Question

Given three subsets A, BandC of the universal set ζ. Suppose n(ζ) =
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:
\[
n(A \cap B \cap C) = 27
\]