Problem 82
Question
In Exercises 79 - 86, solve for \( n \). \( _nP_6 = 12 \cdot _{n - 1} P_5 \)
Step-by-Step Solution
Verified Answer
The value of \( n \) in the equation is 12
1Step 1: Substitute Formula of Permutations
Substitute permutations formulae into \( _nP_6 = 12 \cdot_{n - 1} P_5 \) to obtain \( \frac{n!}{(n-6)!} = 12 \cdot \frac{(n-1)!}{(n-6)!} \)
2Step 2: Simplify
Simplify to remove \( (n-6)! \) from the denominator on both sides, getting \( n! = 12 \cdot (n-1)! \)
3Step 3: Expand Factorial and Solve
Expand \( n! \) and \( (n-1)! \) as follows \( n \cdot (n-1)! = 12 \cdot (n-1)! \). Divide through by \( (n-1)! \) to get \( n = 12 \)
Key Concepts
FactorialAlgebraic ManipulationCombinatorics
Factorial
The concept of a factorial is essential in mathematics, especially in permutations and combinations. A factorial, denoted by an exclamation mark (!), represents the product of an integer and all the integers below it. For example, \( n! \) equals \( n \times (n-1) \times (n-2) \times \ldots \times 1 \). This operation has broad applications:
- Calculates permutations: The number of ways to arrange a set.
- In combinatorics: Helps in finding combinations, which are selections where the order doesn't matter.
- Factorials grow rapidly: Even small numbers result in large values, e.g., \( 5! = 120 \).
Algebraic Manipulation
Algebraic manipulation involves using mathematical operations and properties to transform equations or expressions. This skill is vital for solving complex problems efficiently. Here's how it was applied in our exercise:
- We substituted the permutation formulas into the equation. This allowed us to apply known mathematical properties.
- By simplifying both sides of the equation, we removed the factorial in the denominator: \( \frac{n!}{(n-6)!} = 12 \times \frac{(n-1)!}{(n-6)!} \).
- Further simplification involved cancelling terms, which resulted in the equation \( n! = 12 \cdot (n-1)! \).
Combinatorics
Combinatorics is a branch of mathematics focused on counting, arrangement, and combination of objects. It often uses factorials to determine how many ways items can be arranged or selected. Key concepts include:
- Permutations: These are arrangements where order matters. For example, arranging letters.
- Combinations: These involve selections where order doesn't matter, such as choosing committee members.
- Application: Problems like \( _nP_6 = 12 \cdot _{n-1}P_5 \) involve finding the appropriate arrangements (permutations).
Other exercises in this chapter
Problem 81
In Exercises 79 - 82, use a graphing utility to graph the first \( 10 \) terms of the sequence. (Assume \( n \) that begins with \( 1 \).) \( a_n = 0.2n + 3 \)
View solution Problem 81
In Exercises 77-84, simplify the factorial expression. \( \dfrac{(n + 1)!}{n!} \)
View solution Problem 82
In Exercises 79 - 82, use the Binomial Theorem to approximate the quantity accurate to three decimal places. For example, in Exercise 79, use the expansion \( \
View solution Problem 82
In Exercises 67 - 86, find the sum of the finite geometric sequence. \( \sum_{n=0}^{50}10\left(\dfrac{2}{3}\right)^{n - 1} \)
View solution