Problem 26 - Beginning and Intermediate Algebra

Question

Evaluate each binomial coefficient. $$\left(\begin{array}{l}3 \\\1\end{array}\right)$$

Step-by-Step Solution

Answer

\( \binom{3}{1} = 3 \)

Step 1: Identify n and k values

The binomial coefficient formula requires the values of n and k. In our case, n = 3 and k = 1.

Step 2: Calculate factorials for n, k, and n-k

Using the identified values of n and k, we need to calculate the factorials of n, k, and n-k: \(3! = 3 \times 2 \times 1 = 6\) \(1! = 1\) \((3-1)! = 2! = 2 \times 1 = 2\)

Step 3: Apply the binomial coefficient formula

Now, plug these values into the binomial coefficient formula: \(\binom{3}{1} = \frac{3!}{1!(3-1)!} = \frac{6}{1 \times 2}\).

Step 4: Simplify the expression

Simplify the expression: \(\binom{3}{1} = \frac{6}{1 \times 2} = \frac{6}{2} = 3\) The binomial coefficient is 3.

Key Concepts

Factorial Calculation

The concept of factorial is essential in understanding binomial coefficients. A factorial, denoted by an exclamation point (e.g., \(n!\)), is the product of all positive integers less than or equal to a number \(n\). For instance, \(3!\) is calculated as \(3 \times 2 \times 1 = 6\).
A key property of factorials is that \(0!\) is defined to be 1. This special definition ensures that formulas, like those involving binomial coefficients, hold true even when they involve zero.
When calculating binomial coefficients, you often compute several factorials, like \(n!\), \(k!\), and \((n-k)!\). It’s important to remember:

• \(3! = 6\)
• \(1! = 1\)
• \(2! = 2 \times 1 = 2\)

These calculations provide the basis for plugging values into larger formulas for combinatorial expressions, like the binomial coefficient.

Combinatorics

Combinatorics is the mathematical field that deals with counting, arranging, and finding patterns. The binomial coefficient is a central concept in combinatorics and is used to find the number of ways to choose \(k\) elements from a set of \(n\) elements without considering the order.The binomial coefficient is represented as \(\binom{n}{k}\) and is calculated using the formula:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]This formula expresses the total number of combinations of \(n\) elements taken \(k\) at a time.
For example, the expression \(\binom{3}{1}\) finds the number of ways to choose 1 element from 3, which simplifies to 3 as shown previously. Each term in the formula has its purpose:

• \(n!\) calculates the total arrangements of the \(n\) elements.
• \(k!\) and \((n-k)!\) adjust for overcounting by arranging and then selecting the actual subset.

This concept is not just about simple counts; it forms the backbone of many probability and statistics problems.

Algebra Basics

Algebra involves using symbols and formulas to solve problems. Understanding algebra basics is key for manipulating expressions like binomial coefficients.
Such algebraic expressions leverage factorials and symbols to represent combinations in a concise and readable format. Simplification in algebra involves canceling terms and reducing expressions to their simplest form. For the binomial coefficient \(\binom{3}{1}\), algebraic steps go from plugging values into the formula to simplifying:

• Calculate \(3! = 6\), \(1! = 1\), and \(2! = 2\).
• Substitute into \(\frac{n!}{k!(n-k)!}\), which becomes \(\frac{6}{1 \times 2}\).
• Simplify everything down to \(\frac{6}{2} = 3\).

This systematic reduction in algebra ensures that the component calculations (like cancelling out terms) line up with the theoretical foundation (such as factorial properties), giving you a direct path to the solution.