Problem 33 - Algebra and Trigonometry

Question

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) 16807, (b) 2520, (c) 2401, (d) 2401, (e) 4802.

Step 1: Count words with no conditions

Each of the five positions in a word can be filled using any of the 7 letters: A, B, C, D, E, F, G. Thus, for each position, we have 7 choices.The total number of words formed is calculated as the product of choices for each position:\[7 \times 7 \times 7 \times 7 \times 7 = 7^5 = 16807.\]

Step 2: Count words with no repetition

When no letter can be repeated, we choose a different letter for each position from the remaining unused options.- First position: 7 choices- Second position: 6 choices (one letter used)- Third position: 5 choices- Fourth position: 4 choices- Fifth position: 3 choicesThe total number of such words is:\[7 \times 6 \times 5 \times 4 \times 3 = 2520.\]

Step 3: Count words starting with A

Since each word must begin with A, the first letter is fixed, and we have to arrange the remaining 4 letters.For the other four positions, each can be filled with any of the 7 letters:\[1 \times 7 \times 7 \times 7 \times 7 = 7^4 = 2401.\]

Step 4: Count words with C in the middle

Fix C in the middle position of a 5-letter word. The remaining 4 positions can be any of the 7 letters:The total number of such words is:\[7 \times 7 \times 1 \times 7 \times 7 = 7^4 = 2401.\]

Step 5: Count words with a vowel in the middle

The middle letter must be A or E (the vowels). For each vowel choice in the middle, arrange the remaining 4 positions with any of the 7 letters.- If middle is A: \[7 \times 7 \times 1 \times 7 \times 7 = 7^4 = 2401.\]- If middle is E: \[7 \times 7 \times 1 \times 7 \times 7 = 7^4 = 2401.\]Sum the cases for both vowels:\[2401 + 2401 = 4802.\]

Key Concepts

Permutations

Permutations are at the core of many combinatorial problems. They deal with counting the number of ways to arrange a set of items. In permutation scenarios, the order of arrangement matters. For example, if we are given 5 stickers and need to arrange them on a notebook, each different sequence will count as a unique permutation.

In our exercise, when we consider the scenario where no letter can be repeated, we are effectively calculating a permutation. For each position in the word, we choose from the remaining unused letters, moving from 7 choices for the first position down to 3 for the fifth. Hence, the total permutations without repetition is calculated as the product of choices available at each step: \(7 \times 6 \times 5 \times 4 \times 3\).

Permutation is different from combination, where order does not matter. Remember, every time order is important, you are dealing with a permutation problem.

Combinatorial Counting

Combinatorial counting is essentially about understanding how to count elements, which can be particularly useful when dealing with more complex conditions, such as fixing certain elements or imposing additional constraints upon how items are arranged.

In the provided exercise, one example of combinatorial counting is when we fix a condition like starting with the letter 'A'. Here, we apply combinatorial principles by fixing the first position and freely arranging the other letters in the remaining slots. This results in \(1 \times 7 \times 7 \times 7 \times 7\) possible arrangements.

Learning how to count systematically helps manage vast possibilities. It's important to identify whether repetition is allowed and if specific elements are earmarked for certain positions, which can simplify a task into smaller, more manageable parts.

Arrangements

Arrangements describe organized sequences formed from a set of elements. In combinatorial contexts like this exercise, arrangements involve creating sequences that adhere to specified conditions. For instance, when words must be started by certain letters, or specific elements like a particular letter are placed in predetermined slots, it shapes the way we calculate outcomes.

The exercise provides scenarios such as having the letter 'C' fixed in the middle, which directs us to focus on arranging the remaining letters around this fixed point. By fixing 'C', we limit options for the middle slot, yet continue arranging freely for the others, calculated as \(7 \times 7 \times 1 \times 7 \times 7\).

Understanding arrangements helps simplify counting tasks, distribute possibilities methodically, and ensure all constraints are adequately acknowledged in calculations. This concept is key to solving complex combinatorial problems elegantly and efficiently.