Let's break down the information and solve the problem.

We have a Venn diagram involving three sets $A$, $B$, and $C$ along with their cardinalities (number of elements). The following data is provided:

• $n(A) = 26$
• $n(B) = 30$
• $n(C) = 23$
• $n(A \cap B) = 10$
• $n(B \cap C) = 11$
• $n(A \cap C) = 7$
• $n(A \cap B \cap C) = 4$
• $n(U) = 64$ (The total number of elements in the universal set $U$)

Labeling the regions

The Venn diagram regions are labeled as follows:

• I: Outside of all sets
• II: Only in $A$
• III: Only in $C$
• IV: Only in $B$
• V: In all three sets ($A \cap B \cap C$)
• VI: In $A$ and $B$, but not $C$
• VII: In $A$ and $C$, but not $B$
• VIII: In $B$ and $C$, but not $A$

Using the data to fill in the regions:

• $n(A \cap B \cap C) = 4$ corresponds to region V.
• $n(A \cap B) = 10$, so region VI has $10 - 4 = 6$ elements (since 4 elements are already in region V).
• $n(B \cap C) = 11$, so region VIII has $11 - 4 = 7$ elements.
• $n(A \cap C) = 7$, so region VII has $7 - 4 = 3$ elements.

Now we calculate the remaining elements in sets $A$, $B$, and $C$:

• $n(A) = 26$, so region II has $26 - (6 + 4 + 3) = 13$ elements.
• $n(B) = 30$, so region IV has $30 - (6 + 4 + 7) = 13$ elements.
• $n(C) = 23$, so region III has $23 - (3 + 4 + 7) = 9$ elements.

Finally, we calculate the number of elements in region I (outside all sets):

• $n(U) = 64$, so region I has $64 - (13 + 6 + 3 + 13 + 9 + 7 + 4) = 9$ elements.

Solution for Region II:

• Region II contains 13 elements.

Would you like a breakdown of any specific parts, or do you have additional questions?

Here are some related questions for practice:

1. What is the number of elements in regions VII and VIII combined?
2. How many elements are present in $B \cup C$?
3. What is the number of elements in the set $(A \cup B \cup C)^c$ (outside all sets)?
4. How would the numbers change if $n(A \cap B \cap C)$ increased to 5?
5. How many elements are only in $A \cup C$, but not in $B$?

Tip: In Venn diagrams with three sets, breaking the diagram into distinct regions based on the given intersections simplifies the process of filling in values.