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
 A: Think about it this way:
The RHS counts the number of $(r+1)$-element subsets of $[n+1]$; while the LHS counts the same, though seperated into different cases:  First of all there's ${r\choose r}$ subsets of $[n+1]$ that have $r+1$ elements with largest element $r+1$; then, there're ${r+1\choose r}$ subsets of $[n+1]$ that have $r+1$ elements with largest elements $r+2$; etc.
Therefore ...
A: 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.   
$$\begin{align} \binom{r}{r} + \binom{r+1}{r} + \binom{r+2}{r} + \dotsb + \binom{n}{r} &= \binom{n+1}{r+1} \\
\sum_{\Large{r \le i \le n \text{ or } i \in [r,n]}}  \binom{i}{r}, n \ge r & =\end{align}$$
A: Freestandingly, I dilate on why the OP is a duplicate of Combinatorics-number of permutations of $m$ A's and at most $n$ B's. Incidentally, please observe that all colours in this post are independent of any previous usage.
The OP here has the Fermat Combinatorial Identity: $ \sum_{\Large{r \le i \le n}}  \dbinom{\color{blue}{i}}{\color{#FF4F00}{r}} = \dbinom{\color{brown}{n} + 1}{\color{green}{r + 1}}$
I'll denote with prime the variables in the Hockey Stick Identity that also occurs in the Fermat.
TMM has the Hockey Stick Identity : $\sum_{\Large{0 \le i' \le n'}} \dbinom{\color{blue}{m+i'}}{\color{#FF4F00}{i'}} = {\dbinom{\color{brown}{m + n'} + 1}{\color{green}{n'}}}.$
As already coloured, the changes of variable are :
$(1) \quad \color{blue}{i = m+i'} $
$(2) \quad \color{#FF4F00}{r = i'} $
$(3) \quad \color{brown}{n = m + n'} $
$(4) \quad \color{green}{r + 1 = n'} $
Verify the ranges of summation match:
$r \le i \le n \iff \color{#FF4F00}{ i'} \le \color{blue}{m+i'} \le \color{brown}{m + n'} \iff \color{red}{i' - m} \le i' \le n'. $      
But the $\color{red}{i' - m}$ is supposed to be $\color{red}{0}$.
Could someone please let me know of the discrepancy and error?
A: The left side is the most difficult to see. An easy way to see the equality is the following. Get arbitrarily r+1 elements from your n+1 set. This is 1 possible choice(that is, ${r \choose r}$ = ${r+1 \choose r+1 }$ = 1). Keep this elements. Now pick just one more (anyone) from your n+1-(r+1) that were left behind. Fix this new element in the $(r+1)^{th}$ position and then choose the others r from the initial r+1 that you had chosen in the first step. Repeating this process you will have all possible choices of r+1 elements from a set with n+1 elements, which is the right side of the equation. Try to do this with S={1,2,3,4,5} to see why ${5 \choose 3} = {2\choose 2} + {3\choose 2}+{4\choose 2}$.
