It looks like you're referring to a combination formula, often denoted as $C(n, r)$, which calculates the number of ways to choose $r$ objects from $n$ objects without regard to the order of selection.

The combination formula is:

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

For $C(8, 5)$, this means you are choosing 5 objects from 8. We can calculate it as follows:

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

Now let's calculate this step by step:

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

So,

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

$3! = 3 \times 2 \times 1 = 6$

Now,

$C(8, 5) = \frac{8 \times 7 \times 6}{6} = 8 \times 7 = 56$

Thus, $C(8, 5) = 56$, which matches the value you provided!

Would you like more explanation on combinations or other related topics?

Related Questions:

1. What is the formula for $C(n, r)$, and how does it differ from permutations?
2. How would you calculate $C(10, 2)$?
3. What is the significance of combinations in real-life scenarios, like lottery odds?
4. Can combinations be used for selecting subsets from a set with restrictions?
5. How does the concept of combinations relate to Pascal's Triangle?

Tip:

When dealing with combinations, remember that $C(n, r) = C(n, n - r)$, meaning that choosing $r$ objects from $n$ is the same as choosing $n - r$ objects from $n$.