Explanation about identity $\sum\limits_{R=0}^N\binom{R}r\binom{N-R}{n-r} = \binom{N+1}{n+1}$ in Jaynes' Probability theory chapter 6 Equation from Jaynes Probability Theory chapter 6.
$$\sum_{R=0}^N\binom{R}{r} \times \binom{N-R}{n-r} = \binom{N+1}{n+1}.$$
The following is my proof.
For
$$\sum_{R=0}^N\binom{R}{r} \times \binom{N-R}{n-r}$$
$$\binom{R1}{r} \times \binom{N-R1}{n-r} = \binom{N}{n}$$
Then get
$$(N+1) \times \binom{N}{n}$$
I don't know which step is wrong. Help, thanks very much.
 A: The identity in the book is correct. Let us prove it. 
Start with identity
$$\tag{1}\label{1}
\dfrac{x^r}{(1-x)^{r+1}}\dfrac{x^{n-r}}{(1-x)^{n-r+1}}x = \dfrac{x^{n+1}}{(1-x)^{n+2}}.
$$
We need the Taylor series for all three fractions in (\ref{1}). After we intend to compare some coefficients in both sides. Start with Taylor series (for $|x|<1$)
$$
\frac{1}{1-x}=\sum_{i=0}^\infty x^i
$$
and find the $r$th derivative: 
$$
\frac{d^r}{dx^x}\left[\frac{1}{1-x}\right]=\frac{r!}{(1-x)^{r+1}}=\frac{d^r}{dx^x}\left[\sum_{i=0}^\infty x^i\right]=\sum_{i=r}^\infty \frac{i!}{(i-r)!} x^{i-r}
$$
Multiply by $x^r$ and divide both parts by $r!$ to obtain the following identity:
$$\tag{2}\label{2}
\frac{x^r}{(1-x)^{r+1}}=\sum_{i=r}^\infty {i\choose r}x^i=\sum_{i=0}^\infty {i\choose r}x^i,
$$
where ${i\choose r}=0$ for $i<r$.
With is Taylor series for first term in l.h.s. of (\ref{1}). For the second term we have
$$\tag{3}\label{3}
\frac{x^{n-r}}{(1-x)^{n-r+1}}=\sum_{j=0}^\infty {j\choose n-r}x^j,
$$
and for r.h.s. 
$$\tag{4}\label{4}
\frac{x^{n+1}}{(1-x)^{n+2}}=\sum_{k=0}^\infty {k\choose {n+1}}x^k,
$$
Replace all fractions in (\ref{1}) by Taylor series:
$$
\text{l.h.s.}=x \sum_{i=0}^\infty {i\choose r}x^i \sum_{j=0}^\infty {j\choose n-r}x^j = \sum_{i=0}^\infty \sum_{j=0}^\infty {i\choose r}{j\choose n-r}x^{i+j+1},
$$
$$\text{r.h.s.}=\sum_{k=0}^\infty {k\choose {n+1}}x^k.
$$
Take $k=N+1>n+1$ in the last sum and compare coefficients of $x^{N+1}$ in l.h.s. and in r.h.s. In r.h.s. the coefficient of $x^{N+1}$ is equal to
$$
{N+1\choose {n+1}}
$$
In l.h.s. we should sum all the terms with indices $i+j+1=N+1$ or $i+j=N$. The inner sum in l.h.s. disappears since $j=N-i$. The coefficient of $x^{N+1}$ in l.h.s. equals to
$$
\sum_{i=0}^N {i\choose r}{N-i\choose n-r} 
$$
and we get an identity
$$
\sum_{i=0}^N {i\choose r}{N-i\choose n-r} ={N+1\choose {n+1}}
$$
Replace $i$ by $R$ and get the initial correct identity from the book:
$$
\sum_{R=0}^N {R\choose r}{N-R\choose n-r} ={N+1\choose {n+1}}.
$$
Where, as mentioned above, all binomial coefficients ${k\choose s}$ are zero if $k<s$.
A: Consider the following. We have $N+1$ balls and we have to choose $n+1$ balls from them. Consider these balls are in an array indexed from $0\cdots N$. Total number of ways of choosing is 
$\mathbb {N+1\choose n+1}$. This is right hand side of equation to be proven.
Now do this thought experiment. At every time we will choose $ith$ ball, and choose $r$ balls from $0\cdots i-1$ and $n-r$ balls from $i+1\cdots N$ balls.  Cardinality of $0\cdots i-1$ which is $R$ lies in $[0,N]$ and so is the case for $i+1\cdots N$ (which will equal $N-R$).
This exhausts all possible conditions, and total number of such cases will be equal to how we choose $n+1$ balls from $N+1$ balls. Total number of such combinations is 
$$\sum_{R=0}^{N} {R \choose r }{N-R \choose n-r} $$ which is the left side.
Hence, 
$$\sum_{R=0}^{N} {R \choose r }{N-R \choose n-r} = {N+1\choose n+1}$$.
