Happy birthday to me (part 2)

Suppose there are $p­$ objects (integer quantities) to be distributed to $n­$ rooms (variables). To separate $n­$ rooms, it is needed $n-1­$ "walls" between these rooms. Adding up these walls and objects, we'll have $n­+p-1$ elements. Since we want to distribute $p­$ objects performing simple combinations with the $n­+p-1$ elements, the number $\binom{n+p-1}{p}=\binom{n+p-1}{n-1}$ represents the number of integer and non-negative solutions of an equation with $n­$ variables and $p­$ objects or integer quantities on the right-hand side. With some notation, we have:

$S=\# \{{x_1},{x_2},...,{x_{n}} \in \mathbb{N_{0}} \mid {x_1}+{x_2}+...+{x_{n}}=p\}=\binom{n+p-1}{n-1}$ .

Thus, an inequality is the sum of all the anterior results or number of solutions to the previous equations. Another way to determine the number of solutions to an inequality is to add up the quantity in the left-hand side that should make the inequality becomes equal to (and so, becomes an equality) a referred value in the right-hand side. This is equivalent to determine a new variable on the left-hand side, so that the number of the solutions to an inequality becomes:

$S=\# \{{x_1},{x_2},...,{x_{n}} \in \mathbb{N_{0}} \mid {x_1}+{x_2}+...+{x_{n}}<{q}\}$

$S=\# \{{x_1},{x_2},...,{x_{n+1}} \in \mathbb{N_{0}} \mid {x_1}+{x_2}+...+{x_{n+1}}=q\}$

$S=\binom{(n+1)+q-1}{q}=\binom{n+q}{q}=\binom{n+q}{n}$ .

Compute $S=\# \{{x_1},{x_2},...,{x_{18}} \in \mathbb{N_{0}} \mid {x_1}+{x_2}+...+{x_{18}}<{1998}\}$ is equivalent to calculate the number of solutions to every value lesser than or equal to $1997$ , which we must know if this is the part we want to neglect. This number is $\binom{18+1996}{18}=\binom{2014}{18}$ .

Compute $S=\# \{{x_1},{x_2},...,{x_{18}} \in \mathbb{N_{0}} \mid {x_1}+{x_2}+...+{x_{18}}<{2015}\}$ is equivalent to calculate the number of solutions to every value lesser than or equal to $2015$ , which is the general solution from where we'll take down the part to be neglected. This number is $\binom{18+2015}{2015}=\binom{2033}{2015}$ .

Now, we just have to subtract one from another, that is:

$\displaystyle {S=\binom{2033}{2015}-\binom{2014}{18}=\binom{a}{b}-\binom{b-1}{c}}$

Hence, the answer is $\lfloor{\frac{a+b+c}{3}}\rfloor=\lfloor{\frac{2033+2015+18}{3}}\rfloor=\boxed {1335}$ .