Summing combinations with repetition Given $m,n,k\in\mathbb{N}=\{1,2,...\}$, I wonder if it is possible to find a $F:\mathbb{N}^3\to \mathbb{N}$ such that

$$
\binom{m+k-1}{k}+\binom{n+k-1}{k}=\binom{F(m,n,k)+k-1}{k}.
$$

EDIT: A more reasonable version of the problem (inspired by the smart observations below) is to find, for each $k$, a different function, say $F_k(m,n)$, such that $
\binom{m+k-1}{k}+\binom{n+k-1}{k}=\binom{F_k(m,n)+k-1}{k}$. For instance, for $k=1$, we have $F_1(m,n)=m+n$.
Thanks for your help!
See also https://math.stackexchange.com/a/2841171/559615.
 A: Well, if such $F$ exist, then $k=1$ since for $m=n=1$ we have
$$2{k\choose k} = {F(1,1)+k-1\choose k}\implies F(1,1)=2\;\; \wedge\;\; k=1$$
(remember Pascal triangle and only possibility where 2 stands) so $$
\binom{m}{1}+\binom{n}{1}=\binom{F(m,n)}{1},
$$
So $F(m,n) = m+n$.
A: Given for ascertained that it cannot be $F(m,n)$, but the it shall be $F(m,n,k) as in your new edition of the post,
then,using the Falling and Rising Factorial definition of the binomial, your identity can be rewritten as
$$
\eqalign{
  & \left( \matrix{  m + k - 1 \cr   k \cr}  \right) + \left( \matrix{  n + k - 1 \cr   k \cr}  \right)
 = \left( \matrix{  F(m,n) + k - 1 \cr   k \cr}  \right)\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad {{\left( {m + k - 1} \right)^{\,\underline {\,k\,} } } \over {k!}} + {{\left( {n + k - 1} \right)^{\,\underline {\,k\,} } } \over {k!}}
 = {{\left( {F(m,n,k) + k - 1} \right)^{\,\underline {\,k\,} } } \over {k!}}\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad m^{\,\overline {\,k\,} }  + n^{\,\overline {\,k\,} }  = F(m,n,k)^{\,\overline {\,k\,} }  \cr} 
$$
where
$$
x^{\,\overline {\,k\,} }  = \left( {x + k - 1} \right)^{\,\underline {\,k\,} }  = {{\Gamma (x + k)} \over {\Gamma (x)}}
$$
denote respectively the Rising Factorial, the Falling Factorial  and their relation with the Gamma function.
Now, if we want $F$ to be an integer, the last line looks as the analoguous for the Rising Factorial
of the Fermat's Last Theorem identity, and I expect (although I am not in the position to prove it) that it might
subject to the same ban.
Looking more modestly for a real $F$, then we are facing with the inversion of the Gamma function, which does not have
an easy formulation.   
On the other hand $z^{\,\overline {\,k\,} } $ is a polynomial in $z$ of degree $k$
$$
z^{\,\overline {\,k\,} }  = z\left( {z + 1} \right) \cdots \left( {z + k - 1} \right)
  = \sum\limits_{\left( {0\, \le } \right)\,l\,\left( { \le \,k} \right)} {\left[ \matrix{ k \cr   l \cr}  \right]x^{\,l} } 
$$
where the square brackets indicates the Stirling Numbers of 1st kind.
The equation 
$$
z^{\,\overline {\,k\,} }  = a
$$
has always a unique non-negative solution in $z$ for non-negative values of $a$, but to find
it we shall resort to numerical computation, or to the various asymptotics approximations for 
Gamma and Factorials.
A: The short answer is: No, such a function does not exist.
Consider the case of $k=2$. We have $\binom 42=6$. But $\binom n2\neq12$ for all $n$. This means that $F(3,3,2)$ does not exist.
It is easy to construct more similar examples based on the fact that the set of binomial coefficients $S(k)=\{\binom nk\mid n\in\Bbb{N}\}$ is not closed under addition.
May be you wanted to ask some other question?
