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.

  • $\begingroup$ 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


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.


$Q=(Q\backslash P)\cup(Q\cap P)$ 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!


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