Question

use the venn diagram and the given conditions to determine the number of elements in each region, or explain why the conditions are impossible to meet.\
\\( n(u)=18,\\ n(a)=16,\\ n(b)=14,\\ n(c)=13,\\ n(a\cap b)=12,\\ n(a\cap c)=14,\\ n(b\cap c)=6,\\ n(a\cap b\cap c)=5 \\)\
select the correct choice below and fill in any answer boxes within your choice.\
a. the number of elements in regions i, ii, iii, iv, v, vi, vii, viii are \\( \square,\\ \square,\\ \square,\\ \square,\\ \square,\\ \square,\\ \square,\\ \square \\) respectively.\
b. it is impossible to meet the conditions because there are only \\( \square \\) elements in set a but there are \\( \square \\) elements in set a that are also in set b or c. a similar problem exists for set c. (simplify your answers.)\
c. it is impossible to meet the conditions because there are only \\( \square \\) elements in set b but there are \\( \square \\) elements in set b that are also in set a or c. a similar problem exists for set c. (simplify your answers.)

Explanation:

Step1: Analyze \( n(A \cap B) \) and \( n(A \cap B \cap C) \)

The formula for the number of elements in \( A \cap B \) only (region IV) is \( n(A \cap B) - n(A \cap B \cap C) \). Substituting values: \( 12 - 5 = 7 \). But \( n(A \cap C) = 14 \), and \( n(A \cap B \cap C) = 5 \), so the number of elements in \( A \cap C \) only (region V) would be \( 14 - 5 = 9 \). Now, the number of elements in \( A \) only (region I) is \( n(A) - [n(A \cap B) - n(A \cap B \cap C) + n(A \cap C) - n(A \cap B \cap C) + n(A \cap B \cap C)] \). Wait, actually, the correct formula for \( n(A) \) is \( n(\text{I}) + n(\text{II}) + n(\text{IV}) + n(\text{V}) \). Wait, no, let's re - express the regions:

• Region II: \( A \cap B \) only (not in \( C \)): \( n(A \cap B) - n(A \cap B \cap C)=12 - 5 = 7 \)
• Region IV: \( A \cap C \) only (not in \( B \)): \( n(A \cap C)-n(A \cap B \cap C)=14 - 5 = 9 \)
• Region V: \( A \cap B \cap C \): \( 5 \)

Now, the number of elements in set \( A \) is \( n(A)=n(\text{I})+n(\text{II})+n(\text{IV})+n(\text{V}) \). So \( n(\text{I})=n(A)-n(\text{II})-n(\text{IV})-n(\text{V})=16-(7 + 9+5)=16 - 21=- 5 \). But the number of elements in a set can't be negative.

Step2: Analyze the impossibility

We see that when calculating the number of elements in region I (elements only in \( A \)), we get a negative number (\(16-(7 + 9 + 5)=- 5\)). Also, for set \( B \), let's check the intersection with \( A \) and \( C \). The number of elements in \( B \cap C \) only (region VI) is \( n(B \cap C)-n(A \cap B \cap C)=6 - 5 = 1 \). The number of elements in \( B \) only (region III) is \( n(B)-n(\text{II})-n(\text{VI})-n(\text{V})=14-(7 + 1+5)=14 - 13 = 1 \). But the problem with set \( A \) is more critical. The number of elements in \( A \) that are also in \( B \) or \( C \) is \( n(A \cap B)+n(A \cap C)-n(A \cap B \cap C)=12 + 14-5 = 21 \), and \( n(A)=16 \). Since \( 21>16 \), it's impossible because the number of elements in \( A \) that are in \( B \) or \( C \) (21) is greater than the number of elements in \( A \) (16). So option B is correct. The number of elements in set \( A \) that are also in \( B \) or \( C \) is \( 12 + 14-5 = 21 \), and there are only 16 elements in set \( A \). A similar problem exists for set \( C \) (we can check \( n(A \cap C)+n(B \cap C)-n(A \cap B \cap C)=14 + 6-5 = 15 \), and \( n(C)=13 \), \( 15>13 \)).

Answer:

B. It is impossible to meet the conditions because there are only 16 elements in set A but there are 21 elements in set A that are also in set B or C. A similar problem exists for set C. (Simplify your answers.)