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(anb) = 13,
n(anc) = 11, n(bnc) = 14
there are 8 elements in region i.
there are 4 elements in region ii.
there are 6 elements in region iii.
there are 2 elements in region iv.
there are 5 elements in region vi.
there are 7 elements in region vii.
there are \\(\square\\) elements in region viii.

Explanation:

Step1: Recall the principle of inclusion - exclusion for three sets. The formula for \( n(A \cup B \cup C) \) is \( n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C) \). First, we know \( n(A \cap B \cap C) \) is the number of elements in region V, which is 9 (from the Venn diagram).

Step2: Calculate \( n(A \cup B \cup C) \). Substitute the given values: \( n(A)=23 \), \( n(B)=24 \), \( n(C)=23 \), \( n(A \cap B)=13 \), \( n(A \cap C)=11 \), \( n(B \cap C)=14 \), and \( n(A \cap B \cap C)=9 \). So \( n(A \cup B \cup C)=23 + 24 + 23 - 13 - 11 - 14 + 9 \). Calculate step - by - step: \( 23+24 + 23=70 \), \( 13 + 11+14 = 38 \), then \( 70-38 + 9=41 \).

Step3: The total number of elements in the universal set \( n(U)=44 \). The number of elements in region VIII (the complement of \( A\cup B\cup C \)) is \( n(U)-n(A \cup B \cup C) \). So \( 44 - 41 = 3 \)? Wait, no, wait. Wait, the regions: Let's list the regions. Region I: only A, Region II: \( A\cap B \) only, Region III: only B, Region IV: \( A\cap C \) only, Region V: \( A\cap B\cap C \), Region VI: \( B\cap C \) only, Region VII: only C, Region VIII: outside all three.

First, find the number of elements in each sub - region:

- Region II (only \( A\cap B \)): \( n(A\cap B)-n(A\cap B\cap C)=13 - 9 = 4 \)
- Region IV (only \( A\cap C \)): \( n(A\cap C)-n(A\cap B\cap C)=11 - 9 = 2 \)
- Region VI (only \( B\cap C \)): \( n(B\cap C)-n(A\cap B\cap C)=14 - 9 = 5 \)
- Region I (only A): \( n(A)-n(\text{only }A\cap B)-n(\text{only }A\cap C)-n(A\cap B\cap C)=23-(4 + 2+9)=23 - 15 = 8 \)
- Region III (only B): \( n(B)-n(\text{only }A\cap B)-n(\text{only }B\cap C)-n(A\cap B\cap C)=24-(4 + 5+9)=24 - 18 = 6 \)
- Region VII (only C): \( n(C)-n(\text{only }A\cap C)-n(\text{only }B\cap C)-n(A\cap B\cap C)=23-(2 + 5+9)=23 - 16 = 7 \)
- Now, sum up the number of elements in all regions except VIII: \( 8+4 + 6+2+9+5+7=41 \)
- Then, the number of elements in region VIII is \( n(U)-n(A\cup B\cup C)=44 - 41 = 3 \)? Wait, but the problem says "There are \(\square\) elements in region VIII". Wait, maybe I made a mistake. Wait, let's re - check:

Wait, the sum of elements in regions I - VII:

Region I: 8, Region II: 4, Region III: 6, Region IV: 2, Region V: 9, Region VI: 5, Region VII: 7.

Sum: \( 8 + 4=12 \), \( 12+6 = 18 \), \( 18+2 = 20 \), \( 20+9 = 29 \), \( 29+5 = 34 \), \( 34+7 = 41 \). And \( n(U)=44 \), so region VIII: \( 44-41 = 3 \). But wait, maybe the problem has a typo or I misread. Wait, no, let's check the given regions:

Wait, the problem states:

There are 8 elements in region I (which matches our calculation: 8)

There are 4 elements in region II (matches \( 13 - 9 = 4 \))

There are 6 elements in region III (matches \( 24-(4 + 5+9)=6 \))

There are 2 elements in region IV (matches \( 11 - 9 = 2 \))

There are 5 elements in region VI (matches \( 14 - 9 = 5 \))

There are 7 elements in region VII (matches \( 23-(2 + 5+9)=7 \))

Then, sum of regions I - VII: \( 8 + 4+6 + 2+9+5+7 = 41 \)

Since \( n(U)=44 \), the number of elements in region VIII is \( 44-41 = 3 \). Wait, but the box is empty. Wait, maybe I made a mistake in the Venn diagram regions. Wait, the Venn diagram has three circles: A, B, C. Region I: only A, Region II: A and B only, Region III: only B, Region IV: A and C only, Region V: A, B, C, Region VI: B and C only, Region VII: only C, Region VIII: outside all.

Yes, so the calculation is correct. So the number of elements in region VIII is \( 44-(8 + 4+6 + 2+9+5+7)=44 - 41 = 3 \). Wait, but maybe the problem's given values are different? Wait, no, the given \( n(U) = 44 \), \( n(A)=23 \), \( n(B)=24 \), \( n(C)=23 \), \( n(A\cap B)=13 \), \( n(A\cap C)=11 \), \( n(B\cap C)=14 \), and region V (A\cap B\cap C) has 9 elements.

So the sum of all regions except VIII is \( 8+4 + 6+2+9+5+7 = 41 \), so region VIII has \( 44 - 41 = 3 \). Wait, but maybe I miscalculated the sum. Let's add again: 8 (I) + 4 (II) = 12; 12+6 (III)=18; 18+2 (IV)=20; 20+9 (V)=29; 29+5 (VI)=34; 34+7 (VII)=41. Then 44 - 41 = 3. So the number of elements in region VIII is 3.

Answer:

3