Chapter 11: Problem 14
Evaluate. $$\frac{7 !}{(7-2) !}$$
Short Answer
Expert verified
The value is 42.
Step by step solution
01
- Understand the Factorial Notation
Factorial of a number is the product of all positive integers up to that number. For example, \( n! = n \times (n-1) \times (n-2) \times ... \times 1 \).
02
- Simplify the Expression
The given expression is \( \frac{7!}{(7-2)!} \). First, simplify the denominator: \( (7-2)! = 5! \). So, the expression becomes \( \frac{7!}{5!} \).
03
- Expand and Simplify
Expand the factorials in the numerator and denominator: \( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \) and \( 5! = 5 \times 4 \times 3 \times 2 \times 1 \). Cancel the common terms in numerator and denominator: \( \frac{7 \times 6 \times 5!}{5!} = 7 \times 6 \).
04
- Calculate the Result
Finally, multiply the remaining numbers: \( 7 \times 6 = 42 \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factorial Operation
The factorial operation is a mathematical function symbolized by an exclamation mark, such as \( n! \). It represents the product of all positive integers less than or equal to \( n \). For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). Factorials are especially useful in permutations, combinations, and other areas of mathematics. To give another example:
- \( 3! = 3 \times 2 \times 1 = 6 \)
- \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
Simplifying Expressions
Simplifying expressions often involves reducing them to their simplest form. In the given exercise, the goal was to evaluate \( \frac{7!}{(7-2)!} \). By simplifying the denominator first, we have \( (7-2)! = 5! \), which makes our expression \( \frac{7!}{5!} \).
From there, we can expand each factorial: \( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \) and \( 5! = 5 \times 4 \times 3 \times 2 \times 1 \). By canceling out the common terms in the numerator and the denominator, we are left with \( \frac{7 \times 6 \times 5!}{5!} = 7 \times 6 \). So, the simplified result is 42. Reducing expressions in this way makes complex problems easier to solve.
From there, we can expand each factorial: \( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \) and \( 5! = 5 \times 4 \times 3 \times 2 \times 1 \). By canceling out the common terms in the numerator and the denominator, we are left with \( \frac{7 \times 6 \times 5!}{5!} = 7 \times 6 \). So, the simplified result is 42. Reducing expressions in this way makes complex problems easier to solve.
Combinatorics
Combinatorics is a branch of mathematics dealing with combinations, permutations, and counting. Factorials play a crucial role in these calculations. A common example is finding the number of ways to arrange objects, called permutations. The number of permutations of \( n \) objects is calculated as \( n! \).
For example, the number of ways to arrange 4 books on a shelf is \( 4! = 24 \). Another concept in combinatorics is the combination, which determines how many ways you can choose \( k \) objects from a set of \( n \) objects, usually denoted as \( C(n, k) \) or \( \binom{n}{k} \).
This is calculated as \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]. For instance, choosing 2 fruits from an assortment of 7 is calculated as:
\[\binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} \inserting \; factorials,\; we\; get \frac{7\times6}{2\times1} \].
This simplifies to 21 ways. Combinatorics provides tools for solving real-world problems involving arrangements and selections.
For example, the number of ways to arrange 4 books on a shelf is \( 4! = 24 \). Another concept in combinatorics is the combination, which determines how many ways you can choose \( k \) objects from a set of \( n \) objects, usually denoted as \( C(n, k) \) or \( \binom{n}{k} \).
This is calculated as \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]. For instance, choosing 2 fruits from an assortment of 7 is calculated as:
\[\binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} \inserting \; factorials,\; we\; get \frac{7\times6}{2\times1} \].
This simplifies to 21 ways. Combinatorics provides tools for solving real-world problems involving arrangements and selections.