Problem 41 - Algebra 2 and Trigonometry

Question

How many 2-letter patterns can be formed from the alphabet if the first letter is a vowel (A, E, I, O, U) and the second letter is a consonant that occurs later in the alphabet than the vowel in the first position?

Step-by-Step Solution

Answer

There are 77 possible 2-letter patterns.

Step 1: Identify the Vowels

First, list all the vowels: A, E, I, O, U. These will be the only possible choices for the first letter in the pattern.

Step 2: Determine Consonants after Each Vowel

For each vowel, identify the consonants that appear later in the alphabet. - After A: B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z - After E: F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z - After I: J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z - After O: P, Q, R, S, T, U, V, W, X, Y, Z - After U: V, W, X, Y, Z

Step 3: Count Consonants for Each Vowel

Calculate how many consonants are available after each vowel. - After A, there are 21 consonants. - After E, there are 20 consonants. - After I, there are 18 consonants. - After O, there are 13 consonants. - After U, there are 5 consonants.

Step 4: Sum Total Patterns

Add the total number of patterns for each vowel: - A: 21 patterns - E: 20 patterns - I: 18 patterns - O: 13 patterns - U: 5 patterns Sum total: 21 + 20 + 18 + 13 + 5 = 77 patterns.

Key Concepts

Vowel-Consonant Patterns

When creating 2-letter patterns from the alphabet, where the first letter is a vowel and the second letter is a consonant that occurs later in the alphabet, it results in unique rules for selection. This means the placement of vowels and consonants is critical to forming valid patterns.

Vowels serve as the starting point because they are fewer in number compared to consonants. In English, we have five vowels: A, E, I, O, and U. This fact limits our initial choices, making the process manageable.

After selecting a vowel, you need to look at consonants that follow it in the alphabet order. For instance:

• If the first letter is A, the second letter can be any consonant from B to Z.
• If E is selected, choices start from F onwards.
• This pattern continues, progressively reducing the options for each subsequent vowel.

This pattern ensures the second letter is in strict alphabetical order after the first, adhering to the rule set for the exercise.

Alphabet Combinations

Creating alphabet combinations, especially under constraints like ours, is a common application of combinatorics. This means taking a small set, like vowels, and systematically pairing each with eligible consonants.

For each vowel in our problem, only certain consonants are valid as they need to come later in the sequence. Thus, the challenge is in meticulously listing these valid consonants for each case.

Consider what happens with each vowel:

• A: affords 21 possibilities (B to Z).
• E: brings 20 potential pairings (F to Z).
• I offers 18 choices because only 18 consonants follow it.
• O narrows it down to 13.
• U leaves only 5 potential second letters.

Every combination respects the alphabetical constraint, ensuring a systematic approach that leverages the understanding of sequence and limitation.

Counting Techniques

Counting techniques form the backbone of combinatorics. These methods allow us to systematically determine possible outcomes without having to list them all manually.

In this problem, we apply counting techniques by first determining how many valid choices there are for each segment of our problem: first for vowels, and then for possible consonants after each vowel.

To establish the total number of patterns:

• First, we consider each vowel's combinations with subsequent consonants.
• Then, we sum these individual results to get the grand total.

Thus, adding up the number of patterns for each vowel gives:

• A (21) + E (20) + I (18) + O (13) + U (5) = 77 patterns.

This process highlights how counting techniques enable us to handle constraints effectively, guiding us towards an accurate solution while maintaining efficiency.