Question

use the accompanying venn diagram, which shows the number of elements in region ii to answer the following problem.
if ( n(a)=43 ), ( n(b)=36 ), and ( n(u)=71 ), find the number of elements in each of regions i, iii, and iv.
region i has ( square ) element(s).
region iii has ( square ) element(s).
region iv has ( square ) element(s).

Explanation:

Step1: Find the number of elements in region I

Region I is part of set \(A\) but not in the intersection of \(A\) and \(B\).
We know that \(n(A)=n(\text{I}) + n(\text{II})\). Given \(n(A) = 43\) and \(n(\text{II})=17\), then \(n(\text{I})=n(A)-n(\text{II})\)
\(n(\text{I})=43 - 17=26\)

Step2: Find the number of elements in region III

Region III is part of set \(B\) but not in the intersection of \(A\) and \(B\).
We know that \(n(B)=n(\text{III})+n(\text{II})\). Given \(n(B) = 36\) and \(n(\text{II}) = 17\), then \(n(\text{III})=n(B)-n(\text{II})\)
\(n(\text{III})=36 - 17 = 19\)

Step3: Find the number of elements in region IV

The universal set \(U\) is composed of regions I, II, III, and IV. So \(n(U)=n(\text{I})+n(\text{II})+n(\text{III})+n(\text{IV})\)
We know \(n(U) = 71\), \(n(\text{I}) = 26\), \(n(\text{II})=17\), \(n(\text{III}) = 19\)
Substitute the values into the formula: \(71=26 + 17+19+n(\text{IV})\)
First, calculate \(26+17 + 19=62\)
Then \(n(\text{IV})=n(U)-(n(\text{I})+n(\text{II})+n(\text{III}))\)
\(n(\text{IV})=71-62 = 9\)

Answer:

Region I has \(26\) element(s).
Region III has \(19\) element(s).
Region IV has \(9\) element(s).