Problem 9
Question
If \(a\) represents the number of permutations of \((x+2)\) things taken together, \(b\) represents the number of permutations of 11 things taken together out of \(x\) things, and \(c\) represents the number of permutations of \((x-11)\) things taken together so that \(a=182 b c\), then \(x=\) (A) 15 (B) 12 (C) 10 (D) 18
Step-by-Step Solution
Verified Answer
The value of \( x \) is 12.
1Step 1: Understanding the Problem
We need to express the given conditions in mathematical form. We are given that \( a \) represents the number of permutations of \( (x + 2) \) things, \( b \) represents the number of permutations of 11 things from \( x \) things, and \( c \) represents the permutations of \( (x-11) \) things. We also know that \( a = 182b \times c \). Our goal is to find the value of \( x \).
2Step 2: Expressing Permutations
The permutation of \( n \) items is given by \( n! \). Therefore:- \( a = (x + 2)! \)- \( b = \frac{x!}{(x - 11)!} \)- \( c = (x - 11)! \)
3Step 3: Substitute and Simplify
Substitute the expressions for \( a \), \( b \), and \( c \) into the equation:\[(x+2)! = 182 \times \frac{x!}{(x-11)!} \times (x-11)!\]This simplifies to: \[(x+2)! = 182 \times x!\]
4Step 4: Simplify Further
We can cancel \( x! \) from both sides:\[(x+2) \times (x+1) = 182\]Now solve for \( x \).
5Step 5: Solve for x
Expand and rearrange:\[x^2 + 3x + 2 = 182\]Which leads to:\[x^2 + 3x - 180 = 0\]Now, we solve this quadratic equation.
6Step 6: Quadratic Solution
We solve \( x^2 + 3x - 180 = 0 \) using the quadratic formula: \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]Here, \( a = 1, b = 3, c = -180 \). Calculate the discriminant:\[ \sqrt{{3^2 + 720}} = \sqrt{729} = 27 \]Now solve:\[x = \frac{{-3 \pm 27}}{2}\]This gives the solutions \( x = 12 \) and \( x = -15 \), but since \( x \) must be positive, \( x = 12 \).
Key Concepts
FactorialsQuadratic EquationsMathematical Problem Solving
Factorials
Factorials provide us with a way to determine the number of permutations by calculating the total possible arrangements of a set of objects. In simple words, if you have a set of objects and want to know how many different orders you can arrange them in, you use a factorial to find out.
The factorial of a non-negative integer \( n \), denoted as \( n! \), is the product of all positive integers less than or equal to \( n \). Let's take a closer look with examples:
The factorial of a non-negative integer \( n \), denoted as \( n! \), is the product of all positive integers less than or equal to \( n \). Let's take a closer look with examples:
- For \( 3! \): Calculate \( 3 \times 2 \times 1 = 6 \)
- For \( 5! \): Calculate \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
Quadratic Equations
Quadratic equations are a type of polynomial equation with a degree of two. They are generally written in the standard form \( ax^2 + bx + c = 0 \). Quadratic equations are fundamental in various calculations in mathematics and science.
The key methods to solve quadratic equations include:
The key methods to solve quadratic equations include:
- Factoring: Writing the quadratic as a product of two binomials.
- Completing the square: Rewriting the quadratic into a squared binomial form.
- Quadratic formula: Using \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the roots of the equation.
Mathematical Problem Solving
Mathematical problem solving involves a logical sequence of steps that ensue from understanding the problem to arriving at a solution. It is a skill crucial for tackling complex issues that require clear strategies and thoughtful analysis.
Here's a breakdown of effective problem-solving steps:
Here's a breakdown of effective problem-solving steps:
- Understand the Problem: Read the problem carefully and identify what is being asked.
- Devise a Plan: Determine how you will approach solving the problem. This might include selecting formulas or mathematical operations needed.
- Carry Out the Plan: Implement the steps from your plan computationally, strictly following mathematical logic.
- Review and Recheck: Ensure the solution makes sense in context to the problem initially posed.
Other exercises in this chapter
Problem 7
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