Question

How many four-letter "words" (we use "word" to mean any sequence of letters) which begin and end with a vowel may be formed from the letters \(\mathrm{a}, \mathrm{e}, \mathrm{i}, \mathrm{p}, \mathrm{q}\). (a)if no repetitions are allowed? (b) if repetitions are allowed?

Short answer

(a) There are 108 possible four-letter "words" that begin and end with a vowel with no repetitions allowed. (b) There are 225 possible four-letter "words" that begin and end with a vowel with repetitions allowed.

Step-by-step solution

Determine the possible vowels for the first and last positions

We have 3 vowels (a, e, i) to choose from for the first letter position and the last letter position.

Determine the possible letters for the middle two positions

Since no repetitions are allowed, for the second letter position, we have 2 remaining vowels and 2 consonants to choose from, making a total of 4 options. For the third letter position, there are 3 letters left to choose from (the 2 unused vowels and the remaining consonant).

Calculate the number of possible "words" without repetitions

To find the total number of possible "words" in this case, multiply the number of options available for each of the 4 letter positions: Number of "words" = \(3 \times 4 \times 3 \times 3\) Number of "words" = \(108\) (b) Repetitions allowed

Determine the possible vowels for the first and last positions

The first letter position and the last letter position must be filled with one of the vowels (a, e, i), so there are 3 possible choices for each position.

Determine the possible letters for the middle two positions

Since repetitions are allowed in this case, there are 5 options to choose from for each of the middle letter positions (the 3 vowels and the 2 consonants).

Calculate the number of possible "words" with repetitions

To find the total number of possible "words" in this case, again multiply the number of options available for each of the 4 letter positions: Number of "words" = \(3 \times 5 \times 5 \times 3\) Number of "words" = \(225\) To summarize: (a) There are 108 possible four-letter "words" that begin and end with a vowel with no repetitions allowed. (b) There are 225 possible four-letter "words" that begin and end with a vowel with repetitions allowed.

Key concepts

Permutations

In combinatorics, permutations refer to the different ways in which a set of items can be arranged or ordered. When thinking about permutations, consider it as a way of rearranging objects like letters, numbers, or symbols.
Unlike combinations, where the order doesn't matter, permutations emphasize the sequence, meaning the order is crucial. For example, in a password or a code, "1234" is different from "4321". They are different permutations of the same set of numbers.
When we involve permutations in exercises, the idea is to calculate the number of possible ways to arrange letters, especially when forming combinations like words from letters.

Vowels

Vowels are critical in forming words and sequences because they often determine the core pronunciation structure. The vowels involved in our exercise are three specific ones: \( a, e, \text{and} i \). These vowels serve vital roles in word permutations as they must appear at specified positions.
In our exercise, both the first and last positions of the word sequence must be filled with one of these vowels. Hence, understanding which vowels are available to fill these positions significantly impacts the calculation of permutations.

• Vowels available: \( a, e, i \)
• Affect the first and last letters of the sequence
• Form the backbone of the word formation.

The choice of vowels is thus essential, impacting the total number of permutations possible in constructing these sequences.

Repetitions

Repetitions play a significant role in permutation problems because they change the number of possible arrangements of a set of items.
When repetitions are not allowed, each item can only be used once in forming the sequence. This limits the permutations because the pool of available items decreases as you choose each letter.
On the other hand, allowing repetitions means each letter can be used multiple times which increases the total number of possible arrangements.

• No Repetition: Items are used only once, reducing available choices as selections are made.
• Repetition Allowed: Items can be reused, increasing the number of potential combinations.

In our exercise, handling repetitions impacts the calculation of separate cases: when repetition is not allowed versus when it is, showing that the outcome is directly reliant on how repetitions are managed.

Sequence

A sequence refers to an ordered list of items, which, in this context, can be letters in a word. Understanding sequences is important because you need to assess permutations by considering the positions within the sequence.
Each position in the sequence plays a key role. If a sequence must begin and end with specific types of items, like vowels, this constrains how the sequence can form.
In combinatorial problems, sequences are understood by:

• Breaking them down into individual positions or slots
• Determining allowable entries (e.g., vowels or consonants)
• Calculating total permutations based on constraints.

In the given exercise, the sequence length is four, with a rule that the first and last must be vowels. Understanding this helps calculate how many unique arrangements (or permutations) are possible.