Maximizing PC Set Options: Venn Diagram Equation in a Computer Shop

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In a computer shop, there are 33 PC sets sold, categorized by features: 18 include crystal screens, 12 have printers, and 6 come with scanners, with 3 sets including all features. The goal is to determine the maximum value of x, defined as the number of PC sets not included in the union of these categories. The equations derived from the Venn diagram indicate relationships between sets a, b, and c, which represent overlapping features among the PCs. Constraints on a, b, and c are established, leading to the conclusion that x can be expressed as x = a + b + c + 3. The discussion focuses on maximizing x while adhering to these constraints.
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Homework Statement
In a computer shop, there are $33$ PC set that are sold:


1. with 18 sets of PC have crystal screen PC included,

2. with 12 sets of PC have printer included,

3. with 6 sets of PC have scanner system included,

4. with 3 sets of PC that include all(printer, scanner system,

crystal screen pc)
Relevant Equations
x=33-(18-(a+b+3)+12-(a+c+3)+6-(b+c+3) + ( a+b+c +3 ) )
In a computer shop, there are $$33$$ PC set that are sold:

1. with 18 sets of PC have crystal screen PC included,
2. with 12 sets of PC have printer included,
3. with 6 sets of PC have scanner system included,
4. with 3 sets of PC that include all(printer, scanner system,
crystal screen pc)
Screenshot from 2020-01-09 23-53-26.png
We want to find $$x$$ which is $$33-|S_1 \cup S_2 \cup S_3|=x$$ and get $$x$$ that is max

From the graph :
It means that inside $$18$$ crystal screen PC, there is a set that also includes a scanner or printer and, there is a set that contains only crystal screen.

1. ##S_1\implies 18=a+b+3+## crystal screen only
2. ##S_2\implies 12 = a+c+3+## printer only
3. ##S_3\implies 6=b+c +3+## scanner only

$$x=33-(18-(a+b+3)+12-(a+c+3)+6-(b+c+3) + ( a+b+c +3 ) )=x\\
33-(27-a-b-a-c-b-c + (a+b+c+3))=\\
33-(30 -a-b-c)=\\
3+a+b+c =x\\
a+b+c=x-3$$

More conclusions:

- $$a+c \le 9$$
- $$b+c \le 3$$
- $$a+b \le 15$$

What is the relation with $$x$$ and how do I get the actual result?
 
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fiksx said:
Homework Statement:: In a computer shop, there are $33$ PC set that are sold:1. with 18 sets of PC have crystal screen PC included,

2. with 12 sets of PC have printer included,

3. with 6 sets of PC have scanner system included,

4. with 3 sets of PC that include all(printer, scanner system,

crystal screen pc)
Homework Equations:: x=33-(18-(a+b+3)+12-(a+c+3)+6-(b+c+3) + ( a+b+c +3 ) )

In a computer shop, there are $$33$$ PC set that are sold:

1. with 18 sets of PC have crystal screen PC included,
2. with 12 sets of PC have printer included,
3. with 6 sets of PC have scanner system included,
4. with 3 sets of PC that include all(printer, scanner system,
crystal screen pc)
View attachment 255373We want to find $$x$$ which is $$33-|S_1 \cup S_2 \cup S_3|=x$$ and get $$x$$ that is max

From the graph :
It means that inside $$18$$ crystal screen PC, there is a set that also includes a scanner or printer and, there is a set that contains only crystal screen.

1. $$S_1\implies 18=a+b+3+$$$$crystal screen only$$
2. $$S_2\implies 12 = a+c+3+$$ $$printer only$$
3. $$S_3\implies 6=b+c +3+$$ $$scanner only$$

$$x=33-(18-(a+b+3)+12-(a+c+3)+6-(b+c+3) + ( a+b+c +3 ) )=x\\
33-(27-a-b-a-c-b-c + (a+b+c+3))=\\
33-(30 -a-b-c)=\\
3+a+b+c =x\\
a+b+c=x-3$$

More conclusions:

- $$a+c \le 9$$
- $$b+c \le 3$$
- $$a+b \le 15$$

What is the relation with $$x$$ and how do I get the actual result?
In words, what do a, b, c, S1, S2, and S3 represent?
 
Mark44 said:
In words, what do a, b, c, S1, S2, and S3 represent?
a,b,c you can see the diagram i drew,
a means PC set include crystal screen and printer
b means PC set include crystal screen and scanner
c mean PC set include scanner and printer 1. $$S_1$$ means 18=a+b+3+crystal screen only
2. $$S_2$$ means 12 = a+c+3+ printer only
3. $$S_3$$ means 6=b+c +3+ scanner only
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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