<s>[INST] Combinatorics problem involving binomial coefficient I found this interesting problem in a Romanian mathematical magazine while preparing for the USAMO. Let $k$ be a non-zero natural number. Determine $x,y,z \in \Bbb N$ such that
$$\binom {z+k}{x+y} - \binom {z}{x} \le k \space and \space 2x+y \le z.$$ 
 [/INST] I suspect you are looking for less of a brute force argument than those above.
Firstly, note that $(x,y,z)=(0,0,n)$ is a solution for $n \geq 0$ (for any appropriate convention for $0$ choose $0$), since the LHS of the first inequality is just $0$. Therefore, assume $x+y \geq 1$.
A useful identity is obtained as follows: consider partitioning a set of size $z+k$ into two subsets $A$ and $B$ of sizes $z$ and $k$ respectively. To choose $x+y$ objects from the $z+k$, we can independently choose $n$ objects from $A$ and $m$ objects from $B$ such that $n+m=x+y$, then sum over the possible values of $n$:
$$\binom{z+k}{x+y}=\sum_{j=0}^N\binom{k}{j}\binom{z}{x+y-j},$$
where $N = \min\{x+y, k\}\geq 1$. Note that the second given inequality gives $z\geq x+y$ so these are the correct limits. Applying this inequality further, we obtain:
$$\binom{z}{x}\leq \binom{z}{x+y}; \text{and}$$
$$\binom{z}{r}\geq \binom{2x+y}{r}.$$
Consequently,
$$\binom{z+k}{x+y}-\binom{z}{x}\geq \sum_{j=1}^N\binom{k}{j}\binom{2x+y}{x+j}\geq k.$$
Therefore for the first given inequality to hold, we must have $N=1$ and
$$\binom{2x+y}{x+1}=1;$$
that is, $x+1=2x+y$ and so $x+y=1$, which automatically ensures $N=1$ for every $k$.
It is then straightforward to check the two cases:
$x=1$, $y=0 \implies$
$$\binom{z+k}{x+y}-\binom{z}{x}=\binom{z+k}{1}-\binom{z}{1}=k.$$
$x=0$, $y=1 \implies$
$$\binom{z+k}{x+y}-\binom{z}{x}=\binom{z+k}{1}-\binom{z}{0}=z+k-1.$$
In summary the solutions are:
$$\color{red}{(0,0,n); (1,0,n+2) \text{ and } (0,1,1) \text{ for } n\geq 0}$$
as seen in the other answers.
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