Problem: (HMMT) Eli, Joy, Paul, and Sam want to form a company; the company will have 16 shares to split among the 4 people. The following constraints are imposed: - Every person must get a positive integer number of shares, and all 16 shares must be given out. - No one person can have more shares than the other three people combined. Assuming that shares are indistinguishable, but people are distinguishable, in how many ways can the shares be given out? Solution: First, give everyone one share so that we want to distribute $12$ shares among $4$ people so that no one gets more shares than the sum of the others. This translates to everyone gets $\leq 7$ shares (otherwise, someone gets $8$ shares plus the $1$ we gave them earlier, meaning they get $9$ total shares, which must be greater than the sum of the others' shares). So, we want to calculate $ |\{(a,b,c,d)\in \mathbb{N}^4: a+b+c+d = 12: 0 \leq a,b,c,d \leq 7\}|. $ By simple logic, we find that $ \begin{align} & |\{(a,b,c,d) \in \mathbb{N}^4: a+b+c+d = 12: 0 \leq a,b,c,d \leq 7\}| \\ = & |\{(a,b,c,d)\in \mathbb{N}^4: a+b+c+d = 12\}| - 4|\{(a,b,c,d)\in \mathbb{N}^4: a+b+c+d = 12, a \ge 8\}| \\ = & |\{(a,b,c,d)\in \mathbb{N}^4: a+b+c+d = 12\}| - 4|\{(a,b,c,d)\in \mathbb{N}^4: a+b+c+d = 4\}| \\ = & \binom{12+4-1}{4-1} - 4\binom{4+4-1}{4-1} = 315. \end{align} $