the accompanying venn diagram shows the number of elements in region v. use the given cardinalities to…

Explanation:

Step1: Find elements in region IV

Region IV is $n(A\cap B)-n(A\cap B\cap C)$. We know $n(A\cap B) = 8$ and assume $n(A\cap B\cap C)$ is part of the intersections. But since we are not given $n(A\cap B\cap C)$ separately and we know that the elements in the double - intersections are counted in a way that for region IV, we assume the common part with all three sets is already accounted for in the given double - intersection values. So number of elements in region IV is $n(A\cap B)-n(A\cap B\cap C)=8 - 0=8$.

Step2: Find elements in region VI

Region VI is $n(B\cap C)-n(A\cap B\cap C)$. Given $n(B\cap C)=11$, so number of elements in region VI is $11 - 0 = 11$.

Step3: Find elements in region VII

Region VII is $n(A\cap C)-n(A\cap B\cap C)$. Given $n(A\cap C)=9$, so number of elements in region VII is $9 - 0=9$.

Step4: Find elements in region I

Region I is $n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)$. Substituting $n(A) = 19$, $n(A\cap B)=8$, $n(A\cap C)=9$, we get $n(I)=19-(8 + 9)=2$.

Step5: Find elements in region II

Region II is $n(B)-n(A\cap B)-n(B\cap C)+n(A\cap B\cap C)$. Substituting $n(B)=20$, $n(A\cap B)=8$, $n(B\cap C)=11$, we get $n(II)=20-(8 + 11)=1$.

Step6: Find elements in region III

Region III is $n(C)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)$. Substituting $n(C)=23$, $n(A\cap C)=9$, $n(B\cap C)=11$, we get $n(III)=23-(9 + 11)=3$.

Step7: Find elements in region VIII

Region VIII is $n(U)-n(A\cup B\cup C)$. First, find $n(A\cup B\cup C)$ using the principle $n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)$. Substituting values $n(A) = 19$, $n(B)=20$, $n(C)=23$, $n(A\cap B)=8$, $n(A\cap C)=9$, $n(B\cap C)=11$, we get $n(A\cup B\cup C)=19 + 20+23-(8 + 9+11)=34$. Given $n(U)=44$, so $n(VIII)=44 - 34 = 10$.

Answer:

There are 3 elements in region III.