Let \[A = \left\{ {1,2,3,4,5} \right\}\]; \[B = \left\{ {2,3,6,7} \right\}\].Find the number of elements in \[\left( {A \times B} \right) \cap \left( {B \times A} \right)\].
A. 18
B. 6
C. 4
D. 0
Answer
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Hint First we will find \[A \times B\] and \[B \times A\]. To find \[\left( {A \times B} \right) \cap \left( {B \times A} \right)\], we will find the common ordered pairs in \[A \times B\] and \[B \times A\]. Then we will find the number of ordered pairs in \[\left( {A \times B} \right) \cap \left( {B \times A} \right)\].
Formula used
\[A \times B = \left\{ {\left( {a,b} \right):a \in A,b \in B} \right\}\]
Complete step by step solution
Given that, \[A = \left\{ {1,2,3,4,5} \right\}\]; \[B = \left\{ {2,3,6,7} \right\}\].
Now we will calculate \[A \times B\].
\[A \times B = \left\{ {\left( {1,2} \right),\left( {1,3} \right),\left( {1,6} \right),\left( {1,7} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,6} \right),\left( {2,7} \right),\left( {3,2} \right),\left( {3,3} \right),\left( {3,6} \right),\left( {3,7} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,6} \right),\left( {4,7} \right),\left( {5,2} \right),\left( {5,3} \right),\left( {5,6} \right),\left( {5,7} \right)} \right\}\]
Now calculate \[B \times A\].
\[B \times A = \left\{ {\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {3,1} \right),\left( {3,2} \right),\left( {3,3} \right),\left( {3,4} \right),\left( {3,5} \right),\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {7,1} \right),\left( {7,2} \right),\left( {7,3} \right),\left( {7,4} \right),\left( {7,5} \right)} \right\}\]
Now finding the common ordered pairs:
The common ordered pairs are \[\left( {2,2} \right),\left( {2,3} \right),\left( {3,2} \right),\left( {3,3} \right)\].
Now we calculate \[\left( {A \times B} \right) \cap \left( {B \times A} \right)\]:
\[\left( {A \times B} \right) \cap \left( {B \times A} \right) = \left\{ {\left( {2,2} \right),\left( {2,3} \right),\left( {3,2} \right),\left( {3,3} \right)} \right\}\]
The number of ordered pairs in \[\left( {A \times B} \right) \cap \left( {B \times A} \right)\] is 4.
The number of elements of \[\left( {A \times B} \right) \cap \left( {B \times A} \right)\] is 4.
Hence the correct option is C
Note We can write \[\left( {A \times B} \right) \cap \left( {B \times A} \right)\] as \[\left( {A \cap B} \right) \times \left( {B \cap A} \right)\]. Then \[\left( {A \cap B} \right) = \left\{ {2,3} \right\}\] and \[\left( {B \cap A} \right) = \left\{ {2,3} \right\}\]. So, \[\left( {A \times B} \right) \cap \left( {B \times A} \right) = \left\{ {\left( {2,2} \right),\left( {2,3} \right),\left( {3,2} \right),\left( {3,3} \right)} \right\}\]. So the number of elements of \[\left( {A \times B} \right) \cap \left( {B \times A} \right)\] is 4.
Formula used
\[A \times B = \left\{ {\left( {a,b} \right):a \in A,b \in B} \right\}\]
Complete step by step solution
Given that, \[A = \left\{ {1,2,3,4,5} \right\}\]; \[B = \left\{ {2,3,6,7} \right\}\].
Now we will calculate \[A \times B\].
\[A \times B = \left\{ {\left( {1,2} \right),\left( {1,3} \right),\left( {1,6} \right),\left( {1,7} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,6} \right),\left( {2,7} \right),\left( {3,2} \right),\left( {3,3} \right),\left( {3,6} \right),\left( {3,7} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,6} \right),\left( {4,7} \right),\left( {5,2} \right),\left( {5,3} \right),\left( {5,6} \right),\left( {5,7} \right)} \right\}\]
Now calculate \[B \times A\].
\[B \times A = \left\{ {\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right),\left( {3,1} \right),\left( {3,2} \right),\left( {3,3} \right),\left( {3,4} \right),\left( {3,5} \right),\left( {6,1} \right),\left( {6,2} \right),\left( {6,3} \right),\left( {6,4} \right),\left( {6,5} \right),\left( {7,1} \right),\left( {7,2} \right),\left( {7,3} \right),\left( {7,4} \right),\left( {7,5} \right)} \right\}\]
Now finding the common ordered pairs:
The common ordered pairs are \[\left( {2,2} \right),\left( {2,3} \right),\left( {3,2} \right),\left( {3,3} \right)\].
Now we calculate \[\left( {A \times B} \right) \cap \left( {B \times A} \right)\]:
\[\left( {A \times B} \right) \cap \left( {B \times A} \right) = \left\{ {\left( {2,2} \right),\left( {2,3} \right),\left( {3,2} \right),\left( {3,3} \right)} \right\}\]
The number of ordered pairs in \[\left( {A \times B} \right) \cap \left( {B \times A} \right)\] is 4.
The number of elements of \[\left( {A \times B} \right) \cap \left( {B \times A} \right)\] is 4.
Hence the correct option is C
Note We can write \[\left( {A \times B} \right) \cap \left( {B \times A} \right)\] as \[\left( {A \cap B} \right) \times \left( {B \cap A} \right)\]. Then \[\left( {A \cap B} \right) = \left\{ {2,3} \right\}\] and \[\left( {B \cap A} \right) = \left\{ {2,3} \right\}\]. So, \[\left( {A \times B} \right) \cap \left( {B \times A} \right) = \left\{ {\left( {2,2} \right),\left( {2,3} \right),\left( {3,2} \right),\left( {3,3} \right)} \right\}\]. So the number of elements of \[\left( {A \times B} \right) \cap \left( {B \times A} \right)\] is 4.
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