Problem 16 - College Algebra

Question

In Exercises \(9-16,\) use the formula for \(_{n} C_{r}\) to evaluate each expression. $$ _{6} C_{0} $$

Step-by-Step Solution

Answer

The evaluated result of \( _{6} C_{0} \) is 1.

Step 1: Calculate n!

First, calculate the factorial of n, that is 6!. The factorial of a number is the product of all positive integers less than or equal to the number. Therefore, calculate 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720.

Step 2: Calculate r! and (n-r)!

Next, calculate the factorial of r, which is 0!, and the factorial of (n-r), which is also 6!. The factorial of 0! is defined as 1. Therefore, calculate 0! = 1 and (6-0)! = 6! = 720 as previously calculated.

Step 3: Evaluate the expression

Finally, substitute these values into the formula for \( _{6} C_{0} \). So, \( _{6} C_{0} = \frac{720}{1 \times 720} = 1.

Key Concepts

Factorial Notation

Understanding factorial notation is crucial when working with permutations and combinations. The factorial of a natural number n, denoted as n!, is the product of all positive integers from n down to 1. For example, the factorial of 5, written as 5!, is calculated as 5 × 4 × 3 × 2 × 1, which equals 120.

It's important to note that the factorial of zero, 0!, is defined to be 1. This might seem counter-intuitive, but it's a convention that ensures the formulas for permutations and combinations work correctly, even when we choose zero items from a set.

In our exercise, the notation 6! represents the factorial of 6, which is equal to 720, calculated as follows: \( 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\). This step is fundamental for determining the number of ways to arrange or choose elements in algebraic problems.

Permutations and Combinations

Two foundational concepts in combinatorics are permutations and combinations, which refer to the different ways of arranging or selecting items from a set without replacement. Permutations concern arrangements where order matters, while combinations focus on selections where order is irrelevant.

Take, for example, a fruit bowl with an apple, a banana, and a cherry. If we were to choose two fruits, the order in which we pick them is inconsequential for combinations; picking an apple then a banana is considered the same as picking the banana followed by the apple. However, for permutations, these would be two distinct outcomes.

The key formula for combinations is \( _nC_r = \frac{n!}{r!(n-r)!} \), which tells us the number of ways to choose r items from a set of n without regard to the order. In our exercise with \( _6C_0 \), we're essentially asking in how many ways we can choose zero items from six, which logically is just one way - to choose nothing.

Binomial Coefficient

The binomial coefficient, as seen in exercises like ours with \( _6C_0 \), reflects the number of ways to pick r unordered outcomes from n possibilities, known as combinations. It is represented by the formula \( _nC_r = \frac{n!}{r!(n-r)!} \) and is often read as 'n choose r.'

The binomial coefficient appears in the binomial theorem, which is used to expand expressions raised to a power. It also crops up in probability theory, where it's employed to find the likelihood of achieving a certain number of successes in a series of independent trials.

In simpler terms, if you were given 6 different books and wanted to know how many ways you can align 0 books on your shelf, the binomial coefficient would tell you that, irrespective of the books, there's only one way to arrange zero books - by not placing any books at all. Hence, applying the formula in our case where r equals 0, we get \( _6C_0 = 1 \), because \( r! = 0! = 1 \) and \( (n - r)! = 6! \), as computed before.