Question

The English alphabet contains 26 letters, including five vowels. In each case determine how many words of length five are possible provided that: (a) Words contain at most two distinct vowels (b) Words contain at most one letter that is a vowel (c) Words contain at least four distinct vowels

Short answer

(a) Add total from 0, 1, and 2 vowels. (b) Add cases of 0 and 1 vowel. (c) Sum 4-vowel and 5-vowel cases.

Step-by-step solution

Understanding the Problem

We are asked to find how many five-letter words can be formed under different constraints involving vowels. We know the alphabet has 26 letters comprising 5 vowels (A, E, I, O, U) and 21 consonants.

At Most Two Distinct Vowels

For this part, we can have 0, 1, or 2 distinct vowels in our word. We need to count each case separately and add the results.- 0 vowels: We use only consonants, so choose any 5 letters from 21 consonants: \( 21^5 \).- 1 vowel: Choose 1 vowel and 4 consonants. There are \( \binom{5}{1} \) ways to choose the vowel and \( \binom{21}{4} \) ways to choose the consonants. So the number of arrangements is: \( \binom{5}{1} \times \binom{21}{4} \times 5! \).- 2 vowels: Choose 2 vowels and 3 consonants. There are \( \binom{5}{2} \) ways to choose the vowels and \( \binom{21}{3} \) ways to choose the consonants. The number of arrangements is: \( \binom{5}{2} \times \binom{21}{3} \times 5! \).Add all cases together for the final count.

At Most One Vowel

In this case, we either have 0 vowels or 1 vowel.- 0 vowels: Only consonants, which is \( 21^5 \).- 1 vowel: Choose 1 vowel and 4 consonants. Choose the vowel in \( \binom{5}{1} \) ways and the consonants in \( \binom{21}{4} \) ways. The number of arrangements is \( \binom{5}{1} \times \binom{21}{4} \times 5! \).Add the two cases for the total count.

At Least Four Distinct Vowels

For this part combine the possibilities of using 4 or 5 vowels in the word.- 4 vowels: Choose 4 vowels and 1 consonant. Select vowels in \( \binom{5}{4} \) and consonants in \( \binom{21}{1} \) ways. The number of arrangements is \( \binom{5}{4} \times \binom{21}{1} \times 5! \).- 5 vowels: Use all 5 vowels. Arranging them gives us \( 5! \) possibilities.Add the two cases for the final result.

Key concepts

Vowels and Consonants

In the world of combinatorics, a key concept when forming words is distinguishing between vowels and consonants. The English alphabet has 26 letters with five being vowels: A, E, I, O, and U. The remaining 21 letters are consonants. This distinction is crucial when solving problems related to word formation as it helps in setting constraints such as choosing only vowels, only consonants, or a specific combination of both.

When forming words under particular rules, we often count the number of vowels and consonants separately. For instance, a problem may ask us to create words with no more than two distinct vowels. Here, first identify the available vowels and calculate combinations using them. Simultaneously, consider the consonants and how they fill the remaining positions in the word. By understanding these two groups distinctly, solving word problems becomes more structured and manageable.

• Vowels: A, E, I, O, U
• Consonants: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Y, Z

Word Formation

Word formation in combinatorics deals with creating words or sequences of letters that adhere to specific criteria. Constraints can be about the inclusion of certain letters, like vowels, or limits on repetition of letters.

To find the number of possible words of a given length, one must often create a strategy that accounts for all permissible combinations. For instance, if we need a five-letter word with no more than two distinct vowels, consider cases like having no vowels, one vowel, or two vowels.

• No vowels: Use only consonants in all positions.
• One vowel: Choose the vowel first, then fill other positions with consonants.
• Two vowels: Select both vowels initially, followed by consonants.

This structured approach ensures all cases are considered, and no possible word formations are missed. Such methods are pivotal when deducing outcomes in combinatorics.

Arrangements in Discrete Mathematics

Arrangements in discrete mathematics involve organizing a set number of items according to certain rules or constraints. It is a fundamental topic within combinatorics, a branch of mathematics that explains counting, arrangement, and probability principles.

With word formation problems, the focus is often on arranging vowels and consonants in a sequence to form valid words under specific conditions. This involves calculating permutations and combinations to determine possible configurations.
For example, the factorial function, denoted as \(!\), is crucial for understanding permutations. It represents the product of all positive integers up to a given number, e.g., \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\), which counts the number of ways to arrange five distinct items.
When arranging vowels and consonants, such as forming a word with exactly four distinct vowels, you calculate the potential permutations of these letters to get the total number of arrangements. By exploring the capabilities of combinations and permutations in varying conditions, discrete mathematics offers a systematic means to resolve word formation puzzles.