Problem 20 - Precalculus: Mathematical for Calculus

Question

Evaluate the expression. $$\left(\begin{array}{l} 5 \\ 0 \end{array}\right)-\left(\begin{array}{l} 5 \\ 1 \end{array}\right)+\left(\begin{array}{l} 5 \\ 2 \end{array}\right)-\left(\begin{array}{l} 5 \\ 3 \end{array}\right)+\left(\begin{array}{l} 5 \\ 4 \end{array}\right)-\left(\begin{array}{l} 5 \\ 5 \end{array}\right)$$

Step-by-Step Solution

Answer

The expression evaluates to 0.

Step 1: Understanding the Binomial Coefficient

The binomial coefficient \( \binom{n}{k} \) is a way to count the number of ways to select \( k \) elements from a set of \( n \) elements. In this problem, we have several binomial coefficients to evaluate: \( \binom{5}{0} \), \( \binom{5}{1} \), \( \binom{5}{2} \), \( \binom{5}{3} \), \( \binom{5}{4} \), and \( \binom{5}{5} \).

Step 2: Calculating Each Binomial Coefficient

Calculate each of the binomial coefficients:- \( \binom{5}{0} = 1 \) since any number \( n \) choose 0 is always 1.- \( \binom{5}{1} = 5 \) as choosing 1 element from a set of 5 can be done in 5 ways.- \( \binom{5}{2} = 10 \) calculated by \( \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \).- \( \binom{5}{3} = 10 \) by symmetry, or \( \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \).- \( \binom{5}{4} = 5 \) since \( \frac{5!}{4!(5-4)!} = \frac{5}{1} = 5 \).- \( \binom{5}{5} = 1 \) as choosing all elements from a set has exactly 1 way.

Step 3: Applying the Given Operations

Substitute the binomial coefficients into the expression and apply the operations:\[ \binom{5}{0} - \binom{5}{1} + \binom{5}{2} - \binom{5}{3} + \binom{5}{4} - \binom{5}{5} \]which becomes:\[ 1 - 5 + 10 - 10 + 5 - 1 \].

Step 4: Simplifying the Expression

Step-by-step, simplify the expression:1. \( 1 - 5 = -4 \)2. \( -4 + 10 = 6 \)3. \( 6 - 10 = -4 \)4. \( -4 + 5 = 1 \)5. \( 1 - 1 = 0 \)

Step 5: Final Result

The expression evaluates to 0 after completing the simplification of all operations.

Key Concepts

Combinatorics

Combinatorics is a branch of mathematics dealing with combinations and arrangements of elements into sets. It provides tools to count how these arrangements can be made. One fundamental concept in combinatorics is the binomial coefficient. The binomial coefficient, denoted as \( \binom{n}{k} \), represents the number of ways to choose \( k \) elements from a set of \( n \) elements, irrespective of the order. This coefficient is key to solve many combinatorial problems and can be calculated using the formula:

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

Here, \( n! \) ("n factorial") is the product of all positive integers up to \( n \). This formula helps us find the different possible selections or groupings from a set, making it extremely useful in permutations and combinations.
In the given problem, the series of binomial coefficients \( \binom{5}{0} \), \( \binom{5}{1} \), etc., helps derive specific values used to evaluate the algebraic expression.

Pascal's Triangle

Pascal's Triangle is a triangular array of numbers. Each number is the sum of the two numbers directly above it in the previous row. One of the most fascinating properties of Pascal's Triangle is its relationship with binomial coefficients. Each row corresponds to the coefficients of an expanded binomial expression. For example, the third row \([1, 2, 1]\) represents \( (a + b)^2 \), showing how coefficients align with binomial expansions.

• The entry in row \( n \) and column \( k \) of Pascal's Triangle is \( \binom{n}{k} \).
• This means the triangular structure combines combinatorial properties with simple addition.

Pascal's Triangle simplifies the calculation of binomial coefficients, providing a quick reference to values especially for small integers.
In our exercise, referring to the row for \( n = 5 \) in Pascal's Triangle would provide the coefficients: \([1, 5, 10, 10, 5, 1]\). These values directly support the solution by offering a visual representation of the coefficients needed to simplify the expression.

Algebraic Expressions

An algebraic expression is a mathematical phrase that can include numbers, variables, and operation symbols. Expressions are essential in formulating and solving mathematical problems. They allow representation of complex operations in a simplified form. In our exercise, we deal with an algebraic expression composed solely of numbers and arithmetic operations:

• \( 1 - 5 + 10 - 10 + 5 - 1 \)

This expression exemplifies how binomial coefficients are used to represent mathematical relationships and calculations.
To evaluate, one simply follows arithmetic rules, breaking down the process:

• First, subtraction and addition operations are performed from left to right.
• Keep track of positive and negative values to ensure simplification is correct.

The outcome of the expression illustrates both combinatorial reasoning and algebraic simplification techniques.
By exploring these elements, students reinforce their understanding of how algebra and combinatorics intersect in problem solving.