Problem 67 - Calculus: One and Several Variables

Question

Evaluate. \(\frac{8 !}{3 ! 5 !}\).

Step-by-Step Solution

Answer

The answer is 56.

Step 1: Understand Factorials

The term \(8!\) means multiplying 8 by all the numbers less than it down to 1 i.e. \(8*7*6*5*4*3*2*1\). Similarly, \(3!\) is \(3*2*1\) and \(5!\) is \(5*4*3*2*1\).

Step 2: Factorization

This step involves writing out the factorials as products of numbers. Hence, the given expression becomes \(\frac{8*7*6*5*4*3*2*1}{(3*2*1)*(5*4*3*2*1)}\).

Step 3: Simplification

Next, simplify the expression by cancelling out common factors from both the numerator and denominator. In this example, \(5*4*3*2*1\) from the numerator and denominator cancels out leaving \(\frac{8*7*6}{3*2*1}\).

Step 4: More simplification

The numerator of the fraction can be reduced further to \(8*7*6=336\) and the denominator remains as \(3*2*1=6\). This simplifies to \(\frac{336}{6}\).

Step 5: Obtain Final Solution

Finally, divide 336 by 6 to get the final answer.

Key Concepts

Factorials

In mathematics, a factorial, denoted as a number followed by an exclamation point (e.g., \(8!\)), represents the product of all positive integers up to that number. For instance, \(8!\) is calculated as \(8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40,320\). Understanding factorials is crucial in fields like combinatorics, where we often need to determine the number of ways to arrange or select items.

• For \(n!\), where \(n\) is a non-negative integer, the simplest way to think of it is counting down by one from that integer until reaching 1, then multiplying all those numbers together.
• Special case: \(0!\) is defined to be 1. This may seem strange, but it's important for consistency in mathematical formulas, especially in combinatorial expressions.

Factorials can grow quite large very quickly. Therefore, understanding and performing operations with factorials efficiently, like in this example, often involves using properties like cancellation rather than direct computation.

Simplification

Simplification is all about reducing mathematical expressions to their simplest form. It's a crucial step in solving equations that involve complex operations. Here, we demonstrate simplification by tackling the expression \(\frac{8!}{3!5!}\).
To begin the simplification, we express the factorial terms as full products:

• The numerator, \(8!\), is \(8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\).
• For the denominator, \(3!5!\) expands to \((3 \times 2 \times 1) \times (5 \times 4 \times 3 \times 2 \times 1)\).

Next, we look for common factors that appear in both the numerator and denominator. By canceling these out, particularly \(5 \times 4 \times 3 \times 2 \times 1\) here, the expression simplifies significantly. This leaves us with the simplified result of \(\frac{8 \times 7 \times 6}{3 \times 2 \times 1}\). Further reducing those remaining terms results in simplified calculations and provides the final value efficiently.

Mathematical Notation

Mathematical notation, like factorials and fractions, provides a concise way to express mathematical ideas and operations. In combinatorics and probability, clear notation is essential for effective communication and understanding.

• Factorial notation \(n!\) conveys a powerful concept compactly, enabling us to resolve advanced combinatorial problems without writing lengthy products.
• Fractions, like \(\frac{8!}{3!5!}\), help us understand operations involving ratios and division, and they are often simplified through common factor cancellation.
• Careful notation with parentheses ensures we interpret the operations correctly, especially in more complex expressions where multiple operations are involved.

By mastering mathematical notation and its interpretation, students can approach all types of equations with confidence, unlocking the ability to tackle and simplify otherwise daunting expressions efficiently.