Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

Answer:

$10$ (given in solutions)

$15$ (as per me )

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

share|improve this question
1  
Have you tried drawing a Venn diagram? –  Git Gud Feb 2 '13 at 20:41

2 Answers 2

up vote 2 down vote accepted

There seems to be something wrong here.

Consider this Venn diagram and its labeled parts.

enter image description here

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:

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

share|improve this answer
    
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

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.