Problem 33 - Algebra for College Students

Question

In how many ways can \(x^{4} y^{2}\) be written without using exponents? [Hint: One way is xxxxyy.] 15

Step-by-Step Solution

Answer

15 ways.

Step 1: Understand the problem

We need to find the arrangements of the letters in the expression of \(x^4 y^2\) without using exponents. This means the problem is about permutations of the set \{x, x, x, x, y, y\}.

Step 2: Identify the total elements

There are a total of 6 letters: 4 x's and 2 y's. So, the total number of elements to arrange is 6.

Step 3: Calculate total permutations

Calculate the number of distinct permutations of the letters using the formula for permutations with repeated elements: \[ \frac{n!}{n_1! \cdot n_2!} \] where \(n\) is the total number of items, and \(n_1\), \(n_2\) are the frequencies of each repeated item.

Step 4: Apply the formula

Substitute the known values into the formula: \(n = 6\), \(n_1 = 4\) for x's, and \(n_2 = 2\) for y's. The formula becomes \[ \frac{6!}{4! \cdot 2!} \]

Step 5: Simplify the expression

Calculate each factorial: \(6! = 720\), \(4! = 24\), and \(2! = 2\). Substitute into the expression: \[ \frac{720}{24 \times 2} = \frac{720}{48} = 15 \]

Step 6: Conclusion

Thus, there are 15 distinct ways to arrange the letters in \(x^4 y^2\) without using exponents.

Key Concepts

Permutations

When dealing with permutations, we're essentially talking about arranging items in different orders. Imagine you have a set of distinct objects, and you want to know how many different ways you can order these objects. In algebra, permutations are used in various ways, especially when dealing with problems that involve arrangements or sequences.

Permutations become even more interesting when you have repeated elements. This means some of the items in the set are indistinguishable from each other, like having multiple 'x's or 'y's in our problem. In cases of repeated permutations, you use a special formula to calculate the number of distinct arrangements. This formula is given by:

• \[ \frac{n!}{n_1! \cdot n_2! \cdot n_3! \cdots} \]

where \(n\) is the total number of items, and \(n_1, n_2, n_3, \ldots \) are the frequencies of each type of indistinguishable object. By applying this concept, we can solve problems like arranging the letters in \(x^4 y^2\) effectively.

Factorials

Factorials are a fundamental concept you often encounter while dealing with permutations and combinatorics. The factorial of a non-negative integer \(n\), denoted as \(n!\), is the product of all positive integers less than or equal to \(n\). For example, to find \(6!\), you calculate

• \(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\)

Factorials grow rapidly with increasing values of \(n\). They're particularly useful when you need to compute the number of permutations or combinations in a given scenario.

In problems involving repeated permutations, you'll often see factorials in the numerator and the frequencies of repeated objects in the denominator. This effectively counts the distinct permutations by dividing the total permutations by the cellular permutations of the indistinguishable objects.

Combinatorics

Combinatorics is a branch of mathematics that focuses on counting, arrangement, and combination of objects. It's the backbone of problems involving arrangements like the one we've solved.

In the problem concerning \(x^4 y^2\), we used combinatorial thinking to determine how these symbols can be arranged without exponents. This involves understanding how permutations with repetition work and neatly leverages the factorial concept to find an answer efficiently.

Combinatorics often poses questions like:

• "How many ways can these objects be ordered?"
• "How many ways can a subset of items be chosen from a larger set?"

Using these foundational tools in mathematics, one can tackle a wide range of problems beyond just arranging letters or numbers, encompassing realms of probability, logic, and algebraic analysis.