https://file.aiquickdraw.com/mathbot/file/1729436037393w5wwpn5u.jpg

The problem in the image appears to be written as:

\[
\text{If } {}^nP_3 = 8! , \text{ find } n
\]

This refers to a permutation problem where \( {}^nP_3 \) represents the number of ways to arrange 3 objects out of \( n \), which is given by the formula:

\[
{}^nP_3 = \frac{n!}{(n - 3)!}
\]

We are also given that \( {}^nP_3 = 8! \). So, we set up the equation:

\[
\frac{n!}{(n - 3)!} = 8!
\]

Now, we simplify the left-hand side by expanding \( n! \) up to \( (n-3)! \):

\[
\frac{n!}{(n-3)!} = n \times (n-1) \times (n-2)
\]

Thus, the equation becomes:

\[
n \times (n - 1) \times (n - 2) = 8!
\]

We know that \( 8! = 40320 \), so:

\[
n \times (n - 1) \times (n - 2) = 40320
\]

Now, we solve this equation to find the value of \( n \).

Do you want the full step-by-step calculation for finding \( n \)? Let me know!

Here are 5 related questions to further explore this concept:
1. What is the formula for combinations and how does it differ from permutations?
2. Can you express \( n! \) as a product of consecutive numbers?
3. What is the value of \( {}^{10}P_3 \) and how does it compare to \( {}^{10}C_3 \)?
4. How does the concept of factorial relate to probability problems?
5. Can we use this permutation formula in circular arrangements?

**Tip:** When solving permutation problems, remember that the order of arrangement matters, unlike combinations where order doesn't matter.

Learn how to solve the permutation problem where nP3 = 8! with a detailed step-by-step solution. This problem covers key concepts of permutations and factorials. Suitable for students in Grades 9-12.

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