/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Evaluate each expression. $$\f... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 8

Evaluate each expression. $$\frac{8 !}{1 ! 7 !}$$

Short Answer

Expert verified
The expression evaluates to 8.

Step by step solution

01

Understand the Problem

We need to evaluate the expression \ \( \frac{8!}{1! \cdot 7!} \). Factorials are the product of all positive integers up to a given number.
02

Simplify Factorials

Factorials can be simplified by canceling out terms. \ \( 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \) and \ \( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \). Notice that 7! appears in both the numerator and denominator.
03

Cancel Common Terms

Cancel the common terms \ \( 7! \) from both the numerator and the denominator: \ \( \frac{8!}{7! \cdot 1!} = \frac{8 \times 7!}{7! \cdot 1} = 8 \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factorial Simplification
Factorial simplification helps in managing large numbers resulting from factorials. A factorial, denoted as \( n! \), is the product of all positive integers up to \( n \). For instance, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \). Factorials grow rapidly, but often, we don't need the full factorials. Simplifying them by canceling common terms makes calculations easier. In the given problem, \( \frac{8!}{7!} \), we observed that 7! appears in both numerator and denominator. Canceling them simplifies \( \frac{8!}{7!} \) to just 8. Simplification keeps things manageable.
Permutations
Permutations are about arranging objects in different orders. Think of arrangements where order matters, like seating people in a row. If we need to find the number of ways to arrange 8 objects where we choose 1 at a time, we use factorials. The trick here is that the formula for permutations, \( P(n, k) \), is the ratio of factorials: \( P(n, k) = \frac{n!}{(n-k)!}\). In our example, we only want to pick and arrange 1 object out of 8: \( \frac{8!}{1!7!} = 8 \). Permutations help answer 'how many ways can we order?' backed by factorials.
Combinatorial Mathematics
Combinatorial mathematics deals with counting combinations and arrangements. A combination is selecting items where order doesn't matter. For example, choosing 2 fruits out of 3. Combinations use the binomial coefficient formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \). The problem we solved uses factorials, a core concept in combinatorics. While solutions often involve permutations (order matters) or combinations (order doesn't), both rely on factorials for their calculations. Understanding how to simplify factorials and apply them in different scenarios is key to mastering combinatorial problems.

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