Problem 10 - College Algebra with Corequisite Support

Question

For the following exercises, evaluate the binomial coefficient. \(\left(\begin{array}{l}25 \\ 11\end{array}\right)\)

Step-by-Step Solution

Answer

The binomial coefficient \( \binom{25}{11} \) is 326,876,512.

Step 1: Understand the Binomial Coefficient Formula

The binomial coefficient \( \binom{n}{k} \) can be calculated using the formula: \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n! \) (n factorial) is the product of all positive integers up to \( n \).

Step 2: Identify the Variables

In this problem, \( n = 25 \) and \( k = 11 \). Plug these values into the binomial coefficient formula.

Step 3: Calculate the Factorials

First, calculate \( 25! \), \( 11! \), and \( (25-11)! = 14! \). Factorials can be very large. We simplify calculations later by canceling out terms.

Step 4: Simplify the Formula

Write the formula as \( \frac{25!}{11! \, 14!} \) and simplify by canceling terms. This means you only need to compute from \( 25 \) to \( 15 \) (the non-cancelled part of 25!) divided by \( 11! \).

Step 5: Compute the Calculation

Calculate \( \frac{25 \times 24 \times 23 \times 22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15}{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \). This results in 326,876,512.

Key Concepts

Factorial Calculation

Factorial calculation involves a sequence where you multiply a series of descending natural numbers. For any positive integer n, its factorial is represented as \( n! \) and is defined by the product \( n \times (n-1) \times (n-2) \times \, \ldots \, \times 3 \times 2 \times 1 \). So, if you wanted to calculate the factorial of a number like 5, you would multiply | 5 x 4 x 3 x 2 x 1 = 120 |.

Factorials grow fast, very fast. Even for small numbers, the results can be quite large. For instance, 10! is already 3,628,800.

• Understanding factorials: It’s a way to compute quantities involving ordered arrangements.
• Use in combinations: Factorials are key in counting the number of ways things can be ordered or arranged, which is central to solving combinatorial problems.

Knowing how to calculate factorials quickly and correctly is crucial when working on problems involving binomial coefficients, like in our original problem.

Combinatorics in Algebra

Combinatorics is a branch of mathematics dealing with counting, arrangement, and combination of elements within sets. In algebra, combinatorics helps in solving problems that revolve around counting and arranging numbers or objects.

The binomial coefficient, \( \binom{n}{k} \), often arises in this context. It represents the number of ways to choose \( k \) elements from a set of \( n \) elements, disregarding the order of selection. This concept is foundational in understanding how different arrangements and selections are possible within a particular set.

• Binomial coefficients in practice: These coefficients are used extensively in solving problems related to probabilities and distributions.
• Connection to Pascal's Triangle: Every entry in Pascal's Triangle represents a binomial coefficient, which beautifully demonstrates the symmetry and recursive relationships of combinations.

Understanding combinatorics in algebra allows students to solve complex counting problems by breaking them down into simpler steps, using the power of factorial calculations and binomial theorem.

Simplifying Expressions

In algebra, simplifying expressions refers to reducing an expression to its most compact or manageable form. This often involves factoring, applying specific algebraic identities, and canceling like terms.

When evaluating a binomial coefficient such as \( \binom{25}{11} \), simplifying it using factorial cancellation is key. As in the exercise, \( \frac{25!}{11! \, 14!} \) can be daunting, but with simplification, you don’t need to calculate each factorial fully. Instead, noting that \( 25! \) is \( 25 \times 24 \times \ldots \times 15 \times 14! \), you can cancel out the common \( 14! \) in the numerator and denominator.

• Benefits of simplification: It makes computations significantly easier and less error-prone.
• Algebraic manipulation: It helps identify common factors that can be canceled, streamlining the solution.
• Efficiency: Especially useful in problems like binomial coefficients, where direct calculation can be too complex and unwieldy.

Always strive to simplify expressions whenever possible to make mathematical problems more tractable and solutions more elegant.