Combinatorics: How many persons like apples and pears and strawberries? Out of $32$ persons, every person likes to eat at least one of the following type of fruits: Strawberries, Apples and Pears. (Which means that there does not exist any person, who does not like to eat any type of fruit). Furthermore, we know that $20$ persons like to eat apples, $18$ persons like to eat pears and $28$ persons like to eat strawberries.
(a) There are $10$ persons who like apples and pears, $16$ persons who like apples and strawberries, and $12$ persons who like pears and strawberries.
How can I find out how many people like apples as well as pears as well as strawberries?
To structure this a bit:


*

*$32$ persons

*$20 \rightarrow$ apples ($\rightarrow$ meaning "like")

*$18 \rightarrow$ pears

*$28 \rightarrow $ strawberries


And for (a)


*

*$10$ persons $\rightarrow$ (apples & pears)

*$16$ persons $\rightarrow$ (apples  & strawberries)

*$12$ persons $\rightarrow$ (pears & strawberries)


Since we know that the total number of persons is $32$. 
Can I just do the following? 
Because $20$ persons like apples I can just add the following numbers together: 
$10 \rightarrow$ ($10$ apples & $0$ pears) + $16 \rightarrow$ ($6$ apples & $10$ strawberries) $+ 12 \rightarrow$ ($12$ pears & $0$ strawberries). So in total I'd get $10 + 16 + 12 = 28$ people who like apples, pears and strawberries? Is that correct? 
(b) Assume that you don't have the information in (a). Give the preferably  limits for the amount of persons who like to eat all kind of fruits.
Since $18$ person like pears, can I just say that $18$ persons like to eat pears, apples and strawberries? (As $18$ is the minimal amount of fruits).
 A: Part (a) is an ill-posed question (almost in the spirit of this other question). By inclusion/exclusion, using the given values the number of people liking all three fruits is
$$32-(20+18+28-10-16-12)=32-28=4$$
Then the number of people liking apples  and some other fruit(s) is $16+10-4=22$, which is greater than the 20 people who like apples.
But never mind. For (b), since 18 people like pears, the maximum number of people liking all three fruits is 18, but we can't put exactly 18: 4 people must like only apples and 12 people only strawberries, at which point the number of people is greater than the fixed 32. We can have 17 people liking all three fruits, though, in which case


*

*3, 1, 11 like only apples, pears, strawberries respectively

*nobody likes exactly two fruits


For the other extreme, consider the least amount of people who must like at least two fixed fruits:


*

*20 like apples and 18 like pears, so since there are 32 people there must be at least $20+18-32=6$ people liking both apples and pears

*Similarly, at least $20+28-32=16$ like apples and strawberries; at least 14, pears and strawberries


Now place these "forced" people in such a way that nobody likes all three fruits. We find that there are two more people than stipulated who like apples, so at least two people like all three fruits, and we get the same result for the other fruits. Thus we must merge six people into two in the centre where people like all three fruits; fortituously the total number of people becomes exactly 32. Thus the minimum number of people who like all three fruits is 2, with


*

*14 liking only apples/strawberries

*12 liking only pears/strawberries

*4 liking only apples/pears

*nobody likes exactly one fruit.


Both the configurations for 17 and 2 people liking all three are physically valid.
A: What you have actually worked out for part a) is the number of people who don't like all three. So since there are $32$ people, and $28$ don't like all three, the number of people who like Apples, Pears and Strawberries is $4$.
