Compute $\dbinom{14}{11}$.
I recall that $\dbinom{n}{k}$ is the number of ways to choose $k$ objects out of $n$ distinct objects, regardless of order. I also remember that there is a formula for computing $\dbinom{n}{k}$ in terms of factorials: $\dbinom{n}{k} = \frac{n!}{k!(n-k)!}$. So, to compute $\dbinom{14}{11}$, I can plug in $n=14$ and $k=11$ into this formula and simplify. I get $\dbinom{14}{11} = \frac{14!}{11!(14-11)!} = \frac{14!}{11!3!}$. Now, I notice that $14!$ and $11!$ have a lot of common factors, so I can cancel them out and simplify further. I get $\dbinom{14}{11} = \frac{14\times 13\times 12}{3\times 2\times 1} = \frac{14\times 13\times 2}{1} = 14\times 26 = 364$. 