Give the numbers of elements in the regions marked I, II, III, IV, V, VI, VII, VIII in the figure.
|U| = 280, |A| = 20, |B| = 20, |C| = 24 |A�¿B| = 3, |A�¿C| = 3, |B�¿C| = 5, |A�¿B�¿C| = 2
| I | = 1 .
| II | = 2 .
| III | = 3 .
| IV | = 4 .
| V | = 5 .
| VI | = 6 .
| VII | = 7 .
| VIII | = 8 .
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To determine the number of elements in each region, we can use the principle of inclusion-exclusion.
First, let's define the variables:
|U| = 280 (total number of elements in the universal set)
|A| = 20 (number of elements in set A)
|B| = 20 (number of elements in set B)
|C| = 24 (number of elements in set C)
|A∩B| = 3 (number of elements in the intersection of sets A and B)
|A∩C| = 3 (number of elements in the intersection of sets A and C)
|B∩C| = 5 (number of elements in the intersection of sets B and C)
|A∩B∩C| = 2 (number of elements in the intersection of sets A, B, and C)
Now, let's calculate the number of elements in each region:
|I| = |A - (A∩B∩C)| = |A| - |A∩B∩C| = 20 - 2 = 18
|II| = |B - (A∩B∩C)| = |B| - |A∩B∩C| = 20 - 2 = 18
|III| = |C - (A∩B∩C)| = |C| - |A∩B∩C| = 24 - 2 = 22
|IV| = |A∩B - A∩B∩C| = |A∩B| - |A∩B∩C| = 3 - 2 = 1
|V| = |A∩C - A∩B∩C| = |A∩C| - |A∩B∩C| = 3 - 2 = 1
|VI| = |B∩C - A∩B∩C| = |B∩C| - |A∩B∩C| = 5 - 2 = 3
|VII| = |A∩B∩C| = 2
|VIII| = |U - (A∪B∪C)| = |U| - (|A| + |B| + |C| - |A∩B| - |A∩C| - |B∩C| + |A∩B∩C|) = 280 - (20 + 20 + 24 - 3 - 3 - 5 + 2) = 218
So, the numbers of elements in the regions marked I, II, III, IV, V, VI, VII, and VIII are 18, 18, 22, 1, 1, 3, 2, and 218, respectively.