A={a,b,p},B={2,3},C={p,q,r,s} then n[(AUC)×B]is
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AUC = {a,b,p,q,r,s}
(AUC)×B = {(a,2),(b,2),(p,2),(q,2),(r,2),(s,2),(a,3),(b,3),(p,3),(q,3),(r,3),(s,3)}
n[(AUC)×B] = 12
#Secretgirl✌️
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0
Answer:
12
Step-by-step explanation:
Given that
B = {2,3}
A = {a,p,b} and C = {p,q,r,s}
A ∩ B = {p}
⇒ n(A) = 3, n(B) = 2, n(C) = 4, n(A ∩ C) = 1
We know that,
n(A ∪ C) = n(A) + n(C) - n(A ∩ C)
⇒n(A ∪ C) = 3+4-1=6
If n(X) = p and n(Y) = q then we have n(XxY) = pq,
⇒ n[(AUC)×B] = n((AUC)) × n(B) = 6×2 = 12
#SPJ3
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