Problem 34 - Algebra for College Students

Question

In how many ways can \(x^{3} y^{4} z^{3}\) be written without using exponents? 4200

Step-by-Step Solution

Answer

There are 4200 ways to arrange the letters in \(x^3y^4z^3\).

Step 1: Understanding the Problem

We need to find the number of ways to arrange the letters in the expression obtained by expanding the terms without exponents. The expression is \(x^3y^4z^3\), which can be rewritten as \(xxxyyyyzzz\).

Step 2: Counting Total Permutations

The total number of letters in the sequence \(xxxyyyyzzz\) is 10. To find the number of permutations of a set of items where some items are repeated, we use the formula for permutations of a multiset: \(\frac{n!}{n_1!n_2!...n_k!}\), where \(n\) is the total number of items, and \(n_1, n_2,..., n_k\) are the frequencies of each type of item.

Step 3: Applying the Formula

In this problem, we have 3 x's, 4 y's, and 3 z's. Plug these numbers into the formula:\[\frac{10!}{3!4!3!}\]

Step 4: Calculating Factorials

First, calculate the factorials:- \(10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3628800\)- \(3! = 6\)- \(4! = 24\)Now substitute these into the formula.

Step 5: Simplify the Calculation

Substitute the values into the formula:\[\frac{3628800}{6 \times 24 \times 6} = \frac{3628800}{864}\]

Step 6: Final Calculation

Divide to find the number of arrangements:\[\frac{3628800}{864} = 4200\]Thus, there are 4200 ways to arrange the letters in the expression \(xxxyyyyzzz\).

Key Concepts

Multisets

A multiset, also known as a bag, is a generalized concept of a set that allows for multiple occurrences of the same element. Unlike traditional sets, where each element can appear at most once, multisets allow repetitions of elements. This is extremely useful in permutations and combinations, especially when dealing with real-world scenarios where duplicates occur frequently. In our problem with the expression \(x^3y^4z^3\), by expanding it, you obtain a sequence \(xxxyyyyzzz\), which is a classic example of a multiset. Here, each letter can appear more than once:

• "x" appears 3 times.
• "y" appears 4 times.
• "z" appears 3 times.

Understanding multisets is crucial to applying the right permutation formulas, as you need to account for these repetitions in your calculations.

Factorials

Factorials are a fundamental concept in combinatorics and permutations. The factorial of a number \(n\), denoted as \(n!\), is the product of all positive integers less than or equal to \(n\). It is used to calculate the number of ways to arrange or order a set of items. For example, the number of ways to arrange 10 distinct items is \(10!\), which equals 3,628,800. In our exercise, we use factorials to solve for permutations when some items are repeated. For each repeated group in the multiset, we calculate its factorial and use it to adjust the overall permutation count. For the problem at hand:

• Calculate \(10!\) for the total letters.
• Calculate \(3!\) for each of the x's.
• Calculate \(4!\) for the y's.
• Calculate \(3!\) again for the z's.

These factorials help in determining the exact number of unique permutations that can occur.

Combinatorics

Combinatorics is the branch of mathematics concerning the study of finite or countable discrete structures. It plays a vital role in calculating permutations, especially in distinguishing when and how to apply formulas for multisets. The core idea is to understand how different objects can be arranged or combined under specific constraints.When working with permutations of multisets, like \(xxxyyyyzzz\), combinatorics provides tools like the multiset permutation formula: \[\frac{n!}{n_1!n_2!...n_k!}\]. This formula considers the total number of items and the factorials of the frequency of each unique item. It sets the foundation for accurately determining the number of distinct arrangements possible.Applying combinatorics in this problem:

• We calculate the total permutations of all letters using \(10!\).
• Then, we divide by the factorials of the repeat counts, \(3!\), \(4!\), and \(3!\), to eliminate redundant arrangements.

The result: a precise count of 4,200 unique permutations. Combinatorics thus enables us to navigate complex arrangements with detailed precision.