0
$\begingroup$

Is there any way in which the sets $$\{\{b_1,\dots,b_n\}\}\quad(k=0)$$ $$\{\{a_1,b_2,\dots,b_n\},\{b_1,a_2,\dots,b_n\},\dots\{b_1,\dots,b_{n-1},a_n\}\}\quad(k=1)$$ $$\{\{a_1,a_2,b_3,\dots,b_n\},\{a_1,b_2,a_3,\dots,b_n\},\dots,\{b_1,\dots,a_{n-1},a_n\}\}\quad(k=2)$$ can be formed using some briefer notation? In each case we need a set which contains the subsets of size $n$ from $A\cup B$ where $A=\{a_1,a_2,\dots,a_n\}$ and $B=\{b_1,b_2,\dots,b_n\}$ that contain an element with every index in $\{1,\dots,n\}$; $k$ elements from $A$ and $n-k$ elements from $B$.

$\endgroup$
0

2 Answers 2

3
$\begingroup$

Here is one idea. Let $S=\{1,2,\dots,n\}$. Consider the set of all "choice" functions $\phi:S\to\{0,1\}$.

Then consider the map $f_{\phi}:S\to A\cup B$ given by

$$ f_{\phi}(i) = \begin{cases} a_i & \text{if } \phi(i)=0 \\ b_i & \text{if } \phi(i)=1 \end{cases} $$

Finally you can denote the set generated by a given choice function by $f_\phi(S)=\{f_\phi(i):i\in S\}$.

The $k$-th set in your list is then given by $\{f_\phi(S):|\phi^{-1}(0)|=k\}$.

This might be overblowing it. You can drop some of the notation.

$\endgroup$
2
$\begingroup$

I'd write $\cup_{k=0}^n \{X_k \in \mathcal{P}(A\cup B) | \hspace{.1cm} card(X_k) = n \wedge \forall i\in \{1,...,n\}\hspace{.1cm} a_i \in X_k \vee b_i \in X_k \wedge card(X_k \cap A) = k \}$, but i doubt that this is what youre looking for.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .