
If $n\left( A \right)$ denotes the number of elements in set A and if $n\left( A \right) = 4$,$n\left( B \right) = 5$ and $n\left( {A \cap B} \right) = 3$, then $n\left( {A \times B} \right) \cap n\left( {B \times A} \right) = $
A) $8$
B) $9$
C) $10$
D) $11$
Answer
496.5k+ views
Hint: In order to find the value of $n\left( {A \times B} \right) \cap n\left( {B \times A} \right)$ 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 $n\left( A \right) = 4$,$n\left( B \right) = 5$ in solving the question.
Formula used:
Distributive Property: $A\left( {B + C} \right) = AB + AC$.
Commutative Law: \[a + b = b + a\]
Complete step by step answer:
We are given the values $n\left( A \right) = 4$,$n\left( B \right) = 5$ and $n\left( {A \cap B} \right) = 3$.
We need to find the value of $n\left( {A \times B} \right) \cap n\left( {B \times A} \right)$.
From Distributive property, we can expand the equation as:
\[ \Rightarrow n\left( {A \times B} \right) \cap n\left( {B \times A} \right) = n\left( {A \cap B} \right) \times n\left( {A \cap A} \right) \times n\left( {B \cap A} \right) \times n\left( {B \cap B} \right)\] …..(1)
Since, there are two values such as \[n\left( {A \cap A} \right)\] and \[n\left( {B \cap B} \right)\] which means A is intersected to A and will give the result as 1, because there will be all same elements.
Therefore, \[n\left( {A \cap A} \right) = 1\] and \[n\left( {B \cap B} \right) = 1\]
Substituting the values \[n\left( {A \cap A} \right) = 1\] and \[n\left( {B \cap B} \right) = 1\] in the equation 1, we get:
\[ \Rightarrow n\left( {A \times B} \right) \cap n\left( {B \times A} \right) = n\left( {A \cap B} \right) \times 1 \times n\left( {B \cap A} \right) \times 1\]
\[ \Rightarrow n\left( {A \times B} \right) \cap n\left( {B \times A} \right) = n\left( {A \cap B} \right) \times n\left( {B \cap A} \right)\] ………..(2)
From the Commutative property, we know that \[a + b = b + a\], using this property, we can write as:
\[n\left( {A \cap B} \right) = n\left( {B \cap A} \right)\]
Since, we were given $n\left( {A \cap B} \right) = 3$, that implies:
\[n\left( {A \cap B} \right) = n\left( {B \cap A} \right) = 3\]
Substituting these values in the equation 2, we get:
\[ \Rightarrow n\left( {A \times B} \right) \cap n\left( {B \times A} \right) = 3 \times 3\]
\[ \Rightarrow n\left( {A \times B} \right) \cap n\left( {B \times A} \right) = 9\]
Therefore, the value of \[n\left( {A \times B} \right) \cap n\left( {B \times A} \right)\] is equal to \[9\].
Hence, Option (B) is correct.
Note:
Since, there is no use of $n\left( A \right) = 4$ and $n\left( B \right) = 5$ 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.
Formula used:
Distributive Property: $A\left( {B + C} \right) = AB + AC$.
Commutative Law: \[a + b = b + a\]
Complete step by step answer:
We are given the values $n\left( A \right) = 4$,$n\left( B \right) = 5$ and $n\left( {A \cap B} \right) = 3$.
We need to find the value of $n\left( {A \times B} \right) \cap n\left( {B \times A} \right)$.
From Distributive property, we can expand the equation as:
\[ \Rightarrow n\left( {A \times B} \right) \cap n\left( {B \times A} \right) = n\left( {A \cap B} \right) \times n\left( {A \cap A} \right) \times n\left( {B \cap A} \right) \times n\left( {B \cap B} \right)\] …..(1)
Since, there are two values such as \[n\left( {A \cap A} \right)\] and \[n\left( {B \cap B} \right)\] which means A is intersected to A and will give the result as 1, because there will be all same elements.
Therefore, \[n\left( {A \cap A} \right) = 1\] and \[n\left( {B \cap B} \right) = 1\]
Substituting the values \[n\left( {A \cap A} \right) = 1\] and \[n\left( {B \cap B} \right) = 1\] in the equation 1, we get:
\[ \Rightarrow n\left( {A \times B} \right) \cap n\left( {B \times A} \right) = n\left( {A \cap B} \right) \times 1 \times n\left( {B \cap A} \right) \times 1\]
\[ \Rightarrow n\left( {A \times B} \right) \cap n\left( {B \times A} \right) = n\left( {A \cap B} \right) \times n\left( {B \cap A} \right)\] ………..(2)
From the Commutative property, we know that \[a + b = b + a\], using this property, we can write as:
\[n\left( {A \cap B} \right) = n\left( {B \cap A} \right)\]
Since, we were given $n\left( {A \cap B} \right) = 3$, that implies:
\[n\left( {A \cap B} \right) = n\left( {B \cap A} \right) = 3\]
Substituting these values in the equation 2, we get:
\[ \Rightarrow n\left( {A \times B} \right) \cap n\left( {B \times A} \right) = 3 \times 3\]
\[ \Rightarrow n\left( {A \times B} \right) \cap n\left( {B \times A} \right) = 9\]
Therefore, the value of \[n\left( {A \times B} \right) \cap n\left( {B \times A} \right)\] is equal to \[9\].
Hence, Option (B) is correct.
Note:
Since, there is no use of $n\left( A \right) = 4$ and $n\left( B \right) = 5$ 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.
Recently Updated Pages
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Differentiate between action potential and resting class 12 biology CBSE

Two plane mirrors arranged at right angles to each class 12 physics CBSE

Which of the following molecules is are chiral A I class 12 chemistry CBSE

Name different types of neurons and give one function class 12 biology CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

Discuss the various forms of bacteria class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

An example of chemosynthetic bacteria is A E coli B class 11 biology CBSE

