Let $\mathcal{C}$ be a family of subsets of $A=\{1,2,\dots,10\}$ satisfying the following two conditions:
- Every $9$ element subset of $A$ is in $\mathcal{C}.$
- For any non empty subset $C\in\mathcal{C}$ there is $c\in C$ such that $C\setminus\{c\}\in \mathcal{C}.$
What is the least possible value of $|\mathcal{C}|$?
I can find a family $\mathcal{C}$ with $10+5+5+3+3+2+2+2+2+1=35$ members:
Clearly there are $10$ 9-element subsets, now we delete some elements in a sucsession and get:
8-element subsets: $\{1,2,3,4,5,6,7,\color{red}8\}$, $\{1,2,3,4,5,6,9,\color{red}{10}\}$, $\{1,2,3,\color{blue}4,7,8,9,10\}$, $\{1,2,5,\color{blue}6,7,8,9,10\}$, $\{3,4,5,6,7,8,9,10\}$.
7-element subsets: $\{1,2,3,4,5,6,\color{red}{7}\}$, $\{1,2,3,4,5,6,\color{red}9\}$, $\{1,2,\color{blue}3,7,8,9,10\}$, $\{1,2,\color{blue}5,7,8,9,10\}$, $\{3,4,5,6,7,8,\color{green}9\}$.
6-element subsets: $\{\color{red}1,2,3,4,5,6\}$, $\{1,2,7,8,9,\color{blue}{10}\}$, $\{3,4,5,6,7,\color{red}8\}$.
5-element subsets: $\{\color{red}2,3,4,5,6\}$, $\{1,2,7,8,\color{blue}9\}$, $\{3,4,5,6,\color{red}7\}$.
4-element subsets: $\{3,4,5,6\}$, $\{1,2,7,8\}$
3-element subsets: $\{3,4,5\}$, $\{1,2,7\}$
2-element subsets: $\{3,4\}$, $\{1,2\}$
1-element subsets: $\{3\}$, $\{1\}$ and empty set $\{\}$
But, can it be done better?