How i can prove that $\sum_{i=0}^{r+1}\binom{n-i}{r-i+1} = \binom{n}{r}$ I'm trying to prove this binomial coefficient identity. 
$$\sum_{i=0}^{r+1}\binom{n-i}{r-i+1} = \binom{n}{r} $$
The textbook's hint is about using the identity
$$\binom{n}{r} =\binom{n}{n-r}$$
I've only reached this
$$\frac{1}{(n-1-r)!}\left(\frac{(n-2)!}{(r-1)!}+\frac{(n-3)!}{(r-2)!}+\dots+\frac{(n-r-1)!}{0!}\right)$$
 A: You can use the Identity:
$$ \begin{pmatrix}
    n \\
    r
    \end{pmatrix} = 
   \begin{pmatrix}
    n - 1 \\
    r
    \end{pmatrix} + \begin{pmatrix}
    n - 1 \\
    r - 1
    \end{pmatrix}$$
Recursively in the last term {  $C(n - 1, r - 1)$  }. Doing this you will have:
$$ \begin{pmatrix} n \\r \end{pmatrix} = \begin{pmatrix} n - 1 \\r \end{pmatrix} + \begin{pmatrix} n - 1 \\r -1 \end{pmatrix} $$
for the first iteration, then:
$$\begin{pmatrix} n \\r \end{pmatrix} = \begin{pmatrix} n - 1 \\r \end{pmatrix} + \begin{pmatrix} n - 2 \\r -1 \end{pmatrix} + \begin{pmatrix} n - 2 \\r - 2 \end{pmatrix}$$
Recursively this leads to:
$$\begin{pmatrix} n \\r \end{pmatrix} = \begin{pmatrix} n - 1 \\r \end{pmatrix} + \begin{pmatrix} n - 2 \\r -1 \end{pmatrix} + \begin{pmatrix} n - 3 \\r - 2 \end{pmatrix} + \begin{pmatrix} n - 4 \\r - 3 \end{pmatrix} + ... + \begin{pmatrix} n - r + 1 \\r - r \end{pmatrix} $$
For all other coming values the terms vanishes to zero ($ C(n, r) = 0 \hspace{2mm} \forall \hspace{2mm} r > n $)
this can be written is sigma notation as:
$$ \begin{pmatrix} n \\r \end{pmatrix} = \sum_{i=0}^{r} \begin{pmatrix} n - 1 - i \\r - i \end{pmatrix} $$
Changing the index
$$ j = i + 1 $$
We can write the summation as fallows:
$$ \sum_{i=0}^{r} \begin{pmatrix} n - 1 - i \\r - i \end{pmatrix} = 
\sum_{j = 1}^{r + 1} \begin{pmatrix} n - 1 - (j - 1) \\r - (j - 1) \end{pmatrix} 
$$
This is:
$$
\sum_{j = 1}^{r + 1} \begin{pmatrix} n - j \\r - j + 1 \end{pmatrix}
$$
And obviously we can change the dummie index as $ j \longrightarrow i $ .
A: Use the hint to write $\binom{n-i}{r-i+1}=\binom{n-i}{n-r-1}$ and $\binom{n}{r} = \binom{n}{n-r}$, then use the Hockey Stick Identity.
For the proof of the Hockey Stick Identity, I recommend this combinatorial approach if you do not like the algebraic approach in the accepted answer.
