<s>[INST] Fermat's Combinatorial Identity: How to prove combinatorially? $$\binom{r}{r} + \binom{r+1}{r} + \binom{r+2}{r} + \dotsb + \binom{n}{r} = \binom{n+1}{r+1}$$
I don't have much experience with combinatorial proofs, so I'm grateful for all the hints.
(Presumptive) Source: Theoretical Exercise 1.11, P18, A First Course in Pr, 8th Ed, by S Ross
 [/INST] This is analogous to https://math.stackexchange.com/a/357087/53259. The RHS here  imports the number of ways of picking $r+1$ numbers out of $\{1,2,...,\color{magenta}{r}, \color{green}{r +1},...,\underbrace{n}_{= \color{magenta}{r} + (n - r)},\color{green}{n +1}\} \qquad (*)$
Now for any given choice of $r+1$ numbers, call the highest number chosen $k$ and esteem it as the $(r + 1)$th number. Observe :
The "leftmost" subset of $(*)$ containing $r + 1$ elements $ = \{1, 2, ..., r, \color{green}{r +1}\},$
Also, the "rightmost" subset of $(*)$ containing $r + 1$ elements $ = \{\color{green}{n +1} - (r + 1), ..., n , \color{green}{n +1}\}.$
Thus, $\color{green}{r +1} \leq k \leq \color{green}{n +1}$.   
For each $k \in [\color{green}{r +1}, \color{green}{n +1}]$.  , we must select the remaining $r$ numbers to be chosen
 from the $k-1$ numbers smaller than $k$.
For $k = r +1$, must pick $r$ numbers to the left of $r + 1$, out of $\{\color{  #0073CF}{1, 2, ..., r}, r+1\}$.
Since there are $ \color{#0073CF}{r}$ such numbers, so $\color{#0073CF}{r}$ possible choices for $r$.
Thus the total number of choices for $r$ numbers $= \binom{\color{#0073CF}{r}}{r}$. 
For $k = r +2$, must pick $r$ numbers to the left of $r + 2$, out of $\{\color{  #0073CF}{1, 2, ..., r, r +1}, r+2\}$.
Since there are $ \color{#0073CF}{r + 1}$ such numbers, so $\color{#0073CF}{r + 1}$ possible choices for $r$.
Thus the total number of choices for $r$ numbers $= \binom{\color{#0073CF}{r + 1}}{r}$. 
...   
For $k = n$, must pick $r$ numbers to the left of $n$, out of $\{\color{  #0073CF}{1, 2, ..., r, r + 1, ..., n - 1},  n\}$.
Since there are $ \color{#0073CF}{n - 1}$ such numbers, so $\color{#0073CF}{n - 1}$ possible choices for $r$.
Thus the total number of choices for $m$ numbers $= \binom{\color{#0073CF}{n - 1}}{r}$. 
For $k = n + 1$, must pick $r$ numbers to the left of $n + 1$, out of $\{\color{  #0073CF}{1, 2, ..., r, r + 1, ..., n},  n + 1\}$.
Since there are $ \color{#0073CF}{n}$ such numbers, so $\color{#0073CF}{n}$ possible choices for $r$.
Thus the total number of choices for $m$ numbers $= \binom{\color{#0073CF}{n}}{r}$. 
Summing up the number of ways of doing this for $k=r+1,...,r+n+1$ yields the LHS.

Remark : (Presumptive) Source: Theoretical Exercise 1.11, P18, A First Course in Pr, 8th Ed, by S Ross, via: $\color{blue}{\mathsf{i \text{ there }}}$ $ = i + 1$ here, $k$ there $= r + 1$ here, $\color{blue}{\mathsf{n \text{ there }}}$ $ = n + 1$ here.   
$$\begin{align} \binom{r}{r} + \binom{r+1}{r} + \binom{r+2}{r} + \dotsb + \binom{n}{r} &= \binom{n+1}{r+1} \\
\sum_{\Large{r \le i \le n \text{ or } i \in [r,n]}}  \binom{i}{r}, n \ge r & =\end{align}$$
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