Give the combinatorial proof of the identity $\sum\limits_{i=0}^{n} \binom{k-1+i}{k-1} = \binom{n+k}{k}$ Given the identity
$$\sum_{i=0}^{n} \binom{k-1+i}{k-1} = \binom{n+k}{k}$$
Need to give a combinatorial proof 
a) in terms of subsets
b) by interpreting the parts in terms of compositions of integers
I should not use induction or any other ways...
Please help.

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\begin{align}
\color{#f00}{\sum_{j = 0}^{n}{k - 1 + j \choose k - 1}} & =
\sum_{j = 0}^{n}{k - 1 + j \choose j} =
\sum_{j = 0}^{n}{-k + 1 - j + j - 1 \choose j}\pars{-1}^{j} =
\sum_{j = 0}^{n}{-k \choose j}\pars{-1}^{j}
\\[3mm] & =
\sum_{j = -\infty}^{n}{-k \choose j}\pars{-1}^{j} =
\sum_{j = -n}^{\infty}{-k \choose -j}\pars{-1}^{-j} =
\sum_{j = 0}^{\infty}{-k \choose n - j}\pars{-1}^{j + n} =
\\[3mm] & =
\pars{-1}^{n}\sum_{j = 0}^{\infty}\pars{-1}^{j}\oint_{\verts{z} = 1^{-}}
{\pars{1 + z}^{-k} \over z^{n - j + 1}}\,{\dd z \over 2\pi\ic}
\\[3mm] & =
\pars{-1}^{n}\oint_{\verts{z} = 1^{-}}
{\pars{1 + z}^{-k} \over z^{n + 1}}\sum_{j = 0}^{\infty}\pars{-z}^{j}
\,{\dd z \over 2\pi\ic} =
\pars{-1}^{n}\oint_{\verts{z} = 1^{-}}
{\pars{1 + z}^{-k - 1} \over z^{n + 1}}\,{\dd z \over 2\pi\ic}
\\[3mm] & =
\pars{-1}^{n}{-k - 1 \choose n} =
\pars{-1}^{n}{k + 1 + n - 1 \choose n}\pars{-1}^{n} =
{n + k \choose n} = \color{#f00}{n + k \choose k}
\end{align}
