# Cardinalities of finite sets

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Consider a set $A$ and its subsets $B$ and $C$. It is known that $|A-(B\cap C)|=8$, $|B|=5$, $|C-B|=1$ and $|B\cap C|=3$ (here $-$ denotes set difference). How many subsets $X\subseteq A$ are there if $X\cap B\cap C\ne\emptyset$, $|X-(B\cup C)|\ge3$ and $|X\cap (B-C)|=2$?

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Since nobody gave an answer, I am posting a solution.

I will omit $\cap$ and write intersection as multiplication. I will also write $\sqcup$ for disjoint union.

In this problem $A$ is the universal set, and each element of $A$ belongs to exactly one of four classes: it can fall in or outside $B$ and $C$. Since $A-BC=\overline{BC}=\overline{B}\cup\overline{C}=\overline{B}C\sqcup B\overline{C}\sqcup\overline{B}\overline{C}$, $B=BC\sqcup B\overline{C}$ and $C-B=\overline{B}C$, we have the following system of equations.
\begin{align*}
|\overline{B}C|+|B\overline{C}|+|\overline{B}\overline{C}|&=8\\
|BC|+|B\overline{C}|&=5\\
|\bar{B}C|&=1\\
|BC|&=3
\end{align*}
Solving this system, we can fill the following table form of the Venn diagram.
\begin{tikzpicture}[scale=1.3,y={(0cm,-1cm)}]
\draw[step=1cm] (-.3,-.3) grid (2,2);
\node[above] at (.5,0) {$B$};
\node[above] at (1.5,0) {$\overline{B}$};
\node
at (0,.5) {$C$};
\node
at (0,1.5) {$\overline{C}$};
\path[shift={(.5,.5)}] (0,0) node {3} (1,0) node {1} (0,1) node {2} (1,1) node {5};
\end{tikzpicture}

Set $$X$$ is also split into four disjoint classes. The condition $XBC\ne\emptyset$ gives us $2^3-1=7$ variants for $XBC$. The condition $|X-(B\cup C)|\ge3$ means that $X\overline{B}\overline{C}$ can be chosen in $\binom{5}{3}+\binom{5}{4}+\binom{5}{5}=16$ ways. The condition $|X(B-C)|=|XB\overline{C}|=2$ requires that both elements of $B\overline{C}$ are included in $X$. Finally, the problem does not say anything about $X\overline{B}C$, which gives us two variants: the only element of $\overline{B}C$ may or may not be in $X$. Altogether, there are $7\cdot16\cdot2=224$ variants for $X$.​

I would appreciate your opinion on the difficulty level of this problem. Which year in the university is it appropriate for?​

#### Country Boy

##### Well-known member
MHB Math Helper
That looks like a fairly standard algebra or pre-calculus problem. I wouldn't be surprised to see it in a college freshman or even secondary school class.