MHB How many subsets of set A satisfy given conditions?

  • Thread starter Thread starter Evgeny.Makarov
  • Start date Start date
  • Tags Tags
    Finite Sets
AI Thread Summary
The discussion revolves around a combinatorial problem involving set theory, specifically focusing on the subsets of set A under certain conditions. Key conditions include the sizes of sets B and C, as well as their intersections and differences. The problem asks for the number of subsets X that meet specific criteria related to intersections with B and C and the size of the set difference from B and C. The difficulty of the problem is debated, with opinions suggesting it is suitable for college freshmen or even secondary school students. Overall, the problem is considered a standard exercise in algebra or pre-calculus.
Evgeny.Makarov
Gold Member
MHB
Messages
2,434
Reaction score
4
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 $|X\cap (B-C)|=2$?
 
Mathematics news on Phys.org
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?​
 
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.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Back
Top