Intuitive proof of the formula ${}_nC_r + {}_nC_{r-1} = {}_{n+1}C_r$

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 $^{n}C_{r-1}$.

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 $$\binom{n + 1}{r} = \binom{n}{r - 1}\binom{1}{1} + \binom{n}{r} = \binom{n}{r - 1} + \binom{n}{r}$$