Looking for a proof of a weird Combinatoric identity I came across Earlier today I came up with a very complicated "proof" of the probability multiplication rule for two independent events. I used the quotation marks because I ended up obtaining a very complicated formula for the wanted probability, and only verified that it gave the right values for different parameters by using desmos. But, I don't know how to simplify the formula into the more basic form. In short, I want to show that
$$
\sum_{i=1}^R \frac{i}{N}\frac{\frac{N!}{i! (R-i)! (B-i)!(N+i-R-B)!}}{\binom{N}{R} \binom{N}{B}}= \frac{RB}{N^2}
$$
Where $R\le B$, And $R+B\le N$
does anyone have any idea of how to do this?
 A: LHS can be simplified to: $$\sum_{i=1}^R\frac{i}N\frac{\binom{R}i\binom{N-R}{B-i}}{\binom{N}B}$$ 
Then to be explained is the equality:$$\sum_{i=1}^R\frac{i\binom{R}i\binom{N-R}{B-i}}{\binom{N}B}=\frac{RB}N$$ 
The LHS of this equality can be recognized as the expectation of the number of red balls that appear if $B$ balls are taken randomly and without replacement from an urn that contains exactly  $N$ balls of which exactly $R$  are red balls.
Note that the LHS is $\sum_{i=1}^Bip_i$ where $p_i$ denotes the probability that $i$ red balls are drawn.
This expectation can be found on a more elegant way using linearity of expectation and symmetry. 
For $i=1,\dots,B$ let $X_i$ take value $1$ if at the $i$-th draw a red ball is chosen and let $X_i$ take value $0$ otherwise. 
Then to be found is: $$\mathbb E(X_1+\dots+X_B)=B\times\mathbb EX_1=B\times\frac{R}{N}=\frac{RB}{N}$$
So this tells us that the equality is a true statement.

Maybe this does not answer your question and you are after some unraveling of the LHS that ends up in the RHS. In that case I feel a too strong reluctance to think about any answer. 
A: $$\begin {array}{}
\sum\limits_{i=1}^R \frac{i}{N}\frac{\frac{N!}{i! (R-i)! (B-i)!(N+i-R-B)!}}{\binom{N}{R} \binom{N}{B}}
&=\frac{1}{N\binom{N}{B}}\sum\limits_{i=1}^R i\binom{R}i\binom{N-R}{B-i}\\
&=\frac{R}{N\binom{N}{B}}\sum\limits_{i=1}^R\binom{R-1}{i-1}\binom{N-R}{B-i}\\
&=\frac{R}{N\binom{N}{B}}\sum\limits_{i=0}^{R-1}\binom{R-1}{i}\binom{N-R}{B-1-i}\\
&\stackrel{V.I.}{=}\frac{R}{N\binom{N}{B}}\binom{N-1}{B-1}=\frac{RB}{N^2},
\end {array}
$$
where $V.I.$ means Vandermonde's identity.
