Question

the accompanying venn diagram shows the number of elements in region v. use the given cardinalities to determine the number of elements in each of the other seven regions.
n(u) = 44, n(a) = 23, n(b) = 24,
n(c) = 23, n(a∩b) = 13,
n(a∩c) = 11, n(b∩c) = 14
there are 8 elements in region i.
there are \square elements in region ii.

Explanation:

Step1: Recall Inclusion-Exclusion for Three Sets

The formula for the number of elements in the union of three sets \( A \), \( B \), and \( C \) is \( n(A \cup B \cup C)=n(A)+n(B)+n(C)-n(A \cap B)-n(A \cap C)-n(B \cap C)+n(A \cap B \cap C) \). First, we know \( n(A \cap B \cap C) = n(V)=9 \) (from the Venn diagram, region V is the intersection of all three sets).

Step2: Calculate \( n(A \cap B \cap C) \) Contribution

We know \( n(A \cap B) = 13 \), which is the sum of regions II and V (since \( A \cap B \) includes the part only in \( A \) and \( B \) (region II) and the part in all three (region V)). So \( n(II)=n(A \cap B)-n(A \cap B \cap C)=13 - 9 = 4 \)? Wait, no, wait. Wait, the problem says "There are 8 elements in region I". Wait, region I is the part of \( A \) not in \( B \) or \( C \). Let's re - organize.

Let's define the regions:
- Region I: \( A - (B \cup C) \), \( n(I) = 8 \) (given)
- Region II: \( (A \cap B)-C \)
- Region III: \( (B \cap C)-A \)
- Region IV: \( (A \cap C)-B \)
- Region V: \( A \cap B \cap C \), \( n(V)=9 \)
- Region VI: \( (B \cap C)-A \)? Wait, no, standard Venn diagram regions for three sets:
- I: \( A \) only (\( A - B - C \))
- II: \( A \cap B - C \)
- III: \( B \) only (\( B - A - C \))
- IV: \( A \cap C - B \)
- V: \( A \cap B \cap C \)
- VI: \( B \cap C - A \)
- VII: \( C \) only (\( C - A - B \))
- VIII: \( U-(A \cup B \cup C) \)

We know \( n(A)=n(I)+n(II)+n(IV)+n(V) \). Given \( n(A) = 23 \), \( n(I)=8 \), \( n(V)=9 \), \( n(A \cap C)=n(IV)+n(V)=11 \), so \( n(IV)=n(A \cap C)-n(V)=11 - 9 = 2 \).

Now, \( n(A)=n(I)+n(II)+n(IV)+n(V) \). Substitute the known values: \( 23=8 + n(II)+2 + 9 \).

Step3: Solve for \( n(II) \)

Simplify the right - hand side: \( 8 + 2+9=19 \). Then \( 23=19 + n(II) \). Subtract 19 from both sides: \( n(II)=23 - 19 = 4 \)? Wait, no, wait, the problem says "There are 8 elements in region I". Wait, maybe I misread. Wait, the user's problem has "There are 8 elements in region I" and we need to find region II.

Wait, let's use the formula for \( n(A) \):

\( n(A)=n(I)+n(II)+n(IV)+n(V) \)

We know:
- \( n(A)=23 \)
- \( n(I)=8 \) (given)
- \( n(V)=9 \) (from Venn diagram, region V)
- \( n(A \cap C)=n(IV)+n(V)=11 \), so \( n(IV)=11 - 9 = 2 \)

Substitute into the formula for \( n(A) \):

\( 23=8 + n(II)+2 + 9 \)

\( 8 + 2+9 = 19 \)

\( n(II)=23 - 19 = 4 \)? Wait, but maybe the initial given "There are 8 elements in region I" is part of the problem. Wait, the problem is to find the number of elements in region II.

Wait, let's re - check the intersection values:

\( n(A \cap B)=13 \), which is \( n(II)+n(V) \) (since \( A \cap B \) is the union of region II (only \( A \) and \( B \)) and region V (all three)). So \( n(II)=n(A \cap B)-n(A \cap B \cap C)=13 - 9 = 4 \).

Wait, but let's also use the \( n(A) \) formula. \( n(A)=n(I)+n(II)+n(IV)+n(V) \). We have \( n(I) = 8 \), \( n(IV)=n(A \cap C)-n(A \cap B \cap C)=11 - 9 = 2 \), \( n(V)=9 \), \( n(A)=23 \). So:

\( 23=8 + n(II)+2 + 9 \)

\( 8 + 2+9=19 \)

\( n(II)=23 - 19 = 4 \). Wait, but maybe I made a mistake. Wait, the problem says "There are 8 elements in region I". Let's confirm the regions again.

Region I: \( A \) only (not in \( B \) or \( C \)): \( n(I) = 8 \)

Region II: \( A \cap B \) only (in \( A \) and \( B \), not in \( C \))

Region IV: \( A \cap C \) only (in \( A \) and \( C \), not in \( B \))

Region V: \( A \cap B \cap C \): \( n(V)=9 \)

So \( n(A)=n(I)+n(II)+n(IV)+n(V) \)

We know \( n(A)=23 \), \( n(I)=8 \), \( n(V)=9 \), \( n(A \cap C)=n(IV)+n(V)=11 \implies n(IV)=11 - 9 = 2 \)

So:

\( 23=8 + n(II)+2 + 9 \)

\( 8 + 2+9 = 19 \)

\( n(II)=23 - 19 = 4 \)

Wait, but maybe the problem is different. Wait, the user's problem has "There are 8 elements in region I" and we need to find region II. So the calculation gives \( n(II)=4 \)? Wait, no, maybe I messed up the intersection. Wait, \( n(A \cap B)=13 \), which is region II + region V. So region II = 13 - 9 = 4. That matches.

Answer:

4