Elementary Set theory question … it was asked in Exam

There are $21$ people.

$9$ eat dish $A$

$10$ eat dish $B$

$7$ eat dish $C$

$5$ eat dish $A , B$ and $C$

How many people eat at least two dishes?

$10$ (given in solutions)

$15$ (as per me )

Please tell me which one is correct. Also, tell if youu have different answer.

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Have you tried drawing a Venn diagram? –  Git Gud Feb 2 '13 at 20:41

There seems to be something wrong here.

Consider this Venn diagram and its labeled parts.

We are told that: $$a+e+f+g=|A|=9 \qquad b+d+f+g=|B|=10 \qquad c+e+d+g=|C|= 7$$ $$g = 5$$

Thus

\begin{align} a+e+f=4 \\ b+d+f=5 \\ c+e+d=2 \end{align}

so that, by adding all three equations,

$$a+b+c+2d+2e+2f=11 \qquad(1)$$

But, assuming each of the 21 people eats at least one dish, we also know that $$a+b+c+d+e+f+g=21$$

so that (since $g=5$) $$a+b+c+d+e+f=16 \qquad(2)$$

Consequently, by subtracting equation (2) from equation (1), $$d+e+f = -5 \text{ (!)}$$

This problematic result is consistent with @Brian's work, which concludes

\begin{align} |A\cap B| + |A\cap C|+|B\cap C| &= (f+g)+(e+g)+(d+g) \\ &= 3g+(d+e+f) \\ &= 3\cdot 5 + (-5) \\ &= 10 \\ \end{align}

Note, though, that this value does not answer OP's question, as it over-counts the people eating all three dishes. The number of people eating at least two dishes should be given by $d+e+f+g$ (that is, $|A\cap B|+|A \cap C|+|B\cap C|-2|A\cap B\cap C|$), but the computed value here is zero. Weird.

Edit. Perhaps the assumption "each of the 21 people eats at least one dish" is in error. Let $h$ be the number of people who eat nothing. Then we have

$$a+b+c+d+e+f+g+h=21$$

and

$$d+e+f=h-5$$

so that (barring negative people from the dinner party) $h \ge 5$. Moreover,

$$21 = a+b+c+(h-5)+5+h = a+b+c+2h$$

Therefore (also barring fractional people from the dinner party), $h \le 10$, and we can write

$$5 \le d+e+f+g \le 10$$

I don't see the conditions that force us to accept $10$ for the value of $d+e+f+g$; indeed, I've found scenarios $(a,b,c,d,e,f,g,h)$ that give rise to each possible value of the expression:

\begin{align} (0,0,1,1,0,4,5,10) \quad &\implies \quad d+e+f+g = h = 10 \\ (1,2,0,1,1,2,5,9) \quad &\implies \quad d+e+f+g = h = 9 \\ (1,3,1,0,1,2,5,8) \quad &\implies \quad d+e+f+g = h = 8 \\ (3,4,0,1,1,0,5,7) \quad &\implies \quad d+e+f+g = h = 7 \\ (3,4,2,0,0,1,5,6) \quad &\implies \quad d+e+f+g = h = 6 \\ (4,5,2,0,0,0,5,5) \quad &\implies \quad d+e+f+g = h = 5 \end{align}

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very nice approach ... i liked it v.much... d+e+f+g is right ans and i dont understand how is it comming to zero.. if i find out i ll comment.. if any1 finds it out please comment.. thank u guys..:) –  Jay Satish Teli Feb 2 '13 at 22:23
WOW...Nice Solution.. SO the FINAL ANS IS closed interval [5 10].. thanks :) –  Jay Satish Teli Feb 3 '13 at 0:02

Let $A,B$, and $C$ be the sets of people eating dish A, B, and C, respectively. By the inclusion-exclusion formula we know that

$$|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|\;.\tag{1}$$

You’re told that $|A\cup B\cup C|=21$: there are $21$ people altogether. You’re also told that $|A|=9$, $|B|=10$, $|C|=7$, and $|A\cap B\cap C|=5$. Rearranging $(1)$ with a little algebra, we see that

\begin{align*} |A\cap B|+|A\cap C|+|B\cap C|&=|A|+|B|+|C|+|A\cap B\cap C|-|A\cup B\cup C|\\ &=9+10+7+5-21\\ &=10\;. \end{align*}

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Scott: wat u said is correct but dont u think A∩B∩c should also be added to 10. Bcoz : atleast two dishes means 2/more dishes. right...? –  Jay Satish Teli Feb 2 '13 at 21:05
@Jay: It’s already been added. –  Brian M. Scott Feb 2 '13 at 21:06
Oh ..thanks ..i was thinking answer should have been |A∩B|+|A∩C|+|B∩C|+|A∩B∩C| which is 10+5 =15.. –  Jay Satish Teli Feb 2 '13 at 21:11
@Jay: You’re welcome. –  Brian M. Scott Feb 2 '13 at 21:13