Compute $\dbinom{25}{2}$.
I also know that there is a formula for this: $\dbinom{n}{k} = \frac{n!}{k!(n-k)!}$, where $n!$ is the factorial of n, meaning the product of all positive integers up to n. So, to compute $\dbinom{25}{2}$, I just need to plug in n = 25 and k = 2 into this formula and simplify. I get $\dbinom{25}{2} = \frac{25!}{2!(25-2)!} = \frac{25 \times 24 \times 23!}{2 \times 1 \times 23!}$. I can cancel out the 23! in the numerator and denominator, since they are equal. This leaves me with $\dbinom{25}{2} = \frac{25 \times 24}{2 \times 1} = \frac{600}{2} = 300$. 