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?
Answer:

$10$ (given in solutions)
$15$ (as per me )

Please tell me which one is correct. Also, tell if youu have different answer.
 A: 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
$$\begin{align*}
|A\cap B|+|A\cap C|+|B\cap C|&=|A|+|B|+|C|+|A\cap B\cap C|-|A\cup B\cup C|\\
&=9+10+7+5-21\\
&=10\;.
\end{align*}$$
A: 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:
$$\begin{align}
(0,0,1,1,0,4,5,10) \quad &\implies \quad d+e+f+g = h = 10 \\
(1,2,0,1,1,2,5,9) \quad &\implies \quad d+e+f+g = h = 9 \\
(1,3,1,0,1,2,5,8) \quad &\implies \quad d+e+f+g = h = 8 \\
(3,4,0,1,1,0,5,7) \quad &\implies \quad d+e+f+g = h = 7 \\
(3,4,2,0,0,1,5,6) \quad &\implies \quad d+e+f+g = h = 6 \\
(4,5,2,0,0,0,5,5) \quad &\implies \quad d+e+f+g = h = 5
\end{align}$$
