Problem 23 - Algebra and Trigonometry

Question

Evaluate each expression. \(1-\frac{_{3} P_{2}}{_{4} P_{3}}\)

Step-by-Step Solution

Answer

The value of the expression is 0.75.

Step 1: Calculate Permutations

First, calculate the permutations for 3P2 and 4P3. For an expression nPr, it means we are finding the number of ways we can choose r items from n items, order matters. Using the formula for permutations P(n, r) = n! / (n-r)!, we can calculate 3P2 as 3! / (3-2)! and 4P3 as 4! / (4-3)!.

Step 2: Calculate Factorials

Next calculate the factorials in the expressions. The factorial of a number n, designated by n!, is the product of all positive integers less than or equal to n. So, 3! is 3*2*1 = 6, (3-2)! is 1! which is 1, 4! is 4*3*2*1 = 24 and (4-3)! is 1! which is 1.

Step 3: Complete Permutations

Now, substitute the factorial values back into the permutation calculations. Therefore 3P2 becomes 6/1 = 6, and 4P3 becomes 24/1 = 24.

Step 4: Complete Calculation

Finally, substitute the permutation values back into the original expression and complete the arithmetic operation. Therefore the expression 1 - (3P2/4P3) becomes 1 - (6/24) which simplifies to 1 - 0.25 = 0.75

Key Concepts

Factorials

Factorials are an essential building block in many areas of mathematics, especially in permutations and combinations. A factorial is represented by an exclamation mark, such as in "n!". It represents the product of an integer and all the integers below it. For example:

• 3! is the same as 3 × 2 × 1 = 6
• 4! is calculated as 4 × 3 × 2 × 1 = 24
• 1! is simply 1 by definition

Factorials grow very quickly with increasing numbers, making them very useful in determining the number of permutations or combinations possible in a given situation. Whenever you see a factorial in a problem, remember it stands for multiplying all whole numbers from the specified number down to one.

Algebraic Expressions

Algebraic expressions are mathematical phrases that include numbers, variables, and operations. They do not have an equality sign, making them different from equations. In the exercise given, the expression is a mix of arithmetic operations and permutation calculations. Here's how it looks:

• The central expression is: \(1-\frac{_{3} P_{2}}{_{4} P_{3}}\).
• This denotes a subtraction operation between the number 1 and a division outcome.
• The division involves the values calculated from permutations, which are \(_{3} P_{2}\) and \(_{4} P_{3}\).

Understanding each part of an algebraic expression is important to simplify or evaluate them correctly. The sequence of steps to solve such expression includes evaluating each mathematical operation, typically starting from the innermost functions.

Arithmetic Operations

Arithmetic operations, including addition, subtraction, multiplication, and division, form the foundation of mathematical calculations. These operations are built into the expression we evaluated. In the final steps of our problem, we used:

• Subtraction to find a difference: 1 - 0.25
• Division to simplify the fraction: 6 divided by 24 becomes 0.25

Performing these operations involves following the order of operations, often remembered by the acronym PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction). In our exercise, we've handled multiplication while computing factorials, and then managed division and subtraction for the final simplification. Mastery over elementary arithmetic operations ensures accuracy in more complex problems.

Combinatorics

Combinatorics is a field of mathematics focused on counting, arrangement, and combination in structures or sets. Within this field, permutations and combinations are prominent concepts. Our exercise dealt with permutations:

• A permutation is about arranging items where the order is important.
• The notation \(P(n, r)\) or \(_{n} P_{r}\) represents the number of ways to arrange \(r\) items from \(n\) total items.
• We used the permutation formula, which is \(n! / (n-r)!\).

In the exercise, we calculated \(_{3} P_{2}\) and \(_{4} P_{3}\) to find the insights needed for our expression, showcasing the practical use of combinatorial principles. Understanding the differences between permutations and combinations, and how to calculate them, is crucial in solving many algebraic and statistical problems.