1. **State the problem:** In a class of 20 boys, 16 play soccer, 12 play hockey, and 2 are not allowed to play games. We need to find the number of students who play both soccer and hockey and those who play only hockey. 2. **Identify given information:** Total students, $N = 20$ Number playing soccer, $S = 16$ Number playing hockey, $H = 12$ Number not playing any games, $N_0 = 2$ 3. **Calculate number of students playing at least one game:** $$N_{game} = N - N_0 = 20 - 2 = 18$$ 4. **Use the principle of inclusion-exclusion for the total playing games:** $$N_{game} = S + H - (S \cap H)$$ Where $S \cap H$ is the number playing both soccer and hockey. 5. **Solve for $S \cap H$:** $$18 = 16 + 12 - (S \cap H)$$ $$18 = 28 - (S \cap H)$$ $$S \cap H = 28 - 18 = 10$$ 6. **Calculate students playing only hockey:** $$\text{Only Hockey} = H - (S \cap H) = 12 - 10 = 2$$ **Final answers:** - Number playing both soccer and hockey = $10$ - Number playing only hockey = $2$