To solve $\binom{12}{3}$, which is a combination problem, we use the formula:

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

For $\binom{12}{3}$, we have:

$\binom{12}{3} = \frac{12!}{3!(12 - 3)!} = \frac{12!}{3! \cdot 9!}$

Simplifying, we only need to calculate the first three terms of $12!$ in the numerator, as the $9!$ terms will cancel out:

$= \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = \frac{1320}{6} = 220$

So, the correct answer is 220.

It appears that the selected answer of 1,320 is incorrect. Would you like further clarification on this solution?

Here are some additional questions for practice:

1. How would you calculate $\binom{10}{2}$ using the combination formula?
2. What is the difference between a permutation and a combination?
3. How can you apply combinations in probability problems?
4. Calculate $\binom{15}{5}$ using the combination formula.
5. What happens to the value of combinations as $r$ approaches $n$?

Tip: For large combination calculations, consider canceling common factorial terms to simplify the computation.