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Chapter 1, topic 2: Venn Diagrams Venn Diagram – a pictorial representation of sets. The universal set U is presented pictorially by a rectangle and the circles inside represents the subsets. Representations Disjoint set:

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Page 1: venn diagram (sets)

Chapter 1, topic 2: Venn Diagrams

Venn Diagram – a pictorial representation of sets. The universal set U is presented pictorially by a rectangle and the circles inside represents the subsets.

Representations

Disjoint set:

Page 2: venn diagram (sets)

Sets A and B are disjoint as seen in the Venn Diagram. Disjoints sets are characterized by having no two elements on both sets the same. The intesection of this set is ɸ(null/empty).

Joint Set:

In the Venn diagram presented, A and B are joint sets. Joint sets are those sets that has one or more elements both in set A and Set B.

Venn Diagrams

A. Complement

A’ ( A-dash) is also known as complement of A. A’ contains all elements that are not present in Set A. Combining Set A and A’ makes up the Universal Set.

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In the Venn Diagram on the left,

All elememts that are at shaded part are the elements

that (A ∪ B)’ contains.

B. Union

(A ∪ B) The union of set A and B as shown in the Diagram. The union is the combination of all the numbers from set A as well as in set B or simply the elements present in each individual Set.

C. Interaction

The intersection of any two different sets contains atleast one element that exist in both sets. In the Venn Diagram presented below, the shaded part is the Intersection of sets A and B.

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D. Subsets

IF set B is a subset of set A then, all elements of set B are elements of set A. But, not all elements in set A are elements of set B.

Number of Members

If A={ 1, 2, 3, 4} then, n(A)=4 . The representation n(A) means the number of elements a set contains.

Connecting the two sets using their unions and intersection is:

n(A ∪ B)= n(A)+n(B)- n(A ∩ B) where;

n(A ∪ B) is the total elements sets A and B has. n(A) is the number of members that set A alone. n(B) is the the number of members that set B alone. n(A ∩ B) is the number of members A ∩ B alone.

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Application

1. There were 40 students in a classroom. 12 of them chose to ate sandwich for snack, while 19 students ate Burger. 6 of them decided to eat both of the two snacks. How many students ate either spaghetti or burger? How many didn’t eat any of the two snacks?

Given:U=40 studentsn(S) = 18 students eat spaghettin(B) = 25 students eat Burgern(S ∩ B) = 6 students eat both

a. How many students ate either spaghetti or burger?n(S ∪ B)= n(S) + n(B) - n(S ∩ B)

n(S ∪ B)= 18 + 25 – 6

n(S ∪ B)=31 – 6

n(S ∪ B)= 37 Then, there are 37 students who ate either spaghetti or burger.

b. How many didn’t eat any of the two snacks?

Note that the number of elements present in the union of the two elements and its compliment combined, makes the universal set.

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n(S ∪ B) + n(S ∪ B)’= U then,

U - n(S ∪ B) = n(S ∪ B)’40 – 37 = 3 Therefore, there are 3 students who doesn’t eat any of the two snacks.

2. Last valentine’s Day, A flower shop sold custom made bouquets of roses of different color combinations. 30 bouquets were with white roses. 29 of them were with red roses, and 28 were with different colors. 8 bouquets were made with red roses and different colors. 5 bouquets were with the combination of red and white roses, And 5 bouquets were made of white roses and of different colors. There were only 5 bouquets sold with all the colors in it.

a. How many bouquets were sold with only white roses?b. Red roses? c. Of different colors?d. How many all in all were sold?

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Given:

n(W) = 30 bouquets

n(R) = 29 bouquets

n(D) = 28 bouquets

n(R ∩ D) =8

n(W ∩ D) = 5

n(W ∩ R) = 5

n(W ∩ R ∩ D)= 5

a. white only = n(W) - n(W ∩ R) - n(W ∩ D) + n(W ∩ R ∩ D)

white only = 30 – 5 – 5 – 5

=15 Therefore there are 15 bouquets with white roses only.

b. red only = n(R) - n(R ∩ D) - n(W ∩ R) - n(W ∩ R ∩ D)

red only = 29 – 8 – 5 – 5

=11 Therefore, there are 11 bouquets with red roses only.

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c. Different colors only = n(D) - n(R ∩ D) - n(W ∩ D) - n(W ∩ R ∩ D)

Different colors only =28 – 8 – 5 – 5

= 10 Therefore there are 10 bouquets of different colors only,

d. n(W ∪ R ∪ D) = n(W) + n(R) + n(D) – n(R ∩ D) – n(W ∩ D) – n(W ∩ R) + n(W ∩ R ∩ D)

=30 + 29 + 28 – 8 – 5 – 5 + 5

= 74 Therefore, there were 74 bouquets sold in total.

End of Chapter 1