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
$$\binom {z+k}{x+y} - \binom {z}{x} \le k \space and \space 2x+y \le z.$$ 
 A: 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.
A: I would appreciate if someone could poke hole in my argument.  It is my humble try
Let $x+y = n$
Then through vandermonde's identity
$$ \sum_{l=0}^{n} {z\choose l}{k\choose(n-l)} - {z\choose{z-n}} \le k$$
If you expand this
$${z\choose0}{k\choose n} + {z\choose1}{k\choose {n-1}} +\cdots {z\choose n}{k\choose 0} - {z\choose n} $$
$${z\choose0}{k\choose n} + {z\choose1}{k\choose {n-1}} +\cdots {z\choose {n-1}}{k\choose 1} $$
But this expression is $\ge k$
For this to be $\le k$ only the last term should be considered which is 
$${z\choose {n-1}}{k\choose 1}  \le k =>n-1 = 0 and n = 1$$
$ x+y = 1$
Coming back to the original inequality
$${{z+k}\choose {x+y}} - {z\choose x}\le k\tag 1$$
If x = 0 and y = 1, inequality 1 reduces to 
$ z+k-1\le k$ implies z = 1
if x = 1 and y = 0, inequality 1 reduces to 
$z+k-z \le k$ along with $2x+y\le z$ implies $z \ge2$
Thus the solution is (0,1,1), (1,0,n+2) and ofcourse the trivial solution (0,0,n) for which the inequality is $0 \le k$ 
