Problem 1 - Discrete Mathematics with Applications

Question

Evaluate each. $$\frac{5 !}{4 !}$$

Step-by-Step Solution

Answer

The evaluated expression is \(\frac{5!}{4!} = 5\).

Step 1: 1. Factorial Definition

To understand how to evaluate the given expression, let's remember the definition of a factorial. For a non-negative integer n, $$n! = n \times (n - 1) \times (n - 2)\times...\times 3\times 2 \times 1$$ Using this definition, we can expand the factorials in the given expression.

Step 2: 2. Expanding Factorials

Now, let's expand the factorials for both \(5 !\) and \(4 !\) using the definition of factorials. We have: \(5 ! = 5 \times 4 \times 3 \times 2 \times 1\) \(4 ! = 4 \times 3 \times 2 \times 1\)

Step 3: 3. Evaluating the Expression

We are now ready to evaluate the given expression \(\frac{5 !}{4 !}\). By substituting the expanded factorials from step 2 into the expression, we get: $$\frac{5 !}{4 !} = \frac{5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1}$$ Now we can simplify the expression by canceling common factors from the numerator and the denominator. Notice that \(4 \times 3 \times 2 \times 1\) can be canceled out: $$\frac{5 !}{4 !} = \frac{5 \times (4 \times 3 \times 2 \times 1)}{(4 \times 3 \times 2 \times 1)} = 5$$ So, the final result is: $$\frac{5 !}{4 !} = 5$$

Key Concepts

Factorial Definition

The factorial of a number is a fundamental mathematical concept that can be used to solve many problems, especially in combinatorics and probability.
At its core, a factorial is the product of all positive integers up to a specified number.
For any non-negative integer \( n \), the factorial, represented as \( n! \), is defined by the expression:

• \( n! = n \times (n - 1) \times (n - 2) \times \ldots \times 3 \times 2 \times 1 \)

This pattern means that you keep multiplying sequentially until you reach 1.
The factorial operation is used to calculate permutations and combinations, organize large datasets, and solve intricate equations in theoretical mathematics.
For instance, the factorial of 5, written as \( 5! \), equals \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \). It's simple to compute for small numbers, but factorial values grow quickly with larger numbers.

Expanding Factorials

Expanding factorials is vital when simplifying expressions involving factorials.
This involves writing out the entire multiplication chain for given numbers to see each component clearly.
Let's take a look at how this can be done using an example.Suppose we want to expand \( 5! \). We follow the factorial definition:

• Start with the highest number, which is 5 in this case.
• Multiply it by each preceding whole number down to 1.

So, \( 5! \) expands to \( 5 \times 4 \times 3 \times 2 \times 1 \).
Similarly, \( 4! \) can be expanded as \( 4 \times 3 \times 2 \times 1 \).
This process allows us to identify and simplify expressions where factorials appear in numerators and denominators, as it helps to spot common terms.
By expanding, we prepare ourselves for simplification, making it easier to cancel out terms shared by both the numerator and denominator.

Simplifying Expressions

Simplifying expressions with factorials is a key skill in solving problems efficiently.
Once you've expanded the factorials, the next step typically involves canceling out common terms in fractions either within or across expressions.
This makes calculations more manageable and diminishes the risk of errors.Let's use the problem from the exercise as an illustration. We want to simplify \( \frac{5!}{4!} \).

• First, we expand both \( 5! \) and \( 4! \):
• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 \)
• \( 4! = 4 \times 3 \times 2 \times 1 \)

Notice the terms \( 4 \times 3 \times 2 \times 1 \) in both the numerator and the denominator.
These factors cancel each other out:

• \( \frac{5 \times (4 \times 3 \times 2 \times 1)}{4 \times 3 \times 2 \times 1} = 5 \)

Thus, simplifying gives us the result \( 5 \).
This technique reduces work and complexity, especially when dealing with larger factorials. Recognizing patterns and common elements is crucial in streamlining factorial problems.