The notation \( 8P3 \) refers to a **permutation** problem, where we want to determine the number of ways to arrange 3 objects out of 8 distinct objects. The formula for permutations is:

\[
nP_r = \frac{n!}{(n - r)!}
\]

In this case, \( n = 8 \) and \( r = 3 \), so we have:

\[
8P3 = \frac{8!}{(8 - 3)!} = \frac{8!}{5!}
\]

Now, calculate \( 8! \) and \( 5! \):

\[
8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320
\]
\[
5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
\]

Now, divide the two values:

\[
8P3 = \frac{40320}{120} = 336
\]

So, \( 8P3 = 336 \).

Would you like more details or have any questions?

Here are 5 related questions to expand on this:

1. What is the difference between permutations and combinations?
2. How would you solve \( 8C3 \) (combinations)?
3. How is the formula for permutations derived?
4. How can you apply permutations to real-world problems?
5. What happens if all objects are identical? How does that affect permutations?

**Tip**: For larger numbers, using factorial functions on a calculator can save time when computing permutations.

Learn how to solve the permutation problem 8P3, where 3 objects are arranged from a total of 8 distinct objects. Understand the formula nPr = n! / (n - r)! and follow the step-by-step calculation of 8! and 5! to get the solution 336. Suitable for high school students studying combinatorics and permutations.

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