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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?

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

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 $21-10 \geq X \geq 21 - 16 \implies 5 \leq X \leq 11.$
by Boss (33.8k points)
selected by
–1
atleast 2 dishes means  2dishes and all 3 dishes   should be (10+5)????
+2

"at least two out of the three " both  means 2 and 3

red shaded gives both 2 and 3

so no need of again finding for 3

so ans should be only 10

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21−Y  are the persons who dont eat anything

21−Y are the persons who eats at least 2.Is this correct?
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@rahul yes correct
+5

Perfect answer , Without your answer I would have completely ignored fact that "IT IS POSSIBLE THAT OUT OF 21 PEOPLE THERE MAY BE SOME PEOPLE WHO ARE NOT EATING ANY OF THREE ITEMS"

Answer given by you in GO pdf is still need to be update.

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21-Y is the no. of people who dont eat anything.

And 31-Y-2*5 is the no. of people who eat atleast two items...

Then how these are equal??
@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.
by Boss (23.8k points)
+6

I got the same doubt.
Let A∩B∩C = x
then ( A∩B+B∩C+A∩C ), this already contains $3$x. therefore subtracting $2$x from this should result into POSITIVE value, but it is zero.
Moreover, u r right. They are asking for "atleast 2" which means ( A∩B+B∩C+A∩C ) - 2x.

Something wrong with given Data ?

@Arjun sir, plz see this.

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Sir, i can't understand how this is possible?

Minimum value of Y is 10 as 10 people eat fish. Is this possible? Yes.
Maximum value of Y is 21. Is this possible? No. Because 5 person eat all three item.
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I did it in same way.Can some one tell,why is this approach?I dont see any flaw here/
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here, we are assuming that the 21 is given as at-least 1 dishes but 21 includes (at-least 1 + none)
+1

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

Answer should be [5 - 10].

by Junior (585 points)
+1

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..