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)​

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)​

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Ivy 1 month 2021-08-18T19:28:00+00:00 2 Answers 0 views 0

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    0
    2021-08-18T19:29:22+00:00

    Answer:

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

    Step-by-step explanation:

    BOTH SIDE ARE EQUAL

    0
    2021-08-18T19:29:33+00:00

    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.

    : )

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