The expression in the image is as follows:

$\frac{\frac{1}{5!} + \frac{1}{6!} + \frac{1}{7!}}{\frac{1}{6!} - \frac{1}{7!} + \frac{1}{8!}}$

Step 1: Calculate factorials

First, let's compute the factorials:

• $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
• $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$
• $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$
• $8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320$

Step 2: Substitute the factorials into the expression

We now substitute the values of the factorials:

$\frac{\frac{1}{120} + \frac{1}{720} + \frac{1}{5040}}{\frac{1}{720} - \frac{1}{5040} + \frac{1}{40320}}$

Step 3: Simplify each part

Numerator:

$\frac{1}{120} + \frac{1}{720} + \frac{1}{5040} = \frac{42}{5040} + \frac{7}{5040} + \frac{1}{5040} = \frac{50}{5040} = \frac{5}{504}$

Denominator:

$\frac{1}{720} - \frac{1}{5040} + \frac{1}{40320} = \frac{7}{5040} - \frac{1}{5040} + \frac{1}{40320} = \frac{6}{5040} + \frac{1}{40320}$ We can rewrite it as: $\frac{6}{5040} + \frac{1}{40320} = \frac{48}{40320} + \frac{1}{40320} = \frac{49}{40320}$

Step 4: Final result

Now the expression becomes: $\frac{\frac{5}{504}}{\frac{49}{40320}} = \frac{5}{504} \times \frac{40320}{49} = \frac{5 \times 40320}{504 \times 49} = \frac{201600}{24744} = \frac{336}{41}$

Thus, the simplified result is:

$\frac{336}{41}$

Let me know if you need further clarification or have any questions!

Here are 5 related questions:

1. What is the general definition of a factorial?
2. How do factorials grow in size as the number increases?
3. Can factorials be used in probability and statistics? How?
4. How can expressions with factorials be simplified using common factors?
5. What is the role of factorials in combinatorics?

Tip: Factorials grow very quickly, so large factorials can lead to very large numbers!