Compute $\dbinom{22}{19}$.
I know that $\dbinom{n}{k}$ is the number of ways to choose k objects out of n distinct objects, regardless of order. I also know that there is a formula for $\dbinom{n}{k}$ in terms of factorials: $\dbinom{n}{k} = \frac{n!}{k!(n-k)!}$. So I can plug in the values of n and k in this problem and simplify: $\dbinom{22}{19} = \frac{22!}{19!(22-19)!} = \frac{22!}{19!3!}$. I can cancel out the common factors in the numerator and denominator: $\frac{22!}{19!3!} = \frac{22\times 21\times 20}{3\times 2\times 1}$. I can do the arithmetic and get the final answer: $\frac{22\times 21\times 20}{3\times 2\times 1} = \frac{9240}{6} = 1540$. 