To find the number of ordered groups of 3 elements that can be arranged from a set of 8 different elements, we need to calculate the number of permutations of 3 elements taken from a set of 8.

The formula for permutations is:

$P(n, r) = \frac{n!}{(n - r)!}$

Where:

• $n$ is the total number of elements in the set (in this case, 8),
• $r$ is the number of elements to arrange (in this case, 3).

Using the formula:

$P(8, 3) = \frac{8!}{(8 - 3)!} = \frac{8!}{5!}$

Now, calculate the factorials:

$8! = 8 \times 7 \times 6 \times 5!$

So,

$P(8, 3) = \frac{8 \times 7 \times 6 \times 5!}{5!} = 8 \times 7 \times 6 = 336$

Therefore, there are 336 ordered groups of 3 elements that can be arranged from a set of 8 different elements.

Would you like more details or have any other questions?

Here are 5 related questions you could explore:

1. How many unordered groups of 3 elements can be chosen from a set of 8?
2. What is the difference between permutations and combinations?
3. How many ways can 4 elements be arranged from a set of 10 different elements?
4. What if repetition of elements was allowed in the groups of 3?
5. How many different ways can 5 elements be arranged from a set of 8?

Tip: Permutations focus on the order of selection, while combinations do not.