In how many ways can we distribute 80 gold bars to 3 peasants and 3 kings, such that every king gets at least 10 bars, and each peasant gets at most 10 bars?
I tried to reduce the problem to a more combinatorial form. I tried splitting the question to two unrelated parts and then adding them up:
First, I say the following inequality represents each king: $${k_1 + k_2 + k_3 \geq 30}$$ where each ${k_i}$ is variable containing the sum of golden bars for that king. The sum of ${k_1, k_2, k_3}$ however must be at most 80. Because I know each king must have at least 10 golden bars, the number of possible ways to distribute those golden bars is ${{52}\choose{2}}$
Similarly, for the second part I represented each peasant as a variable in the inequality: $${p_1 + p_2 + p_3 \leq 30}$$ such that ${p_1,p_2,p_3 \leq 10}$.
This is where I got stuck. I'm not sure how I can calculate the possibilities for the last inequality. Moreover, I'm not sure how one can then provide an answer to the original question. Unfortunately I couldn't think way to reduce the question to something I'm capable of providing an answer to. Any suggestions?
Give 10 bars to each of the kings. Then each of the kings can receive any number of additional bars, and each of the peasants can receive between $0$ and $10$ bars. Assuming that the kings as well as the peasants have names the number $N$ you are after is the coefficient of $x^{50}$ of the function $$f(x):=\left({1\over 1-x}\right)^3\left({1-x^{11}\over 1-x}\right)^3=(1-x^{11})^3(1-x)^{-6}\ .$$ Now $$f(x)=(1-3x^{11}+3 x^{22}-x^{33})\sum_{k=0}^\infty{-6\choose k}(-x)^k\ .$$ Collecting terms gives you an alternating sum of four binomial coefficients, reflecting an inclusion/exclusion process: $$N={55\choose 5}-3{44\choose 5}+3{33\choose 5}-{22\choose5}=906\,411\ .$$