Let , and be three distinct subsets of the universal set , no two of which are disjoint.
WITHOUT USING VENN DIAGRAMS, find , given: , , , and .
DETAILS AND ASSUMPTIONS
- ( set cardinality ) is the number of elements of set .
- ( set complement ) is the set of all elements in that are not in .
- ( set difference ) is the set of elements in which all elements of that are also in are removed.
- ( set intersection ) is the set of common elements of and .
- NO VENN DIAGRAMS!
- If your get the answer right, you can comment your solution here.
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We decompose set X as follows:
X
= X ∩ U
= ( X ∩ U ) ∩ U
= [ X ∩ ( Y ∪ Y ′ ) ] ∩ ( Z ∪ Z ′ )
= [ ( X ∩ Y ) ∪ ( X ∩ Y ′ ) ] ∩ ( Z ∪ Z ′ ) ( Distributive Property )
= ( Z ∪ Z ′ ) ∩ [ ( X ∩ Y ) ∪ ( X ∩ Y ′ ) ] ( Commutative Property )
= [ ( Z ∪ Z ′ ) ∩ ( X ∩ Y ) ] ∪ [ ( Z ∪ Z ′ ) ∩ ( X ∩ Y ′ ) ] (Distributive Property)
= [ ( X ∩ Y ) ∩ ( Z ∪ Z ′ ) ] ∪ [ ( X ∩ Y ′ ) ∩ ( Z ∪ Z ′ ) ] (Commutative Property)
= [ ( X ∩ Y ∩ Z ) ∪ ( X ∩ Y ∩ Z ′ ) ] ∪ [ ( X ∩ Y ′ ∩ Z ) ∪ ( X ∩ Y ′ ∩ Z ′ ) ] (Distributive Property)
= ( X ∩ Y ∩ Z ) ∪ [ X ∩ ( Y − Z ) ] ∪ [ X ∩ ( Z − Y ) ] ∪ [ ( X − Y ) − Z ] ( Set Difference , defined as A − B = A ∩ B ′ .)
Therefore,
n ( X ) = n ( X ∩ Y ∩ Z ) + n [ X ∩ ( Y − Z ) ] + n [ X ∩ ( Z − Y ) ] + n [ ( X − Y ) − Z ]
n ( X ) = 5 + 2 0 + 2 5 + 5 0 = 1 0 0 .