Problem 16 - Algebra and Trigonometry

Question

Find the number of distinguishable permutations of the given letters. $$ X X Y Y Z Z $$

Step-by-Step Solution

Answer

There are 90 distinguishable permutations of the letters XXYYZZ.

Step 1: Understand the problem

We are asked to find the number of distinguishable permutations for the set of letters \(XXYYZZ\). To solve this, we need to use the formula for permutations of multiset, where certain items are repeated.

Step 2: Apply the Multiset Permutation Formula

The formula for finding the number of permutations of a multiset is \(\frac{n!}{n_1! \times n_2! \times ... \times n_k!}\). Here, \(n\) is the total number of items, and \(n_1, n_2, ..., n_k\) are the counts of each repeated item. For \(XXYYZZ\), we have 6 letters in total, \(X\) repeated 2 times, \(Y\) repeated 2 times, and \(Z\) repeated 2 times.

Step 3: Calculate the Factorials

Calculate each component of the formula: - Total factorial: \(6! = 720\).- Factorial of repeated \(X\): \(2! = 2\).- Factorial of repeated \(Y\): \(2! = 2\).- Factorial of repeated \(Z\): \(2! = 2\).

Step 4: Solve the Formula

Plug the values into the formula: \[\frac{6!}{2! \times 2! \times 2!} = \frac{720}{2 \times 2 \times 2} = \frac{720}{8} = 90.\]

Step 5: Verify the Calculation

Verify the division step: - The result of \(\frac{720}{8}\) is indeed 90, confirming our calculation for the number of distinguishable permutations.

Key Concepts

Multiset Permutation Formula

When dealing with distinguishable permutations in combinatorics, especially when some items in a set are identical, we use the **multiset permutation formula**. Imagine you have a bag of letters, and some of these letters are repeated. If you rearrange these letters, not every arrangement will be unique because of the repetitions. The multiset permutation formula helps us calculate how many of these arrangements are actually distinguishable.

• The formula is: \( \frac{n!}{n_1! \times n_2! \times \ldots \times n_k!} \)
• Where \( n \) is the total number of items.
• \( n_1, n_2, \ldots, n_k \) represent the frequency of each distinct item.

This way, by dividing by the factorials of the repeated items, we account for all the repeats, ensuring each unique arrangement is counted only once.

Factorial Calculation

Factorials are a fundamental part of permutations and combinations; they're used extensively in the multiset permutation formula. Simply put, a factorial of a number \( n! \) is the product of all positive integers less than or equal to \( n \). So, for example, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \).Factorials grow very fast as the number increases, which can sometimes make calculations challenging by hand. To ease calculations, especially when calculating permutations of letters, consider:

• Total tally: Always count the total number of items involved for the top of the multiset formula.
• Count repeat: For each set of repeated items, compute their factorial.
• Apply: Use these factorials in the multiset formula to find the number of distinguishable permutations.

Combinatorics

Combinatorics is the fascinating branch of mathematics focused on counting, arranging, and understanding patterns of numbers. It helps us solve problems related to permutations and combinations, making it crucial for tasks like determining distinguishable permutations.In our example with letters like \(XXYYZZ\), combinatorics allows us to not just list every possible arrangement (which can be cumbersome for large sets) but efficiently count the distinct ones. By leveraging formulas such as permutations, students can understand how to methodically approach these problems.

• **Permutations** involve arranging all the members of a set into some sequence or order.
• **Combinations** (although not directly used here) focus on selecting items from a set without concern for order.

Understanding these concepts is key to solving diverse problems in mathematics.

Permutations of Letters

Let's dive deeper into permutations of letters. When letters are different, each rearrangement of the set can be considered a distinct permutation. However, this changes when some letters are identical, as in the case of \(XXYYZZ\).The goal is to find the number of unique ways to arrange these letters. Here’s the process broken down:

• Count the total letters: Here, 6 letters in total.
• Identify repeats: \(X\), \(Y\), and \(Z\) each occur twice.
• Apply the multiset formula: Plug these values into \( \frac{6!}{2! \times 2! \times 2!} \). This calculation acknowledges the repetitions by effectively grouping same letters, ensuring each counted permutation is unique.

This way, we find that there are 90 distinguishable permutations of these letters, thanks to the formula simplifying the factorial division by accounting for identical items.