2n (A setminus B) = n (B setminus A) and 5n ( | vedclass

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(A) Given $2n(A \setminus B) = n(B \setminus A)$. Since $n(A \setminus B) = n(A) - n(A \cap B)$ and $n(B \setminus A) = n(B) - n(A \cap B)$,we have:$2

Vedclass · Accepted Answer

$63$

Vedclass · Answer

(A) Given $2n(A \setminus B) = n(B \setminus A)$. Since $n(A \setminus B) = n(A) - n(A \cap B)$ and $n(B \setminus A) = n(B) - n(A \cap B)$,we have:$2(n(A) - n(A \cap B)) = n(B) - n(A \cap B)$$2n(A) - 2n(A \cap B) = n(B) - n(A \cap B)$$n(A \cap B) = 2n(A) - n(B) \quad \dots(1)$Also given $5n(A \cap B) = n(A) + 3n(B) \quad \dots(2)$Substitute $(1)$ into $(2)$:$5(2n(A) - n(B)) = n(A) + 3n(B)$$10n(A) - 5n(B) = n(A) + 3n(B)$$9n(A) = 8n(B) \implies \frac{n(A)}{n(B)} = \frac{8}{9}$.Let $n(A) = 8k$ and $n(B) = 9k$. Then $n(A \cap B) = 2(8k) - 9k = 7k$.We know $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 8k + 9k - 7k = 10k$.Given $n(A \cup B) \leq 10$,so $10k \leq 10 \implies k = 1$.Thus,$n(A) = 8, n(B) = 9, n(A \cap B) = 7$.The required value is $\frac{8 \cdot 9 \cdot 7}{8} = 9 \cdot 7 = 63$.

$2n (A \setminus B) = n (B \setminus A)$ and $5n (A \cap B) = n (A) + 3n (B)$,where $P \setminus Q = P \cap Q^C$. If $n (A \cup B) \leq 10$,then the value of $\frac{n(A) \cdot n(B) \cdot n(A \cap B)}{8}$ is:

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