In two different societies, there are some school going students inclu

1.	In two different societies, there are some school going students including girls as well as boys. Satish forms two sets with these students for his college project.Let = {a1, a2, a3, a4, a5 } and B= {b1, b2, b3, b4 } where ai's and bi's are school going students of first and second societies respectively.Satish decides to explore these sets for various types of relations and functions.With the help of above information answer the following question:Satish and his friend Rajat are interested to know the number of symmetric relations defined on both the sets and , separately. Satish decides to find the symmetric relation on set , while Rajat decides to find the symmetric relations on set . What is the difference between their results ?
Answer» We know that total number of symmetric relation on set having n elements is \(2^{\cfrac{n(n+1)}2}\). Given = {a₁, a₂, a₃, a₄, a₅ } and B = {b₁, b₂, b₃, b₄ }. Therefore, A has 5 elements and B has 4 elements. Therefore, total number of symmetric relation on set A is ^{\(2^\cfrac{5(5+1)}2\)} = 2¹⁵ . And total number of symmetric relation on set B is ^{\(2^\cfrac{4(4+1)}2\)} = 2¹⁰. Difference between their results is 2¹⁵ − 2¹⁰ = 2¹⁰(2⁵ − 1) = 2¹⁰(32 − 1) = 31 × 1024 = 31774.

In two different societies, there are some school going students including girls as well as boys. Satish forms two sets with these students for his college project.Let = {a1, a2, a3, a4, a5 } and B= {b1, b2, b3, b4 } where ai's and bi's are school going students of first and second societies respectively.Satish decides to explore these sets for various types of relations and functions.With the help of above information answer the following question:Satish and his friend Rajat are interested to know the number of symmetric relations defined on both the sets and , separately. Satish decides to find the symmetric relation on set , while Rajat decides to find the symmetric relations on set . What is the difference between their results ?

Answer»

We know that total number of symmetric relation on set having n elements is

\(2^{\cfrac{n(n+1)}2}\).

Given = {a₁, a₂, a₃, a₄, a₅ } and B = {b₁, b₂, b₃, b₄ }.

Therefore, A has 5 elements and B has 4 elements.

Therefore, total number of symmetric relation on set A is ^{\(2^\cfrac{5(5+1)}2\)} = 2¹⁵ .

And total number of symmetric relation on set B is ^{\(2^\cfrac{4(4+1)}2\)} = 2¹⁰.

Difference between their results is 2¹⁵ − 2¹⁰ = 2¹⁰(2⁵ − 1) = 2¹⁰(32 − 1)

= 31 × 1024 = 31774.