1
$\begingroup$

the questions i have to ask i believe have a similar process which is why i have grouped them together:

  1. Sets P and Q; |P|= 6, |Q|= 15 and |P 'AND' Q| = 5; what's |Q\P|

    I know the answers is 10, I'm just not sure that my method was correct as i was trying to figure out the process from the answer; at first i did 15-5 then 15-6+1 but I'm sure its incorrect

  2. each set X and Y contain 19 elements, then the maximum number of elements in the set

    (X 'OR' Y)\Y (answer 19)

  3. each set P and Q contains 40 elements; the maximum number of elements in the set

    P 'OR' (Q\P) (answer 80)

If anyone can help that would be appreciated; i think i am confusing myself by trying to think of the elements themselves rather than how many of them there are and I am not sure how to adapt counting principles to these questions.

$\endgroup$
4
  • $\begingroup$ Have you tried drawing Venn diagrams? $\endgroup$ Nov 18, 2012 at 4:32
  • $\begingroup$ Thanks for your suggestion :)! I tried for one question but I'm not particularly good at them; these questions are multiple choice for an exam that we have limited time on so i was wondering more-so if there was a quicker way to answer the questions rather than Venn diagrams ;) $\endgroup$
    – Z Oj
    Nov 18, 2012 at 4:36
  • $\begingroup$ I think Venn diagrams are pretty quick. I submit the answer by @Andre as evidence. $\endgroup$ Nov 18, 2012 at 4:48
  • $\begingroup$ No worries, thanks ;) $\endgroup$
    – Z Oj
    Nov 18, 2012 at 4:52

2 Answers 2

3
$\begingroup$

Draw a picture (Venn diagram).

For the first problem, draw two intersecting circles. There are $5$ items in the part the two circles have in common, so in $Q\setminus P$, that is, in the part of $Q$ which is outside $P$, there are $15-5$ objects.

For the second problem, $(X\cup Y)\setminus Y$ is biggest if we are taking away nothing from $X$, that is, if $X$ and $Y$ have nothing in common. In that case, $(X\cup Y)\setminus Y$ has $19$ elements.

For the third problem, our union $P\cup (Q\setminus P)$ is biggest if $Q\setminus P=Q$, that is, if $P$ and $Q$ have nothing in common.

$\endgroup$
0
$\begingroup$

$Q=(Q\backslash P)\cup(Q\cap P)$ which are $2$ disjoint sets. (You actually have to draw a Venn diagram to see it)

Hence $|Q|=|Q \backslash P|+|Q \cap P| $ and $|Q \backslash P| =|Q| -|Q \cap P|=15-5=10 $

Even I think that Venn diagrams are better to understand such questions!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.