Problem 24 - Algebra and Trigonometry

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 evaluated expression is 0.

Step 1: Understanding Binomial Coefficients

First, recognize that \( \binom{n}{k} \) is a binomial coefficient, which represents the number of ways to choose \( k \) elements from a set of \( n \) elements. It is calculated using the formula: \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\] where \( n! \) denotes the factorial of \( n \).

Step 2: Calculate \( \binom{5}{0} \)

Using the formula for binomial coefficients, calculate \( \binom{5}{0} \): \[\binom{5}{0} = \frac{5!}{0! \times 5!} = 1\] because any number to the power of zero is 1.

Step 3: Calculate \( \binom{5}{1} \)

Now calculate \( \binom{5}{1} \): \[\binom{5}{1} = \frac{5!}{1! \times 4!} = 5\] because there are 5 ways to select 1 element from a set of 5.

Step 4: Calculate \( \binom{5}{2} \)

Next, calculate \( \binom{5}{2} \): \[\binom{5}{2} = \frac{5!}{2! \times 3!} = 10\] since there are 10 ways to select 2 elements from a set of 5.

Step 5: Calculate \( \binom{5}{3} \)

Calculate \( \binom{5}{3} \): \[\binom{5}{3} = \frac{5!}{3! \times 2!} = 10\] because selecting 3 out of 5 is symmetrical to selecting 2 out of 5.

Step 6: Calculate \( \binom{5}{4} \)

Calculate \( \binom{5}{4} \): \[\binom{5}{4} = \frac{5!}{4! \times 1!} = 5\] because there are 5 ways to leave out 1 element from a set of 5.

Step 7: Calculate \( \binom{5}{5} \)

Finally, calculate \( \binom{5}{5} \): \[\binom{5}{5} = \frac{5!}{5! \times 0!} = 1\] since there is only 1 way to select all elements from a set.

Step 8: Substitute Values into the Expression

Substitute the calculated binomial coefficients back into the original expression: \[1 - 5 + 10 - 10 + 5 - 1\]

Step 9: Simplify the Expression

Now simplify the expression step-by-step:1. Calculate \( 1 - 5 = -4 \).2. Add 10: \( -4 + 10 = 6 \).3. Subtract 10: \( 6 - 10 = -4 \).4. Add 5: \( -4 + 5 = 1 \).5. Subtract 1: \( 1 - 1 = 0 \).

Step 10: Final Result

After performing and simplifying all operations, the result of the given expression is 0.

Key Concepts

Factorials

A factorial, denoted by the symbol "!", is a fundamental concept in mathematics, especially in combinatorics. It represents the product of all positive integers up to a certain number. For instance, the factorial of 5 is calculated as follows: \[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]Factorials are crucial when calculating permutations and combinations, which is why they are also central in the computation of binomial coefficients.

• 0 Factorial: Understanding that \(0!\) equals 1 is key, as this ensures consistency in mathematical equations and concepts like combinations.
• Application in Binomial Coefficients: Factorials are used to determine possible ways of arranging or selecting items.

Remember, each step down in a factorial is a multiplication by one less than the previous step, showcasing exponential growth quickly.

Combinatorics

Combinatorics is the field of mathematics that studies the counting, arrangement, and combination of objects. It plays a significant role in probability and algebra. One of its most practical applications is through the concept of binomial coefficients.Binomial coefficients are expressed using the notation \( \binom{n}{k} \), which illustrates the number of ways to choose \( k \) items from a total of \( n \) items without regard to the order of selection. The calculation of these coefficients is critical to solving many problems that involve selections and arrangements.

• Real-Life Applications: Combinatorics can be applied to solve problems involving seating arrangements, lottery probabilities, and decision-making processes.
• Relation to Algebra: Through algebraic expressions involving binomial expansions, combinatorics provides deeper insights into mathematical structures.

Thus, understanding the principles of combinatorics empowers students to solve complex problems involving large sets and intricate arrangements.

Algebra

Algebra is a branch of mathematics dealing with symbols and the rules for manipulating these symbols. It forms the language through which we express mathematical relationships, patterns, and structures. When performing operations with binomial coefficients, algebra plays an essential role in simplifying expressions and solving equations.

• Expression Simplification: In the exercise, algebra helps in substituting values back into the original expression, which is crucial in simplifying and manipulating the given arithmetic to reach the final result.
• Equations and Formulas: Algebra facilitates the manipulation of binomial expressions and the use of various properties, such as symmetry in binomial coefficients, to solve problems efficiently.

Algebra is thus the foundation that supports the application of combinatoric concepts, enabling us to solve complex mathematical expressions and make rational interpretations.