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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 $iin {1,2,3}$. So, we need to find the number of solution of the equation $$x_1+x_2+x_3=11,~~~ 1le x_1,x_2,x_3le 5$$
Which is equivalent to finding coeff. of $x^{11}$ in the expression $(x+x^2+x^3+x^4+x^5)^3$.
$$begin{align}[x^{11}](x+x^2+x^3+x^4+x^5)^3 &= [x^{11}]~big(x^3(1+x+x^2+x^3+x^4)^3big)\&=[x^{8}]~(1+x+x^2+x^3+x^4)^3\ &=[x^{8}]~left( frac{1-x^5}{1-x}right)^3\&=[x^8]~(1-x^5)^3(1-x)^{-3}\&= [x^8]~(1-3x^5+3x^{10}-x^{15})left(sum_{r=0}^{infty}binom{-3}{r}x^r right)\ &= [x^8]~ (1-3x^5+3x^{10}-x^{15})left(sum_{r=0}^{infty}binom{3+r-1}{r}x^rright)\&=binom{3+8-1}{8}-3cdotbinom{3+3-1}{3}=15 end{align}$$

Generalization

We can generalize for any perimeter $ninmathbb{N}$. We need to find the coeff. of $x^n$ in the expression $(x+x^2+cdots +x^s)^3, s=lfloorfrac{n}{2}rfloor$.

Note: It is equivalent to finding all triangles with integer sides, which has perimeter $11$.

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 $$1 le a le b le 11-a-b < a+b.$$ So the rightmost inequality implies $a+b > 11/2$, 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 assumption that the items are distinguishable and the people are distinct too)

Simply put, no person can pick 6 or more items. Let's try to count "wrong" combinations, i.e. the number of combinations where one of the persons gets $kge6$ items. The first person can pick $k$ items in $binom{11}{k}$ ways, and the remaining items can be split between the second and the third person in $2^{11-k}$ different ways (each remaining item can be picked by either the second or the third person). You have two repeat the same process for the second person taking 6 or more items and than for the third. So the total number of "bad" combinations is:

$$3sum_{k=6}^{11}binom{11}{k}2^{11-k}$$

The total number of combinations is $3^{11}$ (because each item can be picked by any of the three persons involved).

So to total number of "good" combinations is:

$$3^{11}-3sum_{k=6}^{11}binom{11}{k}2^{11-k}=3^{11}-3sum_{k=6}^{11}binom{11}{11-k}2^{11-k}= \ 3^{11}-3sum_{i=0}^{5}binom{11}{i}2^{i}=112266$$

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 $$a(n)= begin {cases} {frac {n-2}2 choose 2}& n text { even}\
{frac {n-1}2 choose 2}& n text { odd}end {cases}$$