generalization of multisets 
Prove that for any $c,d \in \mathbb{R}$ and $k\in\mathbb{N}, \left({c+d\choose k}\right) = \sum_{j=0}^k \left({c\choose j}\right) \left({d\choose k-j}\right).$

I know how to show that ${a+b\choose k} = \sum_{j=0}^k {a\choose j}{b\choose k-j}$ for $a,b\in \mathbb{R}$ using an algebraic proof, but I'm not sure how to show the multiset version of this. I know that $\left({n\choose k}\right) = {n+k-1\choose k}$. But if we insisted that $c,d\in\mathbb{N},$ I think I might be able to come up with a combinatorial proof. Let $S$ denote the set of $j$-multisets (i.e. of size $j$) of $[1,\cdots, c+d]$. Let $C_j$ denote the set of multisets of size $j$ from $[1,\cdots, c]$ and $D_{k-j}$ denote the set of multisets of size $k-j$ from $[c+1,\cdots, c+d]$. Let $E_j$ denote the set of $k$-multisets from $[1,\cdots, c+d]$ with $j$ elements from $[1,\cdots, c].$ Observe that each $E_j$ is disjoint, and $S = \cup_{j=0}^k E_j\Rightarrow |S| = \sum_{j=0}^k |E_j|\tag{1}.$ Also, it is not difficult to define a bijection $f : E_j \to C_j \times D_{k-j}.$ Since $|C_j| = \left({c\choose j}\right)$ and $|D_{k-j}| = \left({d\choose k-j}\right)$ and $ |E_j| = |C_j||D_{k-j}|$, substituting these results into $(1)$ gives the desired equality. But of course, this only works for $c,d\in \mathbb{N}.$
 A: For fixed $k\in\Bbb N$ the expression
$$p(c,d)=\left(\!\!\binom{c+d}k\!\!\right)-\sum_{j=0}^k\left(\!\!\binom{c}j\!\!\right)\left(\!\!\binom{d}{k-j}\!\!\right)$$
is a polynomial in $c$ and $d$. If we fix $c\in\Bbb N$, it becomes a polynomial in $d$. Either this polynomial is identically $0$, or it has only finitely many zeroes. Since it is $0$ for each $d\in\Bbb N$, it must be identically $0$. Thus, $p(n,d)=0$ for each $n\in\Bbb N$ and $d\in\Bbb R$. But we can now hold $d$ fixed and view $p(c,d)$ as a polynomial in $c$, and by the same argument that polynomial must be identically $0$. Thus, $p(c,d)=0$ for all $c,d\in\Bbb R$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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$\ds{\bbox[5px,#ffd]{\mbox{Prove that for any}\ c,d \in \mathbb{R}\ \mbox{and}\ k\in\mathbb{N}, \left(\!{c + d \choose k}\!\right) =
\sum_{j = 0}^{k}\left(\!{c\choose j}\!\right)
\left(\!{d\choose k - j}\!\right)}:\ {\Large ?}}$.

\begin{align}
&\bbox[#ffd,5px]{\sum_{j = 0}^{k}\left(\!{c\choose j}\!\right)
\left(\!{d\choose k-j}\!\right)} =
\sum_{j = 0}^{k}{c^{\,\large\overline{j}} \over j!}
{d^{\,\overline{k - j}} \over \pars{k - j}!}
\\[5mm] = &\
{\pars{c + d + k - 1}! \over \pars{c - 1}!\pars{d - 1}!}\,{1 \over k!}
\sum_{j = 0}^{k}{k! \over j!\pars{k - j}!}
{\Gamma\pars{c + j}\Gamma\pars{d + k - j} \over \Gamma\pars{c + d + k}}
\\[5mm] = &\
{\pars{c + d + k - 1}! \over \pars{c - 1}!\pars{d - 1}!}\,{1 \over k!}
\sum_{j = 0}^{k}{k \choose j}\int_{0}^{1}t^{c + j - 1}
\pars{1 - t}^{d + k - j - 1}\,\dd t
\\[5mm] = &\
{\pars{c + d + k - 1}! \over \pars{c - 1}!\pars{d - 1}!}\,{1 \over k!}\int_{0}^{1}t^{c - 1}\pars{1 - t}^{d + k - 1}
\sum_{j = 0}^{k}{k \choose j}\pars{t \over 1 - t}^{j}\,\dd t
\\[5mm] = &\
{\pars{c + d + k - 1}! \over \pars{c - 1}!\pars{d - 1}!}\,{1 \over k!}
\int_{0}^{1}t^{c - 1}\pars{1 - t}^{d + k - 1}\,
\pars{1 + {t \over 1 - t}}^{k}\,\dd t
\\[5mm] = &\
{\pars{c + d + k - 1}! \over \pars{c - 1}!\pars{d - 1}!}\,{1 \over k!}
\int_{0}^{1}t^{c - 1}\pars{1 - t}^{d - 1}\,\dd t
\\[5mm] = &\
\require{cancel}
{\pars{c + d + k - 1}! \over \cancel{\pars{c - 1}!\pars{d - 1}!}}
\,{1 \over k!}
\bracks{\cancel{\pars{c - 1}!\pars{d - 1}!} \over \pars{c + d - 1}!} =
{c + d + k - 1 \choose k}
\\[5mm] = &\
\bbx{\large\left(\!{c + d \choose k}\!\right)} \\ &
\end{align}
