Given three subsets A, BandC of the universal set ζ. Suppose n(ζ) =

Question

68; n(A ∪ B ∪ C) = 64; n(A) = 50; n(B) = 49; n(C) = 46; n(A ∩ B) =
39; n(C ∩ B) = 33 and n(A ∩ C) = 36.
Use Venn diagram to �nd n(A ∩ B ∩ C

Answer 1

To find \( n(A \cap B \cap C) \), we will use the principle of inclusion-exclusion for three sets:

\[ n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C) \]

We're given the following values:

• \( n(A \cup B \cup C) = 64 \)
• \( n(A) = 50 \)
• \( n(B) = 49 \)
• \( n(C) = 46 \)
• \( n(A \cap B) = 39 \)
• \( n(B \cap C) = 33 \)
• \( n(A \cap C) = 36 \)

We want to find \( n(A \cap B \cap C) \). Let's denote it as \( x \). Plugging in the values into the inclusion-exclusion formula:

\[ 64 = 50 + 49 + 46 - 39 - 36 - 33 + x \]

Now let's do the arithmetic step-by-step:

1. Calculate the sum of \( n(A) + n(B) + n(C) \):

   • \( 50 + 49 + 46 = 145 \)

2. Calculate the sum of \( n(A \cap B) + n(A \cap C) + n(B \cap C) \):

   • \( 39 + 36 + 33 = 108 \)

3. Substitute these sums back into the equation: \[ 64 = 145 - 108 + x \]

4. Simplify this: \[ 64 = 37 + x \]

5. Solve for \( x \): \[ x = 64 - 37 = 27 \]

Thus, the value of \( n(A \cap B \cap C) \) is \( 27 \).

To summarize: \[ n(A \cap B \cap C) = 27 \]