To solve for $n(B)$, we need to interpret and use the given information. Let’s go over it step-by-step.

Given Information:

1. $n(A) = 17$
2. $n(U) = 42$ (where $U$ is the universal set)
3. $n(C) = 22$
4. $n(B) = 2 \times n(A \cap C) = 4 \times n(A \cap B \cap C)$
5. $n(A \cap B) = 8$
6. $n(B \cap C) = 9$
7. $n((A \cup B \cup C)^c) = 5$
8. $n(A \cup B \cup C) = 37$

Step 1: Calculate $n(A \cap C)$ and $n(A \cap B \cap C)$ from $n(B)$

From the fourth given condition: $n(B) = 2 \times n(A \cap C) = 4 \times n(A \cap B \cap C)$

This tells us that $n(A \cap C)$ is half of $n(B)$ and $n(A \cap B \cap C)$ is a quarter of $n(B)$.

Let’s denote $n(B)$ by $x$. Then: $n(A \cap C) = \frac{x}{2}$ $n(A \cap B \cap C) = \frac{x}{4}$

Step 2: Use the Union Formula for Three Sets

The formula for the union of three sets is: $n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)$

Substitute the known values and expressions: $37 = 17 + x + 22 - 8 - \frac{x}{2} - 9 + \frac{x}{4}$

Step 3: Simplify the Equation

Combine like terms: $37 = 31 + x - \frac{x}{2} + \frac{x}{4}$

Combine terms involving $x$ on the right side: $37 = 31 + \left( x - \frac{x}{2} + \frac{x}{4} \right)$

Rewrite $x - \frac{x}{2} + \frac{x}{4}$ as a single fraction: $x - \frac{x}{2} + \frac{x}{4} = \frac{4x - 2x + x}{4} = \frac{3x}{4}$

So now we have: $37 = 31 + \frac{3x}{4}$

Step 4: Solve for $x$

Subtract 31 from both sides: $6 = \frac{3x}{4}$

Multiply both sides by $\frac{4}{3}$: $x = 8$

Conclusion:

$n(B) = 8$

Would you like further details on any of these steps, or have any questions?

Related Questions:

1. How would the solution change if $n(A \cap B)$ were different?
2. What if the value of $n(U)$ changed? How would it affect the calculation?
3. How does the formula for $n(A \cup B \cup C)$ apply if only two sets are involved?
4. Can we calculate $n(A \cap C)$ and $n(A \cap B \cap C)$ directly with $n(B) = 8$?
5. What would happen if $n((A \cup B \cup C)^c)$ were larger?

Tip:

Always substitute values and simplify expressions carefully to avoid algebraic errors, especially when using formulas with multiple terms like the union of sets.