11. How many solutions does the equation $x+y+z=2016$ have, where $x, y$ and $z$ are integers with $x>1000$, $y>600$, and $z>400$ ?
Answer: 105
First, note that the equation is equivalent to $(x-1001)+(y-601)+(z-401)=13$. Letting $x^{\prime}=x-1001$, $y^{\prime}=y-601$, and $z^{\prime}=z-401$, we can instead count the number of solutions of $x^{\prime}+y^{\prime}+z^{\prime}=13$, where $x^{\prime}, y^{\prime}, z^{\prime}$ are nonnegative integers. Now note that each solution $\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ to this equation corresponds to a way of choosing 13 items from a pile of objects of 3 types; the number of such choices is
$$
\binom{13+3-1}{13}=\frac{15!}{13!2!}=105
$$