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Chapter 9: Problem 82

In Exercises 79 - 86, solve for \( n \). \( _nP_6 = 12 \cdot _{n - 1} P_5 \)

Short Answer

Expert verified
The value of \( n \) in the equation is 12

Step by step solution

01

Substitute Formula of Permutations

Substitute permutations formulae into \( _nP_6 = 12 \cdot_{n - 1} P_5 \) to obtain \( \frac{n!}{(n-6)!} = 12 \cdot \frac{(n-1)!}{(n-6)!} \)
02

Simplify

Simplify to remove \( (n-6)! \) from the denominator on both sides, getting \( n! = 12 \cdot (n-1)! \)
03

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 \( n = 12 \)

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Key Concepts

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

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., \( 5! = 120 \).
Understanding factorials is crucial for solving permutation problems, as seen in this exercise, where manipulating factorials helped us simplify the equation to find the value of \( n \).
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 \( n! = 12 \cdot (n-1)! \).
This involved expanding factorials and cancelling common terms to solve for \( n = 12 \). Mastery of algebraic manipulation makes tackling such problems more intuitive.
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).
To solve such combinatorial problems, understanding factorials and algebraic manipulations is essential. This field not only enhances problem-solving but also builds logical thinking skills, useful in mathematics and beyond.

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Most popular questions from this chapter

A shipment of 25 television sets contains three defective units. In how many ways can a vending company purchase four of these units and receive (a) all good units, (b) two good units, and (c) at least two good units?

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