Hi Somenath! The relation given you can be explained as, Using this relation, you can solve various problems. Cheers! Posted by Sunanta Meitei(MeritNation Expert)on 12/11/10 This conversation is already closed by Expert

Hi Somenath! The relation given you can be explained as, Using this relation, you can solve various problems. Cheers! Posted by Sunanta Meitei(MeritNation Expert)on 12/11/10 This conversation is already closed by Expert

Hi Somenath! The relation given you can be explained as, Using this relation, you can solve various problems. Cheers! Posted by Sunanta Meitei(MeritNation Expert)on 12/11/10 This conversation is already closed by Expert

n(AuB) =n(A)+n(B) minus n(A intersection B) suppose,A={1,2,3,4} B={4,5,6,7} thus,(AuB)={1,2,3,4,5,6,7} and (A intersection B)={4} n(A)=4 n(B)=4 n(AuB)=7 n(A intersection B)=1 therefore,7=4+4-1=7 hence proved that,n(AuB)=n(A)+n(B)-n(A intersection B) Posted by Srijan Gupta(student)on 8/11/10

n(AuB) =n(A)+n(B) minus n(A intersection B) suppose,A={1,2,3,4} B={4,5,6,7} thus,(AuB)={1,2,3,4,5,6,7} and (A intersection B)={4} n(A)=4 n(B)=4 n(AuB)=7 n(A intersection B)=1 therefore,7=4+4-1=7 hence proved that,n(AuB)=n(A)+n(B)-n(A intersection B) Posted by Srijan Gupta(student)on 8/11/10

n(AuB) =n(A)+n(B) minus n(A intersection B) suppose,A={1,2,3,4} B={4,5,6,7} thus,(AuB)={1,2,3,4,5,6,7} and (A intersection B)={4} n(A)=4 n(B)=4 n(AuB)=7 n(A intersection B)=1 therefore,7=4+4-1=7 hence proved that,n(AuB)=n(A)+n(B)-n(A intersection B) Posted by Srijan Gupta(student)on 8/11/10