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Evaluate each number. 8 ! / 6 !

Short Answer

Expert verified
The expression \(\frac{8!}{6!}\) simplifies to \(\frac{8 \times 7}{1}\), which is equal to \(56\).

Step by step solution

01

Calculate 8!

To find 8!, which is the factorial of 8, we need to multiply all positive integers from 1 up to 8. So, the calculation is as follows: 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320
02

Calculate 6!

Now, we need to find 6!, which is the factorial of 6. This is done by multiplying all positive integers from 1 up to 6: 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
03

Divide 8! by 6!

Now that we have the values for 8! and 6!, we can divide 8! by 6!. The expression we need to evaluate is: \(\frac{8!}{6!} = \frac{40320}{720}\) Notice that both 8! and 6! have some common factors (6 × 5 × 4 × 3 × 2 × 1) because of the way factorials are calculated. So, we can cancel out these common factors to simplify the division: \(\frac{8!}{6!} = \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1}\) After canceling the common factors, we are left with: \(\frac{8!}{6!} = \frac{8 \times 7}{1}\) Now, perform the multiplication: \(\frac{8!}{6!} = 8 \times 7 = 56\) So, the value of the expression is 56.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Permutation
Permutations play a crucial role in understanding factorials and division in mathematical expressions. In essence, a permutation is an arrangement of objects in a specific order. For example, if you're trying to figure out in how many different ways a group of items can be ordered, you're dealing with permutations. The factorial function, denoted as "!", serves as the heart of permutations, calculating the number of different ways to arrange 'n' distinct objects.

Consider the factorial expressions we worked with: 8! and 6!. These aren't just products of numbers; they represent permutations:
  • 8! gives us the total number of possible ways to arrange eight items.
  • 6! shows us how many ways six items can be arranged.
Understanding permutations and their connection to factorials is an excellent way to familiarize yourself with more complex mathematical concepts and operations.
Mathematical Division
Mathematical division in the context of factorials involves dividing one large product by another. A division between two factorials like \( \frac{8!}{6!} \) allows us to simplify the expression by canceling out common factors. When we say common factors, we refer to numbers that appear in both the numerator and the denominator. This simplification process is pivotal because it reduces the expression to a more manageable form.

In our example, the common factors became apparent:
  • Both 8! and 6! include the factors 6 × 5 × 4 × 3 × 2 × 1.
  • Canceling these leaves us with 8 × 7, which can be easily calculated as 56.
This kind of division helps us avoid calculating enormous products from scratch, making the process more efficient and less error-prone.
Simplification of Expressions
Simplification is a key process when working with factorials and permutations. This doesn't just mean reducing equations to a simpler form but also understanding how to efficiently break down a problem. Simplifying expressions like \( \frac{8!}{6!} \) involves more than just arithmetic; it requires analytical thinking to identify patterns and commonalities.

Here are simplified processes to keep in mind:
  • Identify and cancel out common factors early. This step saves time and can prevent mathematical errors.
  • Consider the larger picture. Factorials often hide repeating patterns that provide insights into shortcuts and easier calculations.
By focusing on simplification, we not only make calculations easier but also enhance our understanding of the relationships between different mathematical concepts. It teaches us the basics of handling complex problems through a systematic approach.

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Most popular questions from this chapter

Minimalist Art You are exhibiting your collection of minimalist paintings. Art critics have raved about your paintings, each of which consists of 10 vertical colored lines set against a white background. You have used the following rule to produce your paintings: Every second line, starting with the first, is to be either blue or grey, while the remaining five lines are to be either all light blue, all red, or all purple. Your collection is complete: Every possible combination that satisfies the rules occurs. How many paintings are you exhibiting?

When is \(n(A \cup B) \neq n(A)+n(B) ?\)

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