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 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 \square elements in region vii.
venn diagram image with regions i, ii, iii, iv, v (with 9), vi, vii, viii

Explanation:

Step1: Recall Venn Diagram Region Formulas

For three sets \(A\), \(B\), \(C\), the number of elements in the intersection of all three (\(A\cap B\cap C\)) is related to the pairwise intersections. The formula for \(n(A\cap B)\) is \(n(A\cap B)=n(A\cap B\cap C)+n(\text{region II})\), but here we know \(n(A\cap B) = 13\), \(n(A\cap C)=11\), \(n(B\cap C)=14\), and region V (the triple intersection) has 9 elements (from the diagram: "region V" has 9). Wait, actually, let's define the regions:
- Region V: \(A\cap B\cap C\), so \(n(A\cap B\cap C)=9\) (given in the diagram as region V has 9? Wait, the problem says "the accompanying Venn diagram shows the number of elements in region V" – maybe region V is \(A\cap B\cap C\) with 9? Wait, no, let's re - express the pairwise intersections:
The formula for \(n(A\cap B)\) is \(n(A\cap B)=n(A\cap B\cap C)+n(\text{region II})\). Similarly, \(n(A\cap C)=n(A\cap B\cap C)+n(\text{region IV})\) and \(n(B\cap C)=n(A\cap B\cap C)+n(\text{region VI})\).
We know \(n(A\cap B) = 13\), \(n(A\cap C)=11\), \(n(B\cap C)=14\), and let \(n(A\cap B\cap C)=x\) (region V). Wait, maybe the diagram shows region V as 9? Wait, the user's diagram has "region V" with 9? Wait, the problem says "the accompanying Venn diagram shows the number of elements in region V" – maybe region V is \(A\cap B\cap C\) with 9. Let's assume that.
So, for region II (part of \(A\cap B\) not in \(C\)): \(n(\text{region II})=n(A\cap B)-n(A\cap B\cap C)=13 - 9=4\)
For region IV (part of \(A\cap C\) not in \(B\)): \(n(\text{region IV})=n(A\cap C)-n(A\cap B\cap C)=11 - 9 = 2\)
For region VI (part of \(B\cap C\) not in \(A\)): \(n(\text{region VI})=n(B\cap C)-n(A\cap B\cap C)=14 - 9=5\)

Step2: Calculate Region I (only in \(A\))

\(n(\text{region I})=n(A)-n(\text{region II})-n(\text{region IV})-n(A\cap B\cap C)\)
We know \(n(A) = 23\), \(n(\text{region II}) = 4\), \(n(\text{region IV})=2\), \(n(A\cap B\cap C)=9\)
So \(n(\text{region I})=23-(4 + 2+9)=23 - 15 = 8\) (matches the given "There are 8 elements in region I")

Step3: Calculate Region III (only in \(B\))

\(n(\text{region III})=n(B)-n(\text{region II})-n(\text{region VI})-n(A\cap B\cap C)\)
\(n(B)=24\), \(n(\text{region II}) = 4\), \(n(\text{region VI})=5\), \(n(A\cap B\cap C)=9\)
\(n(\text{region III})=24-(4 + 5+9)=24 - 18=6\) (matches "There are 6 elements in region III")

Step4: Calculate Region VII (only in \(C\))

