To compute $C(5, 2)$, which represents the number of ways to choose 2 objects from 5 (also known as a combination), we use the formula for combinations:

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

For $C(5, 2)$, where $n = 5$ and $r = 2$:

$C(5, 2) = \frac{5!}{2!(5 - 2)!} = \frac{5!}{2!3!}$

Now, calculate the factorials:

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ $2! = 2 \times 1 = 2$ $3! = 3 \times 2 \times 1 = 6$

Now substitute these into the formula:

$C(5, 2) = \frac{120}{2 \times 6} = \frac{120}{12} = 10$

So, $C(5, 2) = 10$.

Do you want further explanation or examples? Feel free to ask!

Here are 5 related questions:

1. How do you calculate the number of ways to choose 3 objects from 5?
2. What is the difference between combinations and permutations?
3. How does the formula for combinations change for larger values of $n$ and $r$?
4. How do combinations apply to probability problems?
5. What are some real-world examples of using combinations?

Tip: Remember, the formula for combinations involves dividing by the factorials to account for overcounting of equivalent selections.