Question

IF A=(1,2,3,4,5) B = (1,2,3) and C= (5,6) then verify n (AUBUC)+n(ANB)+n(BNC)+n(CNA) = n(a)+n(b)+n(c)+n(AUBUC)​

Answer 1

Hi ,

It is given that ,

1 ) A = { 4 ,5 , 6 }

n( A ) = 3 ,

B = { 5 ,6 , 7 , 8 }

n( B ) = 4 ,

C = { 6 , 7 , 8 , 9 }

n( C ) = 4 ,

A n B = { 4 ,5 , 6 } n { 5 ,6 , 7 , 8 } = { 5 , 6 }

n( A n B ) = 2 ,

B n C = { 5,6,7,8 } n { 6,7,8,9 } = { 6, 7,8 }

n( B n C ) = 3 ,

C n A = { 6,7,8,9 } n { 4 , 5 , 6 } = { 6 }

n( C n A ) = 1

A n B n C = { 4,5,6 } n { 5,6,7,8 }n{ 6 ,7 , 8 }={ 6 }

n( A n B n C ) = 1 ,

A U B U C = { 4,5,6 } U { 5,6,7,8 } U { 6,7,8 }

= { 4,5,6,7,8 }

n( A U B U C ) = 5 ----( i )

n(A)+n(B)+n(C)-n(AnB)-n(BnC)-n(AnC)+n(AnBnC)

= 3 + 4 + 3 - 2 - 3 - 1 + 1

= 11 - 6

= 5 ---( ii )

from ( i ) and ( ii ) , we conclude that ,

( i ) = ( ii )

2 ) A = { a ,b ,c , d , e }

n( A ) = 5 ,

B = { x , y , z }

n( B ) = 3 ,

C = { a , e , x }

n( C ) = 3 ,

A n B = { a,b,c,d,e } n { x , y , z }= { }

n( A n B ) = 0

B n C = { x,y,z } n { a ,e , x } = { x }

n( B n C ) = 1 ,

A n C = { a,b,c,d,e } n { a,e,x } = { a , e }

n( A n C ) = 2 ,

A n B n C = { a,b,c,d,e }n{x,y,z } n { a,e,x }

= { }

n( A n B n C ) = 0

A U B U C = { a,b,c,d,e } U { x,y,z } U { a,e,x }

= { a,b,c,d,e,x ,y,z }

n( A U B U C ) = 8 ---( iii )

n(A)+n(B)+n(C)-n(AnB)-n(BnC)-n(AnC)+n(AnBnC)

= 5 + 3 + 3 - 0 - 1 - 2 + 0

= 11 - 3

= 8 -------( iv )

from ( iii ) and ( iv ) ,

( iii ) = ( iv )

I hope this helps you.

: )

Answer 2

Answer:

n(AUBUC)+N(ANB)+(BNC)+nCNA

Step-by-step explanation:

BOTH SIDE ARE EQUAL