Interpretation of a combinatorial identity involving iterated binomial coefficients I am trying to find an combinatorial interpretation for the following combinatorial identity involving iterated binomial coefficients, which appeared in the November 1980 edition of The American Math Monthly:
$\dbinom{\binom{n}{b}}{2}=\displaystyle\sum_{j=1}^b\dbinom{\binom{b}{j}+e_j}{2}\binom{n+b-j}{2b},$ 
where $e_j=\frac{1+(-1)^{j+1}}{2}$, that is, $e_j=1$ if $j$ is odd, and $0$ otherwise. 
Essentially, the left hand side of the identity can be interpreted as the number of ways to choose 2 $b$-element subsets of $\{1,2,\cdots,n\}$. What I am interested in is the derivation of the right hand side of the identity; I have read the proof in The American Math Monthly, and the author mentioned that the expression $\binom{n+b-j}{2b}$ refers to the selection of $2b$ objects from the original set $\{1,2,\cdots,n\}$, augmented by the adjunction of $b-j$ "jokers", where $0\leq j\leq b-1$, to allow for the fact that the intersection of two distinct $b$-element subsets of $\{1,2,\cdots,n\}$ can have a minimum and a maximum cardinality of $0$ and $b-1$ respectively.
What I am struggling to understand here is the expression $\dbinom{\binom{b}{j}+e_j}{2}$; how does one interpret this expression? Alternatively, is there any other way for which one can prove the identity? Personally, I have proven via a combination of combinatorial and algebraic methods that the identity does hold for small values of $b$, but the expression is not that tractable for large values of $b$.

Here is an  answer using formal power series that  is somewhat simpler
than the first one. We start as before:
$$\frac{1}{2} 
\sum_{j=0}^b \left({b\choose j} + e_j\right)
\left({b\choose j} + e_j-1\right)
{n+b-j\choose 2b}$$
where we have $e_j = [[j\;\text{is odd}]].$
Expanding we get three pieces, the first is
$$A = \frac{1}{2} 
\sum_{j=0}^b {b\choose j}^2
{n+b-j\choose 2b},$$
the second is
$$B = \frac{1}{2} 
\sum_{j=0}^b 
{b\choose j} (2e_j-1)
{n+b-j\choose 2b}
= \frac{1}{2} 
\sum_{j=0}^b 
{b\choose j} (-1)^{j+1}
{n+b-j\choose 2b} 
$$
and the third
$$C = \sum_{j=0}^b 
e_j (e_j-1)
{n+b-j\choose 2b} =0,$$
which leaves  $A$ and $B.$ Starting  with $A$ we observe  that we must
have $n\ge b$ or else the  third binomial coefficient vanishes for all
$j$ in the range, making for a zero contribution. We find
$$\frac{1}{2} 
\sum_{j=0}^b {b\choose j} [z^{b-j}] (1+z)^b
[w^{n-b-j}] (1+w)^{n+b-j}
\\ = \frac{1}{2} [z^b] (1+z)^b [w^{n-b}] (1+w)^{n+b} 
\sum_{j=0}^b {b\choose j} z^j w^j (1+w)^{-j}
\\ = \frac{1}{2} [z^b] (1+z)^b [w^{n-b}] (1+w)^{n+b} 
\left(1+\frac{wz}{1+w}\right)^b
\\ = \frac{1}{2} [z^b] (1+z)^b [w^{n-b}] (1+w)^{n} 
(1+w+wz)^b.$$
Recalling that $n\ge b$ this becomes
$$\frac{1}{2} [z^b] (1+z)^b [w^{n-b}] (1+w)^{n} 
\sum_{q=0}^b {b\choose q} w^q (1+z)^q
\\ = \frac{1}{2} [z^b] (1+z)^b
\sum_{q=0}^b {b\choose q} {n\choose n-b-q} (1+z)^q
\\ = \frac{1}{2}
\sum_{q=0}^b {b\choose q} {n\choose n-b-q} {b+q\choose b}.$$
Factoring the two binomials on the right we get
$$\frac{n!}{(n-b-q)! \times b! \times q!}
= {n\choose b} {n-b\choose q}.$$
We continue with
$$\frac{1}{2} {n\choose b}
\sum_{q=0}^b {b\choose q} {n-b\choose q}
= \frac{1}{2} {n\choose b}
\sum_{q=0}^b {b\choose q} [z^{n-b-q}] (1+z)^{n-b}
\\ = \frac{1}{2} {n\choose b} [z^{n-b}] (1+z)^{n-b}
\sum_{q=0}^b {b\choose q} z^q
= \frac{1}{2} {n\choose b} [z^{n-b}] (1+z)^{n-b} (1+z)^b
\\ = \frac{1}{2} {n\choose b}^2.$$
The piece labeled $B$ is next. We get
$$\frac{1}{2} \sum_{j=0}^b 
{b\choose j} (-1)^{j+1}
[w^{n-b-j}] (1+w)^{n+b-j}
\\ = - \frac{1}{2} [w^{n-b}] (1+w)^{n+b}
 \sum_{j=0}^b 
{b\choose j} (-1)^{j} w^j (1+w)^{-j}
\\ = - \frac{1}{2} [w^{n-b}] (1+w)^{n+b}
\left(1-\frac{w}{1+w}\right)^b
\\ = - \frac{1}{2} [w^{n-b}] (1+w)^{n}
= - \frac{1}{2} {n\choose b}.$$
We have shown that
$$A+B = \frac{1}{2} {n\choose b}^2 - \frac{1}{2} {n\choose b}
= \frac{1}{2} {n\choose b}
\left({n\choose b} - 1\right)$$
as claimed. 
Remark. The piece $B$ was also evaluated in the first answer.
