Venn diagram Puzzle The King informed the Doctor that exactly two-thirds of the men had received a black eye, three-fourths had sprained a wrist, and four-fifths had stubbed a toe. “It follows, then,” the Doctor surmised, “that at least twenty-six of the men have a black eye, a sprained wrist, and a stubbed toe.” How many injured men were in the wood?
Here are some thoughts:
Suppose we have a total of $x$ injured men. Let $E$ be the number of people with blacked-eye, $W$ be sprained wrist, and $T$ be stabbed-toe. So we have

*

*$E= \frac{2}{3}x$

*$W= \frac{3}{4}x$

*$T= \frac{4}{5}x$
From here we can make a Venn diagram.
Here are some information we can figure out:

*

*$E^c=\frac{1}{3}x$, where $E^c$ is the complement of $e$.

*$W^c=\frac{1}{4}x$

*$T^c=\frac{1}{5}x$

*$E-26=E -(E\cap W \cap T ) $

*$W-26=W -(E\cap W \cap T ) $

*$T-26=T -(E\cap W \cap T ) $
From here I played with the Venn diagram, and tried to eliminate some parts. But I failed. Could some please help me to continue?
Thanks!
 A: Hint:
Suppose there are $n$ people injured in the woods, and they are all gathered in a hospital waiting room.

*

*First, ask all people without a black eye to leave.


*From the remaining people, ask those without a sprained wrist to leave.


*Finally, from those who remain, ask those without a stubbed toe to leave.
After these three steps, everyone who remains in the room will have all three injuries. How many people were removed in each step? You cannot answer this question exactly, since there may be overlaps among injuries, but you can give an upper bound for the number of people removed at each step, which leads to a lower bound for the number of people remaining in terms of $n$.
A: Let $(E)$ be the number of people with only blacked-eye, $(EW)$ with only blacked-eye and sprained wrist, $(EWT)$ with all three, so on.
We have:
$$\begin{align}
(E)+(EW)+(ET)+(EWT)=\frac{2}{3}x \qquad(1)\\
(W)+(EW)+(WT)+(EWT)=\frac{3}{4}x\qquad(2)\\
(T)+(ET)+(WT)+(EWT)=\frac{4}{5}x\qquad(3)\\
(E)+(W)+(T)+(EW)+(ET)+(WT)+(EWT)=x\;\;\,\qquad(4)\\
\text{Minimum of }(EWT)=26\;\qquad(5)
\end{align}$$
$$(1)+(2)+(3)-(4): \qquad (EW)+(ET)+(WT)+2(EWT)=\frac{73}{60}x$$
Here, the RHS is fixed, so in order for $(EWT)$ to have its minimum value, $(EW)+(ET)+(WT)$ has to have its maximum value. In $(4)$, $(E),(W),(T) \ge 0$ so the maximum is $x-(EWT)$.
$$\frac{73}{60}x=(EW)+(ET)+(WT)+2(EWT) \le x+(EWT)$$
$$\implies (EWT) \ge\frac{13}{60}x$$
$$\implies x=120$$