\(n(\text{region VII})=n(C)-n(\text{region IV})-n(\text{region VI})-n(A\cap B\cap C)\)
\(n(C)=23\), \(n(\text{region IV})=2\), \(n(\text{region VI})=5\), \(n(A\cap B\cap C)=9\)
\(n(\text{region VII})=23-(2 + 5+9)=23 - 16 = 7\)? Wait, no, wait the problem has "There are \(\square\) elements in region VII". Wait, maybe I made a mistake. Wait, let's use the total number of elements in the universal set.
The total number of elements in all regions: \(n(U)=n(\text{region I})+n(\text{region II})+n(\text{region III})+n(\text{region IV})+n(\text{region V})+n(\text{region VI})+n(\text{region VII})+n(\text{region VIII})\)
We know \(n(U) = 44\), \(n(\text{region I}) = 8\), \(n(\text{region II}) = 4\), \(n(\text{region III}) = 6\), \(n(\text{region IV}) = 2\), \(n(\text{region V}) = 9\), \(n(\text{region VI}) = 5\)
Let \(n(\text{region VII})=x\) and \(n(\text{region VIII})=y\)
So \(44=8 + 4+6 + 2+9 + 5+x + y\)
\(44=34+x + y\)
\(x + y=10\)
But we need to find \(n(\text{region VII})\). Wait, maybe region VIII is the part of the universal set not in \(A\), \(B\), or \(C\). But we can also calculate \(n(\text{region VII})\) as \(n(C)-n(\text{region IV})-n(\text{region V})-n(\text{region VI})\)
\(n(C)=23\), \(n(\text{region IV}) = 2\), \(n(\text{region V}) = 9\), \(n(\text{region VI}) = 5\)
\(n(\text{region VII})=23-(2 + 9+5)=23 - 16 = 7\)? But the problem's blanks: Wait, maybe the diagram's region V is not 9? Wait, the problem says "the accompanying Venn diagram shows the number of elements in region V". Wait, maybe I misread. Let's re - check the given data:
\(n(U)=44\), \(n(A)=23\), \(n(B)=24\), \(n(C)=23\), \(n(A\cap B)=13\), \(n(A\cap C)=11\), \(n(B\cap C)=14\)
The formula for the number of elements in the union of three sets is \(n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)\)
We can find \(n(A\cap B\cap C)\) first? Wait, no, we know the regions. Let's list all regions:
- Region I: only \(A\): \(n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)\) (by principle of inclusion - exclusion for only \(A\))
Wait, the formula for only \(A\) is \(n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)\)
So \(n(\text{region I})=23-13 - 11+9=8\) (matches)
Only \(B\): \(n(B)-n(A\cap B)-n(B\cap C)+n(A\cap B\cap C)=24 - 13-14 + 9=6\) (matches)
Only \(C\): \(n(C)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)=23-11 - 14+9=7\)? But the problem's last blank is for region VII (only \(C\))? Wait, maybe the diagram's region V is \(A\cap B\cap C\) with 9, and we need to find region VII (only \(C\)):
\(n(\text{region VII})=n(C)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)=23-11 - 14 + 9=7\)? But the problem's options (the blanks) have "There are \(\square\) elements in region VII". Wait, maybe I made a mistake. Wait, let's recalculate the total:
Sum of known regions: region I (8) + region II (4) + region III (6) + region IV (2) + region V (9) + region VI (5) = 8 + 4+6 + 2+9 + 5=34
Total \(n(U)=44\), so the sum of region VII and region VIII is \(44 - 34 = 10\)
But region VIII is the part not in \(A\), \(B\), or \(C\). But we need to find region VII (only \(C\)):
\(n(\text{region VII})=n(C)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)=23-(11 + 14 - 9)=23-(16)=7\)
Wait, but maybe the diagram's region V is not 9? Wait, the problem says "the accompanying Venn diagram shows the number of elements in region V". Maybe region V is 9, so the calculation for region VII is 7? But the problem's blanks: "There are \(\square\) elements in region VII". Maybe I made a mistake in the region definitions.
Wait, let's re - define the regions:
- Region I: \(A - (B\cup C)\)
- Region II: \(A\cap B - C\)
- Region III: \(B - (A\cup C)\)
- Region IV: \(A\cap C - B\)
- Region V: \(A\cap B\cap C\)
- Region VI: \(B\cap C - A\)
- Region VII: \(C - (A\cup B)\)
- Region VIII: \(U-(A\cup B\cup C)\)

So \(n(\text{region I})=n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)=23 - 13-11 + 9 = 8\)
\(n(\text{region II})=n(A\cap B)-n(A\cap B\cap C)=13 - 9 = 4\)
\(n(\text{region III})=n(B)-n(A\cap B)-n(B\cap C)+n(A\cap B\cap C)=24 - 13-14 + 9 = 6\)
\(n(\text{region IV})=n(A\cap C)-n(A\cap B\cap C)=11 - 9 = 2\)
\(n(\text{region VI})=n(B\cap C)-n(A\cap B\cap C)=14 - 9 = 5\)
Now, \(n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)=23 + 24+23-13 - 11-14 + 9=41\)
Then \(n(\text{region VIII})=n(U)-n(A\cup B\cup C)=44 - 41 = 3\)
And \(n(\text{region VII})=n(C)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)=23-11 - 14 + 9 = 7\)

Wait, but the problem's last blank is for region VII, so the number of elements in region VII is 7? But maybe the diagram's region V is not 9? Wait, the problem says "the accompanying Venn diagram shows the number of elements in region V". Maybe region V is 9, so the answer for region VII is 7. But let's check the sum again:

Region I:8, II:4, III:6, IV:2, V:9, VI:5, VII:7, VIII:3

Sum:8 + 4+6 + 2+9 + 5+7 + 3=44, which matches \(n(U)=44\).

So the number of elements in region VII is 7. But wait, the problem's blanks: "There are \(\square\) elements in region VII". Maybe I made a mistake in the initial assumption of region V. Wait, the user's diagram has "region V" with 9, so that's correct.

Answer:

7