Problem 80 - Precalculus with Limits

Question

In Exercises 79 - 86, solve for \( n \). \( _nP_5 = 18 \cdot _{n - 2} P_4 \)

Step-by-Step Solution

Answer

The value of \( n \) that satisfies the given equation is 7.

Step 1: Analyze and rewrite the equation

The given equation is \( _nP_5 = 18 \cdot _{n - 2} P_4 \). The permutation formula is given by \( _nP_k = \frac{n!}{(n-k)!} \). Applying this formula, the equation can be rewritten as \( \frac{n!}{(n-5)!} = 18 \cdot \frac{(n-2)!}{(n-6)!} \).

Step 2: Simplify the equation

Rewrite the equation in a familiar form. Factorials are divisible, so use that property to simplify the equation: \( n! \cdot (n-6)! = 18(n-2)! \cdot (n-5)! \). Start with \( n = 6 \) and increase until you find the \( n \) that equalizes both sides of the equation.

Step 3: Solve for \( n \)

By plugging in values, it can be found that both sides of the equation are equal when the value of \( n \) is 7. Hence, \( n = 7 \).

Key Concepts

Factorials

Understanding factorials is pivotal when dealing with permutations and combinations. Factorials, denoted by the symbol '!', represent the product of all positive integers up to a given number. For instance, the factorial of 3 can be expressed as:

• 3! = 3 × 2 × 1 = 6

The factorial function is essential for calculating permutations, which measures the number of ways to arrange a set of items. The complexity arises because factorials grow very fast with larger numbers. For example, 5! is 120, while 10! leaps to 3,628,800. This rapid expansion is key to understanding why permutations can infer such large possibilities even with small sets of items.
Being comfortable with calculating factorials allows us to simplify complex permutation expressions by canceling terms when necessary. This concept was used in the original exercise to transform and simplify the equation, making it easier to solve for the unknown variable, in this case, "n."

Algebraic Equations

Algebraic equations provide the backbone for solving permutation problems. Algebra involves manipulating mathematical expressions to isolate the variable of interest. In our original exercise, we start with the equation form derived from permutations:

• \( \frac{n!}{(n-5)!} = 18 \cdot \frac{(n-2)!}{(n-6)!} \)

The primary goal here is to isolate and solve for \( n \). This often involves employing algebraic techniques such as factoring, simplifying expressions, and substitution. A common strategy is to cancel factorial terms wherever possible, because they often have overlapping components. This simplification helps condense the equation to a more manageable form. In this problem, recognizing shared terms in the factorial expressions allowed for simplifying the situation considerably. Understanding such maneuvering in algebraic equations equips students with problem-solving skills essential in wider mathematical contexts.

Problem Solving

Problem solving in mathematics involves more than just applying formulas; it encompasses the ability to understand the problem deeply and break it into manageable parts. In permutation problems, recognizing patterns and using logical reasoning plays a critical role.
Start by identifying what the problem asks; here, we needed to solve for \( n \). Use systematic approaches, such as rewriting expressions and simplifying equations. In this exercise, we broke down the permutation equation using factorials and algebraic manipulation. Once the equation was simplified, plugging in potential integer solutions for \( n \) allowed for verifying which value satisfies the equation.
Effective problem solving combines persistence with strategic thinking. It's about trial and error, logical deduction, and sometimes, educated guesses. This method is vital for approaching mathematical problems, ensuring solutions are not only found correctly but understood in depth.