In a group of 200 students, 120 study Mathematics, 90 study Physics, and 70 study Chemistry. 40 study Mathematics and Physics, 30 study Physics and Chemistry, and 50 study Chemistry and Mathematics. I

The number of students studying at least one subject is the total number of students minus those who study none.

$$|M \cup P \cup C| = |U| - |M \cup P \cup C|'$$ $$|M \cup P \cup C| = 200 - 20$$ $$|M \cup P \cup C| = 180$$

The formula for the union of three sets is:

$$|M \cup P \cup C| = |M| + |P| + |C| - |M \cap P| - |P \cap C| - |C \cap M| + |M \cap P \cap C|$$

Substitute the known values:

$$180 = 120 + 90 + 70 - 40 - 30 - 50 + |M \cap P \cap C|$$ $$180 = 280 - 120 + |M \cap P \cap C|$$ $$180 = 160 + |M \cap P \cap C|$$ $$|M \cap P \cap C| = 180 - 160$$ $$|M \cap P \cap C| = 20$$

a) The number of students who study all three subjects is $\boxed{\text{20 students}}$.

These values will be used to find the number of students studying only one subject.

$$|M \cap P \text{ only}| = |M \cap P| - |M \cap P \cap C| = 40 - 20 = 20$$ $$|P \cap C \text{ only}| = |P \cap C| - |M \cap P \cap C| = 30 - 20 = 10$$ $$|C \cap M \text{ only}| = |C \cap M| - |M \cap P \cap C| = 50 - 20 = 30$$

To find students who study only Physics, subtract those who study Physics with Mathematics (only), Physics with Chemistry (only), and all three subjects from the total number of Physics students.

$$|P \text{ only}| = |P| - |M \cap P \text{ only}| - |P \cap C \text{ only}| - |M \cap P \cap C|$$ $$|P \text{ only}| = 90 - 20 - 10 - 20$$ $$|P \text{ only}| = 90 - 50$$ $$|P \text{ only}| = 40$$

b) (i) The number of students who study Physics only is $\boxed{\text{40 students}}$.

Similarly, for Chemistry only:

$$|C \text{ only}| = |C| - |C \cap M \text{ only}| - |P \cap C \text{ only}| - |M \cap P \cap C|$$ $$|C \text{ only}| = 70 - 30 - 10 - 20$$ $$|C \text{ only}| = 70 - 60$$ $$|C \text{ only}| = 10$$

b) (ii) The number of students who study Chemistry only is $\boxed{\text{10 students}}$.