Denotemos por n(X) o número de elementos de um conjunto finito X. Sejam A, B e C conjuntos tais que n(A ⋃ B) = 8, n(A ⋃ C) = 9, n(B ⋃ C) = 10, n(A ...

Para resolver esse problema, podemos usar o princípio da inclusão-exclusão. Sabemos que n(A ⋃ B ⋃ C) = 11, que é a união dos três conjuntos. Também sabemos que n(A ⋂ B ⋂ C) = 2, que é a interseção dos três conjuntos. Podemos usar a fórmula da inclusão-exclusão para encontrar o número de elementos em A, B e C: n(A ⋃ B ⋃ C) = n(A) + n(B) + n(C) - n(A ⋂ B) - n(A ⋂ C) - n(B ⋂ C) + n(A ⋂ B ⋂ C) Substituindo os valores conhecidos: 11 = n(A) + n(B) + n(C) - n(A ⋂ B) - n(A ⋂ C) - n(B ⋂ C) + 2 Agora, vamos usar as informações fornecidas: n(A ⋃ B) = 8 n(A ⋃ C) = 9 n(B ⋃ C) = 10 Podemos reescrever essas equações usando a fórmula da inclusão-exclusão: n(A ⋃ B) = n(A) + n(B) - n(A ⋂ B) n(A ⋃ C) = n(A) + n(C) - n(A ⋂ C) n(B ⋃ C) = n(B) + n(C) - n(B ⋂ C) Substituindo os valores conhecidos: 8 = n(A) + n(B) - n(A ⋂ B) 9 = n(A) + n(C) - n(A ⋂ C) 10 = n(B) + n(C) - n(B ⋂ C) Agora, podemos resolver esse sistema de equações para encontrar os valores de n(A), n(B) e n(C). Somando as três equações, temos: 8 + 9 + 10 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) Agora, substituindo o valor de n(A ⋂ B ⋂ C) = 2, temos: 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n(A ⋂ B) + n(A ⋂ C) + n(B ⋂ C)) + 2 27 = 2n(A) + 2n(B) + 2n(C) - (n