NO VENN DIAGRAMS!

Let X X , Y Y and Z Z be three distinct subsets of the universal set U U , no two of which are disjoint.

WITHOUT USING VENN DIAGRAMS, find n ( X ) n(X) , given: n ( X Y Z ) = 5 n(X \cap Y \cap Z) = 5 , n [ X ( Y Z ) ] = 20 n[X \cap (Y - Z)]=20 , n [ X ( Z Y ) ] = 25 n[X \cap (Z - Y)]=25 , and n [ ( X Y ) Z ] = 50 n[(X-Y)-Z]=50 .

DETAILS AND ASSUMPTIONS


- n ( A ) n(A) ( set cardinality ) is the number of elements of set A A .

- A A' ( set complement ) is the set of all elements in U U that are not in A A .

- A B A-B ( set difference ) is the set of elements in which all elements of A A that are also in B B are removed.

- A B A \cap B ( set intersection ) is the set of common elements of A A and B B .

- NO VENN DIAGRAMS!

- If your get the answer right, you can comment your solution here.


The answer is 100.

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1 solution

Jaydee Lucero
Nov 5, 2014

We decompose set X as follows:

X X

= X U =X \cap U

= ( X U ) U = (X \cap U) \cap U

= [ X ( Y Y ) ] ( Z Z ) = [X \cap (Y \cup Y')] \cap (Z \cup Z')

= [ ( X Y ) ( X Y ) ] ( Z Z ) = [(X \cap Y) \cup (X \cap Y')] \cap (Z \cup Z') ( Distributive Property )

= ( Z Z ) [ ( X Y ) ( X Y ) ] = (Z \cup Z') \cap [(X \cap Y) \cup (X \cap Y')] ( Commutative Property )

= [ ( Z Z ) ( X Y ) ] [ ( Z Z ) ( X Y ) ] = [(Z \cup Z') \cap (X \cap Y)] \cup [(Z \cup Z') \cap (X \cap Y')] (Distributive Property)

= [ ( X Y ) ( Z Z ) ] [ ( X Y ) ( Z Z ) ] = [(X \cap Y) \cap (Z \cup Z')] \cup [ (X \cap Y') \cap (Z \cup Z')] (Commutative Property)

= [ ( X Y Z ) ( X Y Z ) ] [ ( X Y Z ) ( X Y Z ) ] = [(X \cap Y \cap Z) \cup (X \cap Y \cap Z')] \cup [(X \cap Y' \cap Z) \cup (X \cap Y' \cap Z')] (Distributive Property)

= ( X Y Z ) [ X ( Y Z ) ] [ X ( Z Y ) ] [ ( X Y ) Z ] = (X \cap Y \cap Z) \cup [X \cap (Y-Z)] \cup [X \cap (Z-Y)] \cup [(X-Y)-Z] ( Set Difference , defined as A B = A B A-B=A \cap B' .)

Therefore,

n ( X ) = n ( X Y Z ) + n [ X ( Y Z ) ] + n [ X ( Z Y ) ] + n [ ( X Y ) Z ] n(X)=n(X \cap Y \cap Z)+n[X \cap (Y-Z)]+n[X \cap (Z-Y)]+n[(X-Y)-Z]

n ( X ) = 5 + 20 + 25 + 50 = 100 n(X)=5+20+25+50=\boxed{100} .

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