Compute $\dbinom{15}{3}$.
\begin{align*}
\dbinom{15}{3} &= \dfrac{15!}{12!3!} \\
&= \dfrac{15\times 14\times 13}{3\times 2\times 1} \\
&= \dfrac{15}{3} \times \dfrac{14}{2} \times \dfrac{13}{1} \\
&= 5\times 7\times 13 \\
&= \boxed{455}.
\end{align*}
455
Level 1
Counting & Probability
Problem: Compute $\dbinom{15}{3}$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To compute $\dbinom{15}{3}$, 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, and $r$ is the number of items to choose. 

In this case, $n = 15$ and $r = 3$. We can substitute these values into the formula:

1. Calculate $15!$, which is the factorial of 15. However, we only need the first few terms since the other terms will cancel out. 
2. The expression becomes $\dbinom{15}{3} = \frac{15!}{3!(15-3)!} = \frac{15!}{3! \cdot 12!}$.
3. We can simplify this by recognizing that $15! = 15 \times 14 \times 13 \times 12!$. Thus, we can rewrite the expression as:
   \[
   \dbinom{15}{3} = \frac{15 \times 14 \times 13 \times 12!}{3! \cdot 12!}
   \]
4. The $12!$ in the numerator and denominator cancels out, leading to:
   \[
   \dbinom{15}{3} = \frac{15 \times 14 \times 13}{3!}
   \]
5. Next, we calculate $3! = 3 \times 2 \times 1 = 6$.
6. Now, we can substitute this back into our expression:
   \[
   \dbinom{15}{3} = \frac{15 \times 14 \times 13}{6}
   \]
7. Calculate the numerator: $15 \times 14 = 210$, and then $210 \times 13 = 2730$.
8. Now divide by 6: 
   \[
   \frac{2730}{6} = 455
   \]

Thus, the final answer is $455$.<|END_OF_LATENT|> The final answer is $455$.