Compute $\dbinom{25}{2}$.
\begin{align*}
\dbinom{25}{2} &= \dfrac{25!}{23!2!} \\
&= \dfrac{25\times 24}{2\times 1} \\
&= 25 \times \dfrac{24}{2} \\
&= 25 \times 12 \\
&= \boxed{300}.
\end{align*}
300
Level 1
Counting & Probability
Problem: Compute $\dbinom{25}{2}$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To compute $\dbinom{25}{2}$, we need to understand the formula for combinations, which is given by $\dbinom{n}{r} = \frac{n!}{r!(n-r)!}$. Here, $n$ is the total number of items to choose from, and $r$ is the number of items to choose.

In this case, $n = 25$ and $r = 2$. We can substitute these values into the formula:

1. Calculate $25!$, but since we only need the first few terms, we can simplify our calculation by recognizing that $25! = 25 \times 24 \times 23!$.
2. Calculate $2! = 2 \times 1 = 2$.
3. Calculate $(25 - 2)! = 23!$.

Now, substituting into the combination formula:

$\dbinom{25}{2} = \frac{25!}{2! \cdot 23!} = \frac{25 \times 24 \times 23!}{2 \times 1 \times 23!}$.

The $23!$ in the numerator and denominator cancels out:

$\dbinom{25}{2} = \frac{25 \times 24}{2 \times 1}$.

Now, perform the multiplication in the numerator:

$25 \times 24 = 600$.

Next, divide by $2$:

$\frac{600}{2} = 300$.

Thus, the final answer is $300$.<|END_OF_LATENT|> The final answer is $300$.