There are many known combinatorial identities involving the binomial coefficient, although essentially all of the well known ones involve some function or constant other than the binomial coefficient itself.
For example, Pascal's identity ${n\choose k} = {n-1\choose k-1} + {n-1 \choose k}$ makes use of subtraction, addition, and the constant $1$.
The identities ${n\choose 0} = 1$ and ${n\choose 1} = n$ use the constants $0$ and $1$.
I am interested in identities which use only variables and the binomial coefficient (in the same way that the associative and commutative laws for multiplication use only variables, multiplication and no other functions).
Notice, for example, that we have the identity ${a\choose a}=1$, which allows us to write the last of the above identities as $${n\choose {a\choose a}} = n$$
Furthermore, since $1=1$ we have the identity $${a\choose a} = {b\choose b}$$
These two identities are built out of only variable and binomial coefficients. Do there exist any other such identities, which don't follow logically from these two above?
In the language of universal algebra, is the smallest variety $V$ containing $\langle\mathbb{N}, B\rangle$ (where $B$ is the 2-ary binomial coefficient function) exactly the class of algebraic structures whose signature is a single 2-ary function $B$ satisfying the identities $B(n,B(a,a)) = n$ and $B(a,a) = B(b,b)$?
If not, how can one describe this variety $V$?
(Note: I take the convention that if $a < b$, then ${a\choose b}=0$)
Best Answer
is the smallest variety $V$ containing $\langle\mathbb{N}, B\rangle$ (where $B$ is the 2-ary binomial coefficient function) exactly the class of algebraic structures whose signature is a single 2-ary function $B$ satisfying the identities $B(n,B(a,a)) = n$ and $B(a,a) = B(b,b)$?
No.
Your algebra $\langle\mathbb{N}, B\rangle$ satisfies the identity
$$B(B(x,x),B(B(y,y),z)) = B(w,w)$$
but this identity is not a consequence of the two identities you identified. To see that it is not a consequence, note that the algebra $\mathbb Z_2 = \langle \{0,1\}; B(x,y)\rangle$ with $B(x,y) = x+y\pmod{2}$ satisfies your two identities, but does not satisfy the more complicated identity I wrote.
Let me mention an interesting and less obvious identity satisfied by the variety $V=V(\mathbf{A})$ that is generated by the binomial algebra $\mathbf{A}:=\langle\mathbb{N}, B\rangle$. (I refer to the identity $p(x,x,z)=q(x,z,z)$ below.) Define terms
It is not too hard to verify that the following identities hold in $\mathbf{A}$ (hence in $V$):
$$p(x,z,z)=x, \quad p(x,x,z)=q(x,z,z), \quad q(x,x,z) = z.$$
These are 'Hagemann-Mitschke identities for $3$-permutability'. These identities imply that, for any two congruences $\alpha, \beta$ on any algebra $\mathbf{B}\in V$ it is the case that $\alpha\circ\beta\circ\alpha = \beta\circ\alpha\circ\beta$. In particular, the existence of HM terms for $3$-permutability implies that $V$ is a congruence modular variety. The first and last of these HM identities follow from your two identities, but the middle one, $p(x,x,z)=q(x,z,z)$, does not follow since $\mathbb Z_2 = \langle \{0,1\}; B(x,y)\rangle$ with $B(x,y)=x+y\pmod{2}$ does not satisfy the middle HM identity.
It is possible to strengthen the conclusion that $V(\mathbf{A})$ to congruence modular to the conclusion that $V(\mathbf{A})$ to congruence distributive, as follows. First, let $x*z$ denote $p(x,x,z) = B(B(x,B(x,z)),B(z,x))$. Then let $x\vee z = (x*z)*(z*x)$. The operation $x\vee z$ is a semilattice operation on $\mathbf{A}$, namely it is the join operation for the semilattice order where $0, 2, 3, 4, \ldots$ are pairwise incomparable and the join of any two of them is $1$. Now, any congruence modular variety which has a semilattice term is congruence distributive.