Problem 2 - College Algebra

Question

In Exercises 1-8, evaluate the given binomial coefficient. $$\left(\begin{array}{l}7 \\ 2\end{array}\right)$$

Step-by-Step Solution

Answer

The value of \( \left(\begin{array}{l}7 \ 2\end{array}\right) \) is 21

Step 1: Identify n and k

n is the total number of elements in the set and k is the number of elements we want to choose. For this problem, n = 7 and k = 2.

Step 2: Substitute n and k into the binomial coefficient formula

Substituting n = 7 and k = 2 into the binomial coefficient formula \( \left(\begin{array}{l}n \ k\end{array}\right) = \frac{n!}{k!(n-k)!} \) gives \( \left(\begin{array}{l}7 \ 2\end{array}\right) = \frac{7!}{2!(7-2)!} \)

Step 3: Simplify the Factorials

7! = 7*6*5*4*3*2*1, 2! = 2*1 , (7-2)! = 5! = 5*4*3*2*1. Substituting these into the equation gives \( \frac{7*6*5*4*3*2*1}{2*1*5*4*3*2*1} \)

Step 4: Simplify the Equation

Simplify the equation by cancelling out the same numbers in the numerator and the denominator to leave only 7*6 on the numerator and 2*1 on the denominator. This simplifies to \( \frac{7*6}{2} = 21 \)

Key Concepts

Factorial

Factorials are a fundamental concept in mathematics that help in counting and solving problems involving permutations and combinations. The factorial of a non-negative integer \(n\) is denoted as \(n!\) and represents the product of all positive integers up to \(n\). This can be defined as:

• \(1! = 1\)
• \(2! = 2 \times 1 = 2\)
• \(3! = 3 \times 2 \times 1 = 6\)
• \(n! = n \times (n-1)! \) for \(n > 1\)

For larger numbers, factorials grow very rapidly. In our example, we calculate the following:

• \(7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040\)
• \(2! = 2 \times 1 = 2\)
• \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)

The factorial helps in determining combinations, which brings us to our next topic.

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, arrangement, and combination of objects. The concept is crucial when we need to determine how different sets of items can be grouped or arranged. A common problem in combinatorics is choosing \(k\) elements from a set of \(n\) elements, known as combinations, which is calculated using the binomial coefficient. The binomial coefficient \( \left(\begin{array}{c}n \ k\end{array}\right) \) is given by the formula:\[\left(\begin{array}{c}n \ k\end{array}\right) = \frac{n!}{k!(n-k)!}\]This formula calculates the number of ways to choose \(k\) elements from \(n\) elements without considering the order.In our example, \(n = 7\) and \(k = 2\). Using the formula, we substitute in our values to get:\[\left(\begin{array}{c}7 \ 2\end{array}\right) = \frac{7!}{2!(7-2)!}\]Simplifying the expression gives us the result 21. This means there are 21 ways to choose 2 items from a set of 7.

Algebraic Expressions

Algebraic expressions are mathematical phrases that use numbers, variables, and operators. Expressions can be manipulated to simplify complex calculations or to solve algebraic equations. In the calculation of binomial coefficients, algebraic manipulation simplifies understanding and solving the expression.In our exercise, the algebraic expression arises from substituting the factorial calculations into the binomial coefficient formula:\[\frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1 \times 5 \times 4 \times 3 \times 2 \times 1}\]Simplifying this expression involves canceling out the common terms in the numerator and the denominator:

• The numbers \(5\), \(4\), \(3\), \(2\), and \(1\) cancel out.
• This leaves us with \(\frac{7 \times 6}{2 \times 1} \)

Finally, this simplifies to \(\frac{42}{2} = 21\), giving the number of combinations. Understanding how to manipulate these expressions is crucial for solving more complex algebraic problems.