Problem 17 - Algebra for College Students

Question

How many three-letter permutations can be formed from the first eight letters of the alphabet (a) if repetitions are not allowed? (b) if repetitions are allowed? (a) 336 (b) 512

Step-by-Step Solution

Answer

(a) 336 permutations without repetition; (b) 512 permutations with repetition.

Step 1: Understanding the Problem

We need to find the number of permutations of three letters that can be formed from the first eight letters of the alphabet. We have two scenarios: (a) repetitions are not allowed, and (b) repetitions are allowed.

Step 1: Calculate without Repetition (a)

Here, we need to find how many permutations can be made without repeating any letter. Each position in the permutation can be filled without repeating the letter. Start from 8 options for the first position, 7 for the second (as one letter is already used), and 6 for the third. The number of permutations is calculated as follows: \[ 8 \times 7 \times 6 = 336 \]

Step 2: Calculate with Repetition (b)

In this case, each position in the three-letter permutation can be filled with any of the eight letters since repetition is allowed. Therefore, each of the three positions has 8 possibilities. The total number of permutations is: \[ 8 \times 8 \times 8 = 8^3 = 512 \]

Step 4: Conclusion

For part (a), where repetitions are not allowed, the number of permutations is 336. For part (b), where repetition is allowed, the number of permutations is 512.

Key Concepts

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, arrangement, and combination of elements within sets. In this context, we explore how different sequences of elements, known as permutations, can be formed. In problems like the one you're facing, combinatorial principles help determine possible arrangements of letters in sequences. By understanding combinatorics, we can systematically count the number of ways to arrange a set of items. This involves analyzing whether each arrangement respects the problem's constraints, such as repetition and specific ordering.

Beyond just listing options, combinatorics provides us with formulas and principles for efficient counting, especially useful with large sets where manual counting is implausible.

Repetition

Understanding repetition in permutations is crucial to solving many combinatorial problems. Repetition refers to the ability to use the same element more than once in an arrangement. It's like allowing a letter in our word puzzle to appear in more than one position.

There are two scenarios:

• **Without repetition**: Here, once you've used an element, it can't be used again. For example, if you pick a letter "A" first, you have to choose from the remaining different letters for subsequent positions.
• **With repetition**: Elements can appear multiple times. This means each position can independently be filled by any of the available options, as shown in part (b) of the original solution.

Both cases significantly impact the count of permutations possible, affecting how combinatorial problems are solved.

Factorial

Factorials are another fundamental concept in combinatorics, especially useful to understanding permutations when repetition is not permitted. The factorial of a number, denoted as \( n! \), is the product of all positive integers up to \( n \). For instance, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \).

Factorials simplify the calculation of permutations by providing a straightforward way to compute the number of ways to order elements. For instance, if ordering all eight letters without repetition, the count would be \( 8! \). In the problem given, because only three positions need filling from eight letters, and the rest four don't need ordering, it's calculated differently without full factorial use: you multiply options decrementally for available positions.

This makes understanding factorials key to mastering permutation problems where order matters.