
• Determine how many elements must be selected simultaneously – for our password, that’s four letters.
• Decide whether the order matters – in this case, it does, thus making it a permutation problem.

Through combinatorics, you simplify complex counting processes, enabling efficient solution finding.

Permutation Formula

The permutation formula is used when the order of elements is important. It calculates how many different sequences can be made from a given set of objects. For our password problem, where no letter repetition is allowed, the formula is:\[ P(n, r) = \frac{n!}{(n-r)!} \]Here, \(n\) is the total number of options, and \(r\) is how many we choose at a time.
Applying the formula to our example:

• \(n = 7\) because we have seven letters (A through G).
• \(r = 4\) because our password is four letters long.

Substitute these into the formula: \[ P(7, 4) = \frac{7!}{(7-4)!} = \frac{7!}{3!} \]
You first calculate the factorial of 7, which is 5040, then the factorial of 3, which is 6. Dividing these gives you 840 different possible passwords without repeating any letters. This demonstrates how the permutation formula is crucial for questions where order and selection are pivotal, like forming secure passwords.