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I'm working on a following question:

Let $S$ be a set of students who solve maths problems. $P$ is the subset of $S$ containing all students who eat pizza while they solve problem; similarly $C$ is the subset of students who drink coffee, and $M$ is the subset of students who listen to music. Given that:

  • there are $225$ students in the set $S$;
  • $100$ of them eat pizza, $75$ drink coffee, and $50$ listen to music;
  • $7$ students eat pizza and drink coffee, $4$ eat pizza and listen to music, while $31$ students drink coffee and listen to music;
  • one student indulges in all three problem-solving stimulants, and is therefore included in all the earlier figures;

How many students are there who solve maths problems without any pizza, coffee or music?

I have calculated how many students do each of them (but only one) and from there the answer the the question as follows: $$\#(P \setminus M \setminus C) = 100 - 6 - 4 +1 = 91\\ \#(C \setminus P \setminus M) = 75 - 6 -30 +1 = 40 \\ \#(M \setminus P \setminus C) = 50 - 3 -30 +1 = 18$$ ($\#(P \setminus M \setminus C) = 100 - 6 - 4 + 1 = 91$ because there are 100 people who eat pizza, 7 of which have pizza and coffee (minus 1 that does all three), 4 that have pizza with music (minus 1 that does all three) and the one that does all).

Then, $\#(S \setminus P \setminus C \setminus M) = 225 - 90 - 40 - 18 - 6 - 3 - 30-1=35$ (#S - #(only P) - #(only C) - #(only M) - ...).

I have a feeling that this is either wrong or done in a horribly wrong way. IS it correct/is there any other way?

share|improve this question
    
You seem to be using exclusively "set minus" in your calculations, or at least as you've written in your notation. The "set minus" operation on sets, say $X \setminus Y$ means $x \in X$ but $x \notin Y$. I think it might make more sense to use unions (X or Y), intersections (X and Y)...perhaps exploring the problem with a Venn diagram? –  amWhy Nov 3 '12 at 21:47
    
When I formulated the problem I was using $\#(P \cap C) = 7$ and similar, I can not imagine how to use union/intersection to indicate P but not M and not C (unless I use compliment). –  Bart Platak Nov 3 '12 at 21:49
1  
P but (and) not M and not C is equivalent to P and not (M or C), which is equivalent to P and not ($M \cup C$) which is equivalent to $P \setminus (M \cup C)$... –  amWhy Nov 3 '12 at 21:58
1  
I transcribed the image to text (for accessibility reasons); I hope you don't mind. Including the image of the problem in your question was understandable; there's nothing wrong with having done so. –  amWhy Nov 3 '12 at 22:25
    
@amWhy: Thanks for doing so. I would've done so in the first instance (images can disappear in future which won't help any1 reading this question later on) but I really need to get on with more questions. Thanks :) –  Bart Platak Nov 3 '12 at 22:29

1 Answer 1

up vote 3 down vote accepted

I generally do small problems of this kind directly on a Venn diagram:

enter image description here

I started with the $1$ in the centre and used it to get the $6,3$ and $30$, then used all of this to get the $90,38$, and $16$. The desired result is then

$$225-(90+38+30+30+6+3+1)=27\;,$$

assuming no computational errors along the way.

share|improve this answer
    
+1 Nice!!...... –  amWhy Nov 3 '12 at 22:08
    
@amWhy: Thanks! –  Brian M. Scott Nov 3 '12 at 22:13
    
+1 Thanks a lot! The visual representation really helps I will make sure I do it from now on! :) –  Bart Platak Nov 3 '12 at 22:20
1  
@norfavrell: You’re welcome. I’m not a particularly visual person, and I’ve always found it helpful, as have my students over the years. –  Brian M. Scott Nov 3 '12 at 22:28

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