Question

1. n(AXB) =10 மற்றும் A = {2,4} எனில் n(B) = 4 அ)1 %3 AF) 5​

Answer 1

Step-by-step explanation:

Let A={a,b,c,d,e}

and B={1,2,3,4}

A×B=(a,1),(a,2),(a,3),(a,4),(b,1),(b,2),(b,3),(b,4),(c,1),(c,2),(c,3),(c,4),(d,1),(d,2),(d,3),(d,4),(e,1),(e,2),(e,3),(e,4)

∴n(A×B)=n(A)n(B)=20

We know that (A×B)∩(B×A)=(A∩B)×(B∩A)

So,(A×B)∩(B×A)=(A∩B)×(B∩A)

Given:Common elements of A and B=3

∴n(A∩B)=3

So,n[(A∩B)×(B∩A)]=n(A∩B)n(B∩A)=3×3=9 elements

∴n[(A∩B)×(B∩A)]=9 elements

Answer 2

Correct Answer:

Step-by-step explanation:

Let A={a,b,c,d,e}

and B={1,2,3,4}

A×B=(a,1),(a,2),(a,3),(a,4),(b,1),(b,2),(b,3),(b,4),(c,1),(c,2),(c,3),(c,4),(d,1),(d,2),(d,3),(d,4),(e,1),(e,2),(e,3),(e,4)

∴n(A×B)=n(A)n(B)=20

We know that (A×B)∩(B×A)=(A∩B)×(B∩A)

So, (A×B)∩(B×A)=(A∩B)×(B∩A)

Given: Common elements of A and B=3

∴n(A∩B)=3

So, n[(A∩B)×(B∩A)]=n(A∩B)n(B∩A)=3×3=9 elements

∴n[(A∩B)×(B∩A)]=9 elements

#SPJ2