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?
Under the assumption that the items are indistinguishable and the people are distinct, let the ordered triple $(a,b,c)$ denote the number of items distributed to the first, second, and third person, respectively. Then determine the number of ordered triples of integers $(a, b, 11-a-b)$ such that $$1 \le a \le b \le 11-a-b &lt; a+b.$$  So the rightmost inequality implies $a+b &gt; 11/2$, which in turn implies $b \ge 3$.  We must also have $a \le 11/3$, or $a \le 3$.  Finally, we must have $b \le 5$ since $2b \le 11 - a \le 11 - 1 = 10$.  So it follows that our candidate triples are $$(a, 3, 8-a), (a, 4, 7-a), (a, 5, 6-a),$$ and the last case can only have $a = 1$ resulting in $(1, 5, 5)$.  The second case can only have $a \in \{2, 3\}$, giving $(2, 4, 5)$ and $(3, 4, 4)$, respectively; and the first case admits $a = 3$, corresponding to $(3, 3, 5)$.  These comprise the only nondecreasing positive integer triples for which the sum is $11$ and the sum of any two elements exceeds the third.  Three of them can be permuted in $3$ ways, and the fourth can be permuted in $6$ ways; hence the total number of triples is $3(3) + 6 = 15$.
Under the assumptions of heropup the sequence is OEIS A008795.  It begins 
$$1, 0, 3, 1, 6, 3, 10, 6, 15, 10, 21, 15, 28, 21, 36, 28, 45, 36, 55, 45, $$
and is given by $$a(n)= \begin {cases} {\frac {n-2}2 \choose 2}&amp; n \text { even}\\
{\frac {n-1}2 \choose 2}&amp; n \text { odd}\end {cases}$$