Problem 32 - Algebra and Trigonometry

Question

How many two-letter codes can be formed using the letters \(A, B, C, D,\) and \(E\) ? Repeated letters are allowed.

Step-by-Step Solution

Answer

25

Step 1 - Identify the number of choices for each position

For the first letter of the code, there are 5 possible choices since any of the letters A, B, C, D, and E can be used.

Step 2 - Determine the choices for the second letter

Since repeated letters are allowed, the second letter also has 5 possible choices from A, B, C, D, and E.

Step 3 - Calculate the total number of two-letter codes

Multiply the number of choices for the first letter by the number of choices for the second letter. This gives us: Number of codes = 5 (choices for the first letter) × 5 (choices for the second letter) = 25.

Key Concepts

Permutations

Permutations are arrangements of items in a specific order. When dealing with permutations, the order in which we arrange items matters. For example, the code 'AB' is different from the code 'BA'. The formula to calculate permutations without repetition is P(n, k) = n! / (n - k)!, where 'n' is the total number of items and 'k' is the number of items to arrange.
In exercises like forming two-letter codes from the letters A, B, C, D, and E, we consider every possible order.

Repetition in Combinations

Repetition in combinations means that items can be used more than once. For the given exercise, we need to form codes allowing repeated letters.
This is a bit different from typical combinations where each item is used only once. In the current problem, since repetition is allowed, each position in the code (first and second) can independently be any of the 5 letters (A, B, C, D, E).
So we apply the counting principle for each position without reducing the choice due to repetition.

Basic Counting Principles

Basic counting principles help in determining the total number of possible outcomes. One important principle is the multiplication rule, which states that if there are 'n' ways to do one thing and 'm' ways to do another, then there are n × m ways to do both.
In our exercise, we see this principle applied directly. There are 5 possible choices for the first letter and, since repetition is allowed, 5 possible choices for the second letter too.
Thus, we multiply these choices together to find the total number of two-letter codes: Number of codes = 5 × 5 = 25.