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Out of a group of $21$ persons, $9$ eat vegetables, $10$ eat fish and $7$ eat eggs. $5$ persons eat all three. How many persons eat at least two out of the three dishes?

$N(A\cup B \cup C) = N(A) + N(B) + N(C) - N(A\cap B) - N(A \cap C) - N(B\cap C) + N(A\cap B\cap C)$

Let $Y$ be the no. of persons who eat at least one item. $21 - Y$ people do not eat anything.

$Y = 9 + 10 + 7 - [N(A\cap B) + N(A\cap C) + N(B\cap C)] + 5$

$[N(A\cap B) + N(A\cap C) + N(B\cap C)] = 31 - Y$.

Now, these include the no. of persons who eat all 3 items thrice. So, excluding those, we get, no. of persons who eat at least two items (by adding the no. of persons eating EXACTLY 2 dishes and the number of persons eating all 3 dishes) as

$31 - Y - 2*5 = 21 - Y$.

The minimum value of Y is 10 as 10 people eat fish. Is this possible? Yes.

The maximum value of Y is 21. Is this possible? No. Because 5 people eat all three items. So, the no. of persons eating at most 2 items $= (9-5) + (10-5) + (7-5) = 11$. And adding 5 we get 16 people who eat at least one item.

So, our required answer is $21-10 \geq X \geq 21 - 16 \implies 5 \leq X \leq 11.$

Why minimum value of Y is 10? It is given that 7 people eat eggs. Won’t 7 be the minimum value of Y? Someone please explain this to me
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@pascua you can see its very simple actually, 10 people are eating something means there are alteast person who eats something so you cant say it is 7 people.
nicely explained! Answer should be [5 - 10].

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### 1 comment

The lower bound is 5, which is very obvious, but the upper bound￼ value won't be 11, it has to be less than Or equal to 10, because if we take it to be 11, then the number of ppl eating exactly 1 item is coming -1, refer the image, and if we take the upperbound to be 10 then the ppl eating exactly 1 item is 0, which is the least it must go.. @Arjun Sir ,I think the data in this question is wrong...

Let A$\cap$B$\cap$C = x

A$\cap$B+B$\cap$C+A$\cap$C contains x 3 times...

And atleast 2 means ( A$\cap$B+B$\cap$C+A$\cap$C ) - 2x

which evaluates to 0 ...which is contradict to the given data i.e.5 persons eat all three items.

IT IS POSSIBLE THAT OUT OF 21 PEOPLE THERE MAY BE SOME PEOPLE WHO ARE NOT EATING ANY OF THREE ITEMS

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whenever you find A∪B∪C it means you have all the elements in A,B,C only once .

Let's say in a class there are 10 students out of which 3 have history , 5 have computer and 8 have science then,

n(H∪C∪S) = n(H)+n(C)+n(S) - (n(H∩C)+n(C∩S)+n(H∩S)) + n(H∩C∩S)

here n(H∪C∪S) :- all the students in a class who have atleast 1 subject from H/C/S. but there might be some student who do not have any of them , he/she might be having Biology or physics who knows. so don't take 10 as whole .

(n(H∩C)+n(C∩S)+n(H∩S)) :- in this case it contain all the students who have atleast 2 subjects(but it contain all 3 subjects thrice , so subtract 2*n(H∩C∩S) from this while calculating

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@Sachin Mittal 1 sir!

@ Sir,

please provide a proper solution for this question.

First let us draw the generalized Venn Diagram of the situation: As per the question the portion whose cardinality we need to find is as shown below in blue: The required cardinality can be given as

$$N=|V\cap F| +|V\cap E|+ |F \cap E| - 2|V\cap E\cap F|$$

But we know :

$$|V\cup E\cup F|=|V|+|E|+|F|-|V\cap F| -|V\cap E|- |F \cap E| +|V\cap E\cap F|$$ $$= 9+7+10 -|V\cap F| -|V\cap E|- |F \cap E| +|V\cap E\cap F|$$ $$= 26 -|V\cap E\cap F| -N = 21- N \text{ [since |V\cap E\cap F|=5]}$$

$$\implies N=21-|V\cup E\cup F|$$

The minimum possible value of $|V\cup E\cup F|$ is $10$ as we could possibly pack the sets $V$ and $E$ in $F$ as shown below (since $F$ is having the largest cardinality among all the three sets): [This is possible because, no explicit values of $|V\cap F|$,$|V\cap E|$ or $|F \cup E|$ are given, so we are free to think of any possible sort of overlapping] So the least value of $|V\cup E\cup F|$ is $10$.

So the max value of $N$ is $11=(21-|V\cup E\cup F|=21-10)$

But the minimum value of $N$ is $5$ since it is given that $|V\cap E\cap F|=5$, and the diagram shall be as shown: So $$5\leq N \leq 11$$

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