Problem 3 - Algebra and Trigonometry

Question

How many three-letter “words” (strings of letters) can be formed using the letters WXYZ if repetition of letters (a) is allowed? (b) is not allowed?

Step-by-Step Solution

Answer

(a) 64 words, (b) 24 words.

Step 1: Understanding the Problem

We are tasked with forming three-letter words using the letters W, X, Y, and Z. Words here mean any combination of these letters in a sequence of three.

Step 1: Calculate with Repetition Allowed

When repetition is allowed, each of the three positions in the word can be independently filled with one of the 4 letters. Thus, for each of the 3 positions, we have 4 choices of letters. The total number of words is calculated as:\[ 4 \times 4 \times 4 = 4^3 \]

Step 2: Compute the Result for Part (a)

Calculating the expression from Step 1 gives us:\[ 4^3 = 64 \]So, 64 different words can be formed when repetition is allowed.

Step 3: Calculate with Repetition Not Allowed

When repetition is not allowed, the choices for each position are reduced after each selection. For the first position, there are 4 options. Once a letter is used, for the second position we have 3 remaining options. Finally, for the third position, 2 options remain. The calculation is:\[ 4 \times 3 \times 2 \]

Step 4: Compute the Result for Part (b)

Calculating the expression from Step 3 gives us:\[ 4 \times 3 \times 2 = 24 \]So, 24 different words can be formed when repetition is not allowed.

Key Concepts

Permutation

Permutations refer to the different ways in which a set of objects can be arranged or ordered. When dealing with problems of arrangement, like forming words from given letters, permutations play a crucial role. The idea is to figure out all the possible ways to position these objects when sequence matters.

• In mathematical terms, permutations of a set of objects can be calculated using factorials, represented as \( n! \) for a set of size \( n \).
• For example, if you have four letters W, X, Y, and Z, the total permutations if you were to arrange all of them at once are \( 4! \), which equals \( 24 \).

Understanding permutations helps tackle more specific counting problems like those involving additional constraints such as restrictions on repetition.

Counting with Repetition

When counting with repetition, each position in an arrangement can be filled with any member from the set of objects repeatedly. This is akin to having a selection process where choices reset with each position.

• In the context of forming words, if you have 3 positions to fill with any of the 4 letters (W, X, Y, Z) and repetition of letters is allowed, each position has 4 possible letters.
• This results in a calculation like \( 4 \times 4 \times 4 = 4^3 \), which for the problem translates to 64 potential combinations or three-letter 'words'.

Counting with repetition is important when evaluating numbers of combinations where elements can appear more than once, such as pin codes or passwords often do.

Counting without Repetition

On the other hand, counting without repetition requires that each choice is unique within the arrangement. This means once an item is used, it cannot appear again in the same sequence.

• This concept applies when forming words where no letters repeat in a sequence. If the word is 3 letters long and uses letters W, X, Y, Z, for the first position you have 4 choices, then 3 for the second, and finally 2 for the third.
• The result is calculated as \( 4 \times 3 \times 2 \), yielding 24 unique combinations. Each position gradually has fewer available options due to prior selections.

This method is crucial when the uniqueness of each item is required, often found in ID codes or unique promotional combinations.