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- Jan 30, 2012

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- Thread starter Evgeny.Makarov
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- Thread starter
- #1

- Jan 30, 2012

- 2,489

- Thread starter
- #2

- Jan 30, 2012

- 2,489

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

\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 $X$.

\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?

- Jan 30, 2018

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