# Binomial coefficient in a binomial coefficient

I am doing some research in combinatorics, and I found that I have to consider the following binomial coefficient :

$$\binom{\binom{i}{j}}{k}$$

(In fact, I have to take the product for fixed $$i,k$$ and odd $$j$$’s, but to make this product, I have to manipulate those coefficients, and I don’t know how.)

Is there a way to write it in terms of other binomial coefficients, power series, or other combinatorial tools?

• We have $\dbinom{\dbinom{n}{j}}{k} = \sum\limits_{i=0}^{jk} a_i \dbinom{n}{i}$, where $a_i$ denotes the number of all $k$-element sets $\left\{S_1, S_2, \ldots, S_k\right\}$ of $j$-element subsets $S_1, S_2, \ldots, S_k$ of $\left\{1,2,\ldots,i\right\}$ satisfying $S_1 \cup S_2 \cup \cdots \cup S_k = \left\{1,2,\ldots,i\right\}$. This is folklore and easy to prove. I don't think the $a_i$ have any explicit algebraic descriptions, though. (Note that I stopped the sum at $i = jk$ simply to make it finite; it is easy to see that $a_i = 0$ for all $i > jk$.) Jan 12 at 0:56

For $$j$$ and $$k$$ (that will remain fixed) in $$\mathbb N$$, and a set $$X$$, denote $$\mathcal F(X):=\mathcal{P}_k\mathcal{P}_j(X)$$, the set of all $$k$$-element sets $$\mathcal U:=\{U_1,\dots,U_k\}$$ of $$j$$-element subsets of $$X$$. Denote $$\mathcal C(X)\subset \mathcal F(X)$$ the set of those elements $$\mathcal U$$ of $$\mathcal F(X)$$ which are coverings of $$X$$, that is $$\displaystyle \bigcup_{U\in \mathcal U}U =X$$. Every $$\mathcal U\in\mathcal F(X)$$ is a covering of exactly one subset $$Y$$ of $$X$$, namely $$Y:=\displaystyle \bigcup_{U\in \mathcal U}U$$. Therefore $$\mathcal F(X)=\bigsqcup_{Y\subset X } \mathcal C(Y)$$ $$\mathcal C(X)=\Big(\bigcup_{x\in X } \mathcal F\big(X\setminus\{x\}\big)\Big)^c,$$ and note also that $$\bigcap_{x\in Y} \mathcal F\big(X\setminus\{x\}\big)= \mathcal F\big(X\setminus Y\big).$$ As to cardinalities, if $$|X|=n$$, we have of course $$\big|\mathcal F(X)\big|=\begin{pmatrix} n\\j \\ k \end{pmatrix} := \bigg( {{n\choose j}\atop k}\bigg)$$, and, if we denote $$\begin{bmatrix} n\\j \\ k \end{bmatrix}:=\big|\mathcal C(X)\big|$$ we have from the above disjoint union $$\begin{pmatrix} n\\j \\ k \end{pmatrix}=\sum_{m=0}^n{n\choose m}\begin{bmatrix} m\\j \\ k \end{bmatrix},$$ which can be inverted giving $$\begin{bmatrix} m\\j \\ k \end{bmatrix}=\sum_{n=0}^m(-1)^{m-n}{m\choose n}\begin{pmatrix} n\\j \\ k \end{pmatrix}.$$ The latter, of course, can be seen as an instance of the inclusion-exclusion formula: $$\big|\mathcal C(X)\big| =\bigg|\Big(\bigcup_{x\in X } \mathcal F\big(X\setminus\{x\}\big)\Big)^c\bigg|=\sum_{Y\subset X}(-1)^{|Y|}\big|\mathcal F\big(X\setminus Y\big)\big|=\sum_{Y\subset X}(-1)^{|X\setminus Y|}\big|\mathcal F(Y)\big|$$