Intuitive proof of the formula ${}_nC_r + {}_nC_{r-1} = {}_{n+1}C_r$ I came across this formula in combination— ${}_nC_r + {}_nC_{r-1} = {}_{n+1}C_r$. Even though I know its rigorous mathematical proof, I want a logical and elegant proof of this.
For example, the famous formula of combination ${}_nC_r = {}_nC_{n-r}$ says selecting $r$ objects out of $n$ objects is same as rejecting $(n-r)$ objects.
So, I am looking for such kind of intuitive proof of the formula ${}_nC_r + {}_nC_{r-1} = {}_{n+1}C_r$, which I am unable to get. 
The thought of the wise man who said "writing a correct equation but not being able to interpret its result is the same as writing a grammatically correct sentence without knowing what it means!!!" is not helping me either!!
Note— ${}_nC_r$ means combination of $n$ objects taken $r$ at a time. 
 A: Here is a combinatorial proof of Pascal's Identity
$$\binom{n + 1}{r} = \binom{n}{r} + \binom{n}{r - 1}$$
where $$\binom{n}{r} = \frac{n!}{r!(n - r)!}$$ is the number of ways of making an unordered selection of $r$ elements from a set of $n$ elements.  
We can select $r \geq 1$ elements from the set $$S_{n + 1} = \{x_1, x_2, \ldots, x_n, x_{n + 1}\}$$ in $\binom{n + 1}{r}$ ways.  Such selections either include $x_{n + 1}$ or they do not.  If a selection of $r$ elements includes $x_{n + 1}$, we must select $r - 1$ elements from the subset $$S_n  = \{x_1, x_2, \ldots, x_n\}$$ and $x_{n + 1}$.  We can select $r - 1$ elements from the subset $S_n$ in $\binom{n}{r - 1}$ ways and select $x_{n + 1}$ in one way.  If the selection does not include $x_{n + 1}$, we must select $r$ elements from $S_n$, which can be done in $\binom{n}{r}$ ways.  Hence, the number of ways of selecting $r$ elements from a set of $n + 1$ elements is 
$$\binom{n + 1}{r} = \binom{n}{r - 1}\binom{1}{1} + \binom{n}{r} = \binom{n}{r - 1} + \binom{n}{r}$$
A: Assume the $n+1$ objects are balls, with only one of them being red and the rest white.
A particular selection of $r$ balls has to either contain the red ball or not.
Those selections omitting the red ball are $^{n}C_r$. Those selections containing the red ball have choice in which $r-1$ white balls have to be chosen, and this is given by $^{n}C_{r-1}$.
