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Olá,
n(AUBUC) = n(AU(BUC)) = n(A) + n(BUC) - n(A inter
(BUC)) = n(A) + n(B) + n(C) - n(B inter C) - n((A inter B) U (A inter C)) = n(A)
+ n(B) + n(C) - n(B inter C) - n(A inter B) - n(A inter C) + n(A inter B inter
C)
deste modo:
n(AUBUC) = n(A) + n(B) + n(C) - n(B inter C) - n(A
inter B) - n(A inter C) + n(A inter B inter C) Seja A o conjunto dos brinquedos de Eduardo, B os
de Felipe e C os de Hugo, entao:
n(AUBUC) = n(A) + n(B) + n(C) - n(B inter C) - n(A
inter B) - n(A inter C) + n(A inter B inter C) = 61
n(AUB) = n(A) + n(B) - n(A inter B) = 45
(a)
n(AUC) = n(A) + n(C) - n(A inter C) = 48
(b)
n(BUC) = n(B) + n(C) - n(B inter C) = 49
(c)
n(A inter C) - n(A inter B inter C) = 5
(d)
n(A inter B) - n(A inter B inter C) = 8
(e)
Fazendo (a) + (b) + (c) - (d) - (e),
temos:
2n(A) + 2n(B) + 2n(C) - 2n(A inter B) - 2n(A inter
C) + 2n(A inter B inter C) - n(B inter C) = 129
2 [ n(AUBUC) + n(B inter C) ] - n(B inter C) =
129
2*61 + n(B inter C) = 129
n(B inter C) = 7
abraços,
Salhab
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