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.
$n(u)=44$, $n(a)=17$, $n(b)=22$,
$n(c)=29$, $n(acap b)=10$,
$n(acap c)=12$, $n(bcap c)=15$
there are $square$ elements in region i.
there are $square$ elements in region ii.
there are $square$ elements in region iii.
there are $square$ elements in region iv.
there are $square$ elements in region vi.
there are $square$ elements in region vii.
there are $square$ elements in region viii.

Explanation:

Step1: Find region II

Region II is \(n(A\cap B)-n(A\cap B\cap C)\). Given \(n(A\cap B) = 10\) and \(n(A\cap B\cap C)=n(V) = 7\).
\(10 - 7=3\)

Step2: Find region IV

Region IV is \(n(A\cap C)-n(A\cap B\cap C)\). Given \(n(A\cap C) = 12\) and \(n(A\cap B\cap C)=7\).
\(12 - 7 = 5\)

Step3: Find region I

Region I is \(n(A)-n(\text{region II})-n(\text{region IV})-n(\text{region V})\). Given \(n(A)=17\), \(n(\text{region II}) = 3\), \(n(\text{region IV})=5\), \(n(\text{region V}) = 7\).
\(17-(3 + 5+7)=17 - 15=2\)

Step4: Find region VI

Region VI is \(n(B\cap C)-n(A\cap B\cap C)\). Given \(n(B\cap C)=15\) and \(n(A\cap B\cap C) = 7\).
\(15 - 7=8\)

Step5: Find region III

Region III is \(n(B)-n(\text{region II})-n(\text{region V})-n(\text{region VI})\). Given \(n(B)=22\), \(n(\text{region II}) = 3\), \(n(\text{region V})=7\), \(n(\text{region VI}) = 8\).
\(22-(3 + 7+8)=22 - 18=4\)

Step6: Find region VII

Region VII is \(n(C)-n(\text{region IV})-n(\text{region V})-n(\text{region VI})\). Given \(n(C)=29\), \(n(\text{region IV}) = 5\), \(n(\text{region V})=7\), \(n(\text{region VI}) = 8\).
\(29-(5 + 7+8)=29 - 20=9\)

Step7: Find region VIII

Region VIII is \(n(U)-n(\text{region I})-n(\text{region II})-n(\text{region III})-n(\text{region IV})-n(\text{region V})-n(\text{region VI})-n(\text{region VII})\). Given \(n(U)=44\), \(n(\text{region I}) = 2\), \(n(\text{region II}) = 3\), \(n(\text{region III})=4\), \(n(\text{region IV}) = 5\), \(n(\text{region V})=7\), \(n(\text{region VI}) = 8\), \(n(\text{region VII}) = 9\).
\(44-(2 + 3+4 + 5+7+8+9)=44 - 38=6\)

Answer:

There are \(2\) elements in region I.
There are \(3\) elements in region II.
There are \(4\) elements in region III.
There are \(5\) elements in region IV.
There are \(8\) elements in region VI.
There are \(9\) elements in region VII.
There are \(6\) elements in region VIII.