Question

Show that there are \(\left(3^{n}+1\right) / 2\) strings of length \(n\) consisting of the letters \(a, b\), and \(x\) in which \(a\) occurs an even number of times.

Short answer

There are \((3^n + 1) / 2\) strings of length \(n\) where \(a\) occurs an even number of times.

Step-by-step solution

Understand the Problem

We need to count the total number of strings of length \(n\) consisting of the letters \(a, b\), and \(x\) such that the letter \(a\) occurs an even number of times in each string.

Total Number of Strings

Calculate the total number of strings of length \(n\) using the three letters \(a, b, x\). Each position in the string has 3 options (either \(a\), \(b\), or \(x\)), and there are \(n\) positions, so there are \(3^n\) possible strings.

Use Parity Argument

For the strings of length \(n\), each string can be classified based on whether the count of \(a\) is even or odd. We find the same number of strings with an even number of \(a\)s as there are with an odd number. This is because for each string with \(a\) appearing an odd number of times, we can find a corresponding string (by changing one \(a\) to a non-\(a\) character) that results in it having \(a\) an even number of times, and vice versa.

Calculate Number of Strings with Even 'a'

Since the number of strings with an even number of \(a\)'s is equal to the number of strings with an odd number of \(a\)'s, we divide the total number of strings by 2: \(\frac{3^n}{2}\). However, because \(3^n\) is odd (as \(3^n\) is not divisible by 2), we use the ceiling function: \(\frac{3^n + 1}{2}\). This ensures the even count covers exactly half (or slightly more by 1 if required in odd cases) of the total strings.

Key concepts

Counting

Counting in mathematics is all about determining the number of elements within a set. In this exercise, we are tasked with counting specific types of strings formed from three possible letters: \(a\), \(b\), and \(x\). These strings are of length \(n\), and we want to find how many of them have an even count of the letter \(a\).
Here’s how we dive into the solution:

• First, calculate how many total strings can be formed. Since there are 3 possible choices for each position within the string, and the string has \(n\) positions, the total number of combinations is \(3^n\).
• Next, consider the parity of the number of \(a\)s in these strings. We split these strings into those with an even count of \(a\) and those with an odd count.

Understanding the concept of parity will guide us in further breaking down this counting problem.

Parity Argument

The parity argument is a clever mathematical tool used to determine whether a number is even or odd. In this context, it helps us determine the number of strings where the letter \(a\) appears an even or odd number of times.
The fundamental idea here is symmetry:

• For any given string with an odd number of \(a\)s, there is a corresponding string with an even number of \(a\)s. This can be achieved by changing one \(a\) to another letter, such as \(b\) or \(x\).
• Conversely, changing a non-\(a\) character in a string with an even number of \(a\)s to \(a\), you get a string with an odd number.

Since these transformations are essentially one-to-one, it indicates that the number of strings with even \(a\)s and odd \(a\)s are equal. Thus, using parity arguments, we can balance these counts.

Combinatorics

Combinatorics explores the art of counting, arranging, and combination formation. Here, we apply combinatorial logic to determine how many strings fit our criteria.
To solve this problem, we relied on a key combinatorial argument:

• We compute the total elements possible (i.e., all possible strings) and dissect them into sub-categories based on a specific condition—the number of \(a\)s being even.
• By dividing the total count of strings, \(3^n\), by 2, we aimed to find the subset meeting our condition (even \(a\)s).
• However, since \(3^n\) is often odd, we need to adjust for rounding up, which leads to using \((3^n + 1)/2\), ensuring adequacy of covering the half plus any necessary adjustment.

This combinatorial approach provides a straightforward yet deep insight into counting even-based occurrences within string arrangements.