The number of students playing at least one game is Total students - Students playing neither = $80 - 8 = 72$. This represents $|C \cup H|$. Using the principle of inclusion-exclusion, $|C \cup H| = |C| + |H| - |C \cap H|$. Substituting the values: $72 = 48 + 36 - |C \cap H|$. $72 = 84 - |C \cap H|$, so $|C \cap H| = 84 - 72 = 12$. These 12 students play both carrom and chess.