Compute $\dbinom{31}{28}$.
I know that $\dbinom{n}{k}$ means the number of ways to choose k objects from n distinct objects, and it can be calculated as $\frac{n!}{k!(n-k)!}$. In this case, n is 31 and k is 28, so I could plug those values into the formula and simplify. Now, using the formula, I get $\dbinom{31}{3}=\frac{31!}{3!(31-3)!}=\frac{31!}{3!28!}$. To simplify this fraction, I can cancel out the common factors in the numerator and the denominator. The largest common factor is 28!, so I can divide both the numerator and the denominator by 28! and get $\frac{31!}{3!28!}=\frac{31\times 30\times 29}{3!}$. Next, I can cancel out the factor of 6 in the numerator and the denominator, since 3!=6. This gives me $\frac{31\times 30\times 29}{3!}=\frac{31\times 5\times 29}{1}$. Finally, I can multiply the remaining numbers in the numerator and get the answer: $\frac{31\times 5\times 29}{1}=4495$. 