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Question: If \(n\left( A \right)\) denotes the number of elements in set A and if \(n\left( A \right) = 4\),\(...

If n(A)n\left( A \right) denotes the number of elements in set A and if n(A)=4n\left( A \right) = 4,n(B)=5n\left( B \right) = 5 and n(AB)=3n\left( {A \cap B} \right) = 3, then n(A×B)n(B×A)=n\left( {A \times B} \right) \cap n\left( {B \times A} \right) =
A) 88
B) 99
C) 1010
D) 1111

Explanation

Solution

In order to find the value of n(A×B)n(B×A)n\left( {A \times B} \right) \cap n\left( {B \times A} \right) expand the equation using the distributive property, then using the commutative law, solve and substitute the values needed and get the results. There is no need to use n(A)=4n\left( A \right) = 4,n(B)=5n\left( B \right) = 5 in solving the question.

Formula used:
Distributive Property: A(B+C)=AB+ACA\left( {B + C} \right) = AB + AC.
Commutative Law: a+b=b+aa + b = b + a

Complete step by step answer:
We are given the values n(A)=4n\left( A \right) = 4,n(B)=5n\left( B \right) = 5 and n(AB)=3n\left( {A \cap B} \right) = 3.
We need to find the value of n(A×B)n(B×A)n\left( {A \times B} \right) \cap n\left( {B \times A} \right).
From Distributive property, we can expand the equation as:
n(A×B)n(B×A)=n(AB)×n(AA)×n(BA)×n(BB)\Rightarrow n\left( {A \times B} \right) \cap n\left( {B \times A} \right) = n\left( {A \cap B} \right) \times n\left( {A \cap A} \right) \times n\left( {B \cap A} \right) \times n\left( {B \cap B} \right) …..(1)
Since, there are two values such as n(AA)n\left( {A \cap A} \right) and n(BB)n\left( {B \cap B} \right) which means A is intersected to A and will give the result as 1, because there will be all same elements.
Therefore, n(AA)=1n\left( {A \cap A} \right) = 1 and n(BB)=1n\left( {B \cap B} \right) = 1
Substituting the values n(AA)=1n\left( {A \cap A} \right) = 1 and n(BB)=1n\left( {B \cap B} \right) = 1 in the equation 1, we get:
n(A×B)n(B×A)=n(AB)×1×n(BA)×1\Rightarrow n\left( {A \times B} \right) \cap n\left( {B \times A} \right) = n\left( {A \cap B} \right) \times 1 \times n\left( {B \cap A} \right) \times 1
n(A×B)n(B×A)=n(AB)×n(BA)\Rightarrow n\left( {A \times B} \right) \cap n\left( {B \times A} \right) = n\left( {A \cap B} \right) \times n\left( {B \cap A} \right) ………..(2)
From the Commutative property, we know that a+b=b+aa + b = b + a, using this property, we can write as:
n(AB)=n(BA)n\left( {A \cap B} \right) = n\left( {B \cap A} \right)
Since, we were given n(AB)=3n\left( {A \cap B} \right) = 3, that implies:
n(AB)=n(BA)=3n\left( {A \cap B} \right) = n\left( {B \cap A} \right) = 3
Substituting these values in the equation 2, we get:
n(A×B)n(B×A)=3×3\Rightarrow n\left( {A \times B} \right) \cap n\left( {B \times A} \right) = 3 \times 3
n(A×B)n(B×A)=9\Rightarrow n\left( {A \times B} \right) \cap n\left( {B \times A} \right) = 9
Therefore, the value of n(A×B)n(B×A)n\left( {A \times B} \right) \cap n\left( {B \times A} \right) is equal to 99.

Hence, Option (B) is correct.

Note:
Since, there is no use of n(A)=4n\left( A \right) = 4 and n(B)=5n\left( B \right) = 5 while solving the above equation, so do not get confused, and do not substitute their values in the middle if not needed.
The letter n outside the brackets of the sets like n(A) and n(B) represents the number of elements in set A or number of elements in set B.