Let P , Q , and R be three finite sets such that n(P) + n(Q) + n(R)...

The number of elements in the symmetric difference of two sets is given by:
$n(P \Delta Q) = n(P) + n(Q) - 2n(P \cap Q)$

Summing the three given symmetric differences yields:
$n(P \Delta Q) + n(Q \Delta R) + n(R \Delta P) = 2(n(P) + n(Q) + n(R)) - 2(n(P \cap Q) + n(Q \cap R) + n(R \cap P))$

Let $S_1 = n(P) + n(Q) + n(R)$ and $S_2 = n(P \cap Q) + n(Q \cap R) + n(R \cap P)$.
Substituting the given values:
$40 + 30 + 50 = 2(110) - 2S_2$
$120 = 220 - 2S_2$
$2S_2 = 100$
$S_2 = 50$

The number of elements present in exactly one of the three sets is given by the formula:
$E_1 = S_1 - 2S_2 + 3S_3$
where $S_3 = n(P \cap Q \cap R) = 5$.

Substituting the values into the formula:
$E_1 = 110 - 2(50) + 3(5)$
$E_1 = 110 - 100 + 15 = 25$

Answer: $25$