Problem 77 - Precalculus with Limits

Question

In Exercises 77-84, simplify the factorial expression. \( \dfrac{4!}{6!} \)

Step-by-Step Solution

Answer

The simplification of the factorial expression \( \dfrac{4!}{6!} \) is \( \dfrac{1}{30} \).

Step 1: Understand factorial notation

Factorial notation is a way of writing a product of decreasing positive integers. For example, \( 4! = 4 \cdot 3 \cdot 2 \cdot 1 \) and \( 6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \). Notice that \( 6! \) contains all the factors of \( 4! \).

Step 2: Simplify the expression

\( \dfrac{4!}{6!} \) can be rewritten as \( \dfrac{4 \cdot 3 \cdot 2 \cdot 1}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \). As \( 4! \) is part of \( 6! \), we can simplify the expression by cancelling out the common terms in the numerator and the denominator. That leaves \( \dfrac{1}{6 \cdot 5} \)

Step 3: Evaluate the left over terms

In this step, evaluate \( \dfrac{1}{6 \cdot 5} \) which equals \( \dfrac{1}{30} \).

Key Concepts

Factorial Notation

When encountering factorial notation in mathematics, understand that it signifies the product of a sequence of descending natural numbers. The symbol for factorial is an exclamation point (!). The factorial of a non-negative integer n, denoted by \(n!\), is the product of all positive integers less than or equal to n. It is defined as \(n! = n \times (n-1) \times (n-2) \times ... \times 3 \times 2 \times 1\), with the factorial of zero defined as \(0! = 1\), a convention that ensures the consistency of various mathematical formulas.For example, \(4! = 4 \times 3 \times 2 \times 1 = 24\). As numbers grow, factorials can become extremely large very quickly—a trait that is both powerful and challenging in computation. In simplifying expressions involving factorials, it's fundamental to notice how factorials of higher numbers include the factorials of their lower counterparts. This recognition is crucial in simplification, such as reducing the fraction \( \frac{4!}{6!} \).

Simplifying Expressions

Simplifying expressions is one of the fundamental skills in precalculus and algebra. The process involves reducing an expression to its simplest form. This often means combining like terms, cancelling common factors, or applying algebraic rules to make the expression more straightforward to work with.
In the context of factorials, simplification often relies on recognizing that a larger factorial contains the product of a smaller one within it. In the example \( \frac{4!}{6!} \), the expression simplifies by cancelling out the common terms. This distinction helps in avoiding unnecessary calculations, particularly with large numbers where the computation of individual factorials can be time-consuming.
When facing such expressions, always break down the factorials to their basic components and look for commonalities. By canceling out these shared factors, you can streamline the original expression to something much more manageable, as with the reduction to \( \frac{1}{6 \cdot 5} = \frac{1}{30} \) in the given exercise.

Precalculus

Precalculus serves as the bridge between algebra and calculus. It encompasses a variety of topics needed to prepare for the study of calculus, including algebraic simplification, functions, and complex numbers. This subject matter is vital for understanding and manipulating expressions necessary for higher-level mathematics.Factorials often appear in precalculus problems, particularly in permutations and combinations, which involve counting and probability. The ability to simplify expressions involving factorials and apply algebraic concepts is thus an integral part of precalculus proficiency.
Furthermore, grasping these foundational concepts in precalculus is critical for future coursework in calculus, where complex problems often require a solid understanding of factorial manipulation. Thus, exercises such as simplifying \( \frac{4!}{6!} \) help students develop the algebraic agility needed for success in more advanced mathematical studies.