In how many ways 11 items can be distributed among 3 peoples such that, sum of items received by any two is more than the third 
In how many ways $11$ items can be distributed among $3$ peoples such that, number of items received by any two person is more than the number of items received by the third person.

I have tried with case by case. But it will be cumbersome when the number of people and no. of items are large. Is there any general strategy to solve this problem, which can be applied for large numbers? 

We have $x_1+x_2>x_3$ or, $x_3<\frac{x_1+x_2+x_3}{2}$. And it holds for $ x_1, x_2$ also. Hence, $x_i$ can be $5$ maximum(as they are integers) for $i\in \{1,2,3\}$. So, we need to find the number of solution of the equation $$x_1+x_2+x_3=11,~~~ 1\le x_1,x_2,x_3\le 5$$
Which is equivalent to finding coeff. of $x^{11}$ in the expression $(x+x^2+x^3+x^4+x^5)^3$. 
$$\begin{align}[x^{11}](x+x^2+x^3+x^4+x^5)^3 &= [x^{11}]~\big(x^3(1+x+x^2+x^3+x^4)^3\big)\\&=[x^{8}]~(1+x+x^2+x^3+x^4)^3\\ &=[x^{8}]~\left( \frac{1-x^5}{1-x}\right)^3\\&=[x^8]~(1-x^5)^3(1-x)^{-3}\\&= [x^8]~(1-3x^5+3x^{10}-x^{15})\left(\sum_{r=0}^{\infty}\binom{-3}{r}x^r \right)\\ &= [x^8]~ (1-3x^5+3x^{10}-x^{15})\left(\sum_{r=0}^{\infty}\binom{3+r-1}{r}x^r\right)\\&=\binom{3+8-1}{8}-3\cdot\binom{3+3-1}{3}=15 \end{align}$$

Generalization

We can generalize for any perimeter $n\in\mathbb{N}$. We need to find the coeff. of $x^n$ in the expression $(x+x^2+\cdots +x^s)^3, s=\lfloor\frac{n}{2}\rfloor$.
Note: It is equivalent to finding all triangles with integer sides, which has perimeter $11$. 
