Problem 3 - Algebra and Trigonometry

Question

Evaluate the given binomial coefficient. $$ \left(\begin{array}{c} {12} \\ {1} \end{array}\right) $$

Step-by-Step Solution

Answer

The given binomial coefficient \(\left(\begin{array}{c}{12} \ {1}\end{array}\right)\) equals 12.

Step 1: Identify the variables

In this case, we have \(n = 12\) and \(k = 1\) in our binomial coefficient.

Step 2: Substitute into the formula

Substitute the values of \(n\) and \(k\) into the formula. The expression thus becomes \(\frac{12!}{1!(12-1)!}\).

Step 3: Simplify the equation

Simplify the equation. The factorial of a number is the product of all positive integers less than or equal to that number. Hence, \(12! = 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\), \(1! = 1\) and \(11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\). Now, \(12! = 12 \times 11!\). Hence, \(\frac{12!}{1!(12-1)!}\) simplifies to \(\frac{12 \times 11!}{1 \times 11!}\). This further simplifies to 12 as \(11!\) cancels out in the numerator and the denominator.

Key Concepts

Understanding Factorials

Factorials are the building blocks of many mathematical concepts, especially in binomial coefficients. A factorial, denoted by an exclamation mark (!), is the product of an integer and all the positive integers below it. For example, to calculate the factorial of 5, you multiply 5 by every positive whole number less than 5:

• 5! = 5 × 4 × 3 × 2 × 1 = 120

Factorials grow very rapidly and are crucial in counting problems, providing a foundation for understanding combinations and permutations. Remember, the factorial of 0 is defined as 1, and this concept is handy when simplifying equations.

Basics of Combinatorics

Combinatorics is a fascinating area of mathematics that focuses on counting, arranging, and classifying objects. It provides tools to answer questions related to how many ways we can choose or arrange a set of items. The binomial coefficient is a fundamental concept in combinatorics, symbolized as \( \binom{n}{k} \).

• This represents the number of ways to choose \( k \) items from \( n \) items without considering the order.
• The formula is: \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).

In our specific problem, \( \binom{12}{1} \), combinatorics helps us understand that we're choosing 1 item from a total of 12, which is straightforward but crucial in larger contexts. It simplifies to the number of objects, which is 12.

Exploring Permutations

Permutations are all about arranging items in specific orders. Unlike combinations, permutations consider the sequence, making them vital in problems where order matters.

• This is why permutations are calculated differently compared to combinations.
• The formula for permutations is \( P(n, k) = \frac{n!}{(n-k)!} \).

Although permutations aren't directly used in our problem, understanding the difference between permutations and combinations broadens the application of binomial coefficients. In situations requiring arrangement, you'd turn to permutations rather than combinations.

Applying Algebra in Combinatorics

Algebra is the language of mathematics, empowering us to manipulate symbols and equations to solve problems. In the realm of combinatorics and binomial coefficients, algebraic skills allow us to simplify expressions and understand complex relationships.

• For instance, simplifying \( \frac{12!}{1!(12-1)!} \) into 12 requires basic algebraic principles.
• Cancelling equal terms in the numerator and denominator showcases how algebra aids in simplifying problems.

Algebra helps us see the patterns and relationships in combinatorial equations, making it easier to tackle similar tasks efficiently. So, a solid grasp of algebra is invaluable across all mathematical disciplines, especially in simplifying and solving equations like these.