a⊂e, b⊂e,\nn(a)+n(b) = 17,\nn(b)+n(a) = 23\n⇒...
a⊂e, b⊂e,\nn(a)+n(b) = 17,\nn(b)+n(a) = 23\n⇒ n(e)=?
Answer
# Answer:
20
# Explanation:
## Step1: Recall set - theory formula
We know that \(n(A)+n(A') = n(E)\) and \(n(B)+n(B')=n(E)\). Also, \(n(A)+n(B') = 17\) and \(n(B)+n(A')=23\).
## Step2: Add the two given equations
\((n(A)+n(B'))+(n(B)+n(A'))=17 + 23\).
## Step3: Rearrange the left - hand side
\((n(A)+n(A'))+(n(B)+n(B'))=40\).
## Step4: Substitute \(n(A)+n(A') = n(E)\) and \(n(B)+n(B')=n(E)\)
\(n(E)+n(E)=40\).
## Step5: Solve for \(n(E)\)
\(2n(E)=40\), so \(n(E)=\frac{40}{2}=20\).