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)=19, n(b)=20, n(c)=23, n(a∩b)=8, n(a∩c)=9, n(b∩c)=11. there are 8 elements in region i. there are 2 elements in region ii. there are elements in region iii.

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.