1

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)

How many PC set that are sold but not contain anything? enter image description here

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?

vedss
  • 343
  • It seems to be worded quite messily but still, it doesn't seem explicit that "with 18 crystal screen PC are included" means that $18$ sets have only the crystal screen PC (and not the other two items), as you have indicated in your diagram. In fact, I'd interpret that to mean "18 sets have a crystal screen PC and possibly other items". I'm also struggling to follow your arithmetic. – Jam Jan 09 '20 at 15:36
  • @Jam I'm sorry, i change the problem a little bit, hope it is now understandable, no in my diagram 18 means include printer and scanner, which mean 18=a+b+c+ itself – vedss Jan 09 '20 at 15:44
  • @TheHolyJoker Sorry the right one is "include all" i edit my questions – vedss Jan 09 '20 at 16:41

1 Answers1

0

The answer is $x=15$.

Notation: $S_1=A,S_2=B,S_3=C$.

Proposition: $|A \cup B \cup C| \geq 18$
proof: $|A \cup B \cup C|\geq|A|=18$

Using $|A \cup B \cup C| \geq 18\implies x\leq 33-18=15$ and by setting:

  1. $B\subset A$
  2. $C\subset A$

For example by $A=\{1,2...,18\},B=\{1,2,..12\},C=\{10,11,12,13,14,15\}$
(which also maintain $\vert A\cap B\cap C\vert=3 $ )

We get $|A \cup B \cup C|=18$, So $15$ is a tight upper bound to $x$.

TheHolyJoker
  • 2,040
  • No, what i want to find is maximum value of x the area outside the three circle.. where PC not contain anything, it is said that the answer is 15, 33 should be $|A \cup B \cup C| $ + x( the area outside) , see my graph . and i really dont know why you can conclude that $|A| + |B| + |C| = 36$ because there is overlap inside of the circle – vedss Jan 09 '20 at 16:54
  • @vedss I edited the answer, does that answer your question? – TheHolyJoker Jan 10 '20 at 21:43
  • Thank you but how do you know $|A \cup B \cup C| \geq 18$? – vedss Jan 10 '20 at 22:02
  • Do you agree $|A\cup B\cup C|\geq|A|=18$? – TheHolyJoker Jan 10 '20 at 22:05
  • i know that we want to minimize $ |A \cup B \cup C|$ but why $ |A \cup B \cup C|\geq|A|=18$? – vedss Jan 10 '20 at 22:09
  • $A$ is the set of PC with a crystal screen. You wrote it is equal to 18 – TheHolyJoker Jan 10 '20 at 22:11
  • Yes but why it has to be 18 not other? and also is my computation a+b+c=x-3 correct? – vedss Jan 10 '20 at 22:48
  • is it also possible to count that a+b+c+3=x , a+b+c=x-3 , then since $a+c \le 9$ and $b+c \le 3$ means $a+b+2c \le 12$ so $a+b+c=12$ and $ x=15$, but im not sure if we can infer from $a+b+2c \le 12$ that maximum for $ a+b+c$ is 12 – vedss Jan 11 '20 at 01:08
  • $a+b+2c\leq12\implies a+b+c\leq12$ If and only if $c\geq 0$ which is the case. – TheHolyJoker Jan 11 '20 at 13:00
  • thankyou so that means my computation also correct ?and it give the maximum for x too? $|A\cup B\cup C|\geq|A|=18$ means that there is 0 printer only and 0 for scanner only but is there relation with inclusion principle? – vedss Jan 12 '20 at 02:05
  • I couldn't follow your computation, sorry, maybe they are correct... For your second question, you could find the maximal value of $x$ using them, assuming they are correct. As for your third question, this inequality is a direct result of the definition of the union of two sets, and the definition of the cardinality of a set. Also, you can not conclude from $a+b+c\leq12$ that $a+b+c=12$. That's why I provided a specific example in which we get equality. I recommend you read my solution again slowly, and maybe edit your question to make your attempt more clear. hope it helped :) – TheHolyJoker Jan 12 '20 at 17:37
  • Thankyou ,but for $a+b+2c\leq12$ in your equation, a=9, b = 3 and c is 0 means we can imply that $a+b+c\leq12$ ? – vedss Jan 13 '20 at 02:57