How to prove this combinatorial identity without using committee argument but direct manipulation? $C(n,r)=C(r,r)\cdot C(n-r,0) + C(r,r-1)\cdot C(n-r,1)+\ldots+ C(r,1)\cdot C(n-r, r-1)+C(r,0)\cdot C(n-r,r)$
Using the committee argument, it is pretty straight forward:
We are dividing $n$ people into two groups $A$ and $B$ each of cardinality $r$ and $n-r$ and then selecting $r$ from these $n$ people in following way:
For $0\leq i \leq r$, We select $i$ from group $A$ in $C(r,i)$ ways  and the remaining $r-i$ from $B$ in $C(n-r,r-i)$ ways. For fixed $i$, we henceforth select $r$ people from the lot in $C(r,i)\cdot C(n-r,r-i$ ways so that $i$ come from $A$ and $r-i$ come from B(using rule of product, as both tasks are compulsory).
Since the task of selecting $r$ people from the lot so that $i$ come from $A$ and $n-i$ come from $B$ is disjoint for each $i$, the rule of sum gives the above identity.
But how do I prove this identity directly using manipulation.
 A: This is a particular case of the Chu--Vandermonde identity.
The Chu--Vandermonde identity (also known as Vandermonde convolution) states that
\begin{align}
\dbinom{x+y}{n} = \sum\limits_{k=0}^n \dbinom{x}{k} \dbinom{y}{n-k}
\label{darij1.eq.chu}
\tag{1}
\end{align}
for any numbers $x$ and $y$ (not necessarily integers) and any nonnegative integer $n$. There are algebraic proofs around for various tastes. See, e.g., §3.3.2 (first proof of Theorem 3.29) in my Notes on the combinatorial fundamentals of algebra, Version of 10 January 2019. This proof (an induction on $n$ using the absorption formula $\dbinom{y}{n} = \dfrac{y}{n} \dbinom{y-1}{n-1}$) has the advantage that it works uniformly for all $x$ and $y$. If you content yourself with a proof that requires $x$ to be a nonnegative integer, then there are simpler ones around, such as the First proof of Theorem 1.3.37 in my Enumerative Combinatorics: class notes (an induction on $x$ essentially, although I frame it as an induction on an auxiliary variable).
To recover your formula, apply \eqref{darij1.eq.chu} to $r$, $n-r$ and $r$ instead of $x$, $y$ and $n$. You obtain
\begin{align}
\dbinom{r+\left(n-r\right)}{r} &= \sum\limits_{k=0}^r \dbinom{r}{k} \dbinom{n-r}{r-k} \\
&= \dbinom{r}{r} \dbinom{n-r}{0} + \dbinom{r}{r-1} \dbinom{n-r}{1} + \cdots + \dbinom{r}{1} \dbinom{n-r}{r-1} + \dbinom{r}{0} \dbinom{n-r}{r}
\end{align}
(here, we have written out the sum from last to first addend). In view of $r + \left(n-r\right) = n$, this rewrites as
\begin{align}
\dbinom{n}{r} &= \dbinom{r}{r} \dbinom{n-r}{0} + \dbinom{r}{r-1} \dbinom{n-r}{1} + \cdots + \dbinom{r}{1} \dbinom{n-r}{r-1} + \dbinom{r}{0} \dbinom{n-r}{r} .
\end{align}
This is precisely your formula.
