We can solve this problem using the principles of set theory and Venn diagrams. Let's denote the following:

• $H$ = Set of students who play hockey.
• $B$ = Set of students who play basketball.
• $C$ = Set of students who play cricket.
• Total number of students = 60.
• Number of students who play none of the three games = 15.

We are given the following:

• $|H| = 23$ (students who play hockey),
• $|B| = 15$ (students who play basketball),
• $|C| = 20$ (students who play cricket),
• $|H \cap B| = 7$ (students who play hockey and basketball),
• $|B \cap C| = 5$ (students who play basketball and cricket),
• $|H \cap C| = 4$ (students who play hockey and cricket),
• Students who do not play any of the three games = 15.

From this, the number of students who play at least one of the games is $60 - 15 = 45$.

(i) How many play hockey, basketball, and cricket?

We need to find $|H \cap B \cap C|$, the number of students who play all three games. We will use the inclusion-exclusion principle:

$|H \cup B \cup C| = |H| + |B| + |C| - |H \cap B| - |B \cap C| - |H \cap C| + |H \cap B \cap C|$ Substituting the known values:

$45 = 23 + 15 + 20 - 7 - 5 - 4 + |H \cap B \cap C|$

Simplifying:

$45 = 54 - 16 + |H \cap B \cap C|$ $45 = 38 + |H \cap B \cap C|$ $|H \cap B \cap C| = 45 - 38 = 7$

Thus, 7 students play all three games.

(ii) How many play hockey but not cricket?

We need to find $|H \setminus C|$, the number of students who play hockey but not cricket. This can be computed by subtracting the students who play both hockey and cricket (including those who play all three games) from the total number of hockey players:

$|H \setminus C| = |H| - |H \cap C|$

We already know that $|H \cap C| = 4$, but this includes those who play hockey, cricket, and basketball. Since we know that $|H \cap B \cap C| = 7$, we subtract this from the 4:

[ |H \cap C| = 4 - 7 = -3