Problem 35 - Algebra and Trigonometry

Question

Five-Letter Words Five-letter "words" are formed using the letters \(A, B, C, D, E, F, G .\) How many such words are possible for each of the following conditions? (a) No condition is imposed. (b) No letter can be repeated in a word. (c) Each word must begin with the letter \(A\) . (d) The letter \(C\) must be in the middle. (e) The middle letter must be a vowel.

Step-by-Step Solution

Answer

(a) 16,807, (b) 2,520, (c) 2,401, (d) 2,401, (e) 4,802.

Step 1: Total Possible Words Without Restrictions

Each position in the five-letter word can be filled by any of the 7 letters: \( A, B, C, D, E, F, G \). So, for a 5-letter word, there are \(7\) choices for each position. The total number of possible words is calculated as follows:\[ 7^5 = 16807 \]Thus, there are 16,807 possible words when no condition is imposed.

Step 2: Words with No Repeated Letters

When no letter is repeated, the first position has \(7\) choices, the second has \(6\) (as one letter is used), the third has \(5\), the fourth has \(4\), and the fifth has \(3\).Calculate the total number of words:\[ 7 \times 6 \times 5 \times 4 \times 3 = 2520 \]There are 2,520 such words without repeated letters.

Step 3: Words Beginning with 'A'

The first position is fixed as 'A', so it has 1 choice. The remaining four positions can each be filled by one of the 7 available letters.Calculate the total number of words:\[ 7^4 = 2401 \]Thus, there are 2,401 words where 'A' is the starting letter.

Step 4: Words with 'C' in the Middle

If 'C' is fixed in the middle position, there are 4 remaining positions that can each be filled by one of the 7 letters.Calculate the total number of words:\[ 7^4 = 2401 \]Therefore, 2,401 words have 'C' in the middle position.

Step 5: Words with a Vowel in the Middle

The vowels available are \(A\) and \(E\). If a vowel is fixed in the middle, there are 2 choices for this position.The remaining four positions can each be filled by one of the 7 letters.Calculate the total number of words:\[ 2 \times 7^4 = 2 \times 2401 = 4802 \]Thus, there are 4,802 words where the middle letter is a vowel.

Key Concepts

Five-Letter Words

In this exercise, we are diving into the realm of constructing five-letter words from a given set of letters. Let's break down what these words entail:

• Our alphabet for these five-letter words includes the letters: \( A, B, C, D, E, F, G \).
• A word is formed by placing one letter in each of the five positions.
• The conditions change how we select which letters can go into each position.

Understanding the constraints, like letter repetition or specific positions for certain letters, is crucial in combinatorial problems.By examining scenarios with and without these constraints, we can discover numerous intriguing combinations!The beauty of combinatorics shines through in these structured yet creative solutions, bringing the simplicity of letter arrangement to life.

Permutations

Permutations are all about arranging a set of items where the order matters. For five-letter words, each unique rearrangement forms a new permutation.

• The concept can be visualized as filling positions one after another, considering the availability and constraints for each spot.
• In the unrestricted scenario, any of the 7 letters can fill each position, leading us to the formula: \(7^5\), which equals 16,807 unique permutations.
• When constraints such as not repeating letters are added, fewer permutations are possible. In this case, the number is reduced to \(7 \times 6 \times 5 \times 4 \times 3\) or 2,520 permutations because each choice reduces the pool of available letters.

Permutations demonstrate the vast number of ways elements can be arranged orderly even in seemingly straightforward scenarios like forming a word.

Vowels

Vowels often play critical roles in combinatorics exercises. Here, our vowels are \( A \) and \( E \). When we place a vowel in a specific position, like the middle of our five-letter word, we shape the word's structure significantly.

• The exercise involves placing either \( A \) or \( E \) in the middle position of the word, giving us 2 choices for this central spot.
• Once a vowel is fixed in the middle, the remaining positions can be filled with any of the 7 given letters, each position being independent of the others.
• This results in the calculation: \(2 \times 7^4\), equaling 4,802 possible words with a vowel in the middle.

Understanding how vowels can impact word structure helps in grasping broader combinatorial principles.

Combinations

Combinations focus on selecting items without involving the order of those items. While not as central in this permutation-heavy exercise, understanding combinations aids in comprehensively tackling any problem.

• Suppose we are choosing which vowels could go into a specific position, like the middle of our word. We're interested in how these choices are combined with other elements rather than their order.
• Though the concept of combinations directly impacts permutations, it subtly underpins decisions like fixing a vowel in the middle position.
• Combinations often serve as the groundwork for defining how limitations affect the problem’s outcomes and enhance the importance of constraints.

Knowing combinations enriches one’s ability to discern how groups of elements act within permutations, fostering a deeper insight into structuring and solving combinatorial challenges.