Question

A={a,b,c,e,f,h} B={c,d,e,f}and C={a,b,c,f} verify n(AUBUC)=n(A)+n(B)+n(C)-n(AintesectionB)-n(BintersectionC)-n(AintersectionC)+n(AINTERSECTIONBINTERSECTIONC)

Answer 1

Answer:  The verification is done below.

Step-by-step explanation:  We are given the following three sets :

A={a,b,c,e,f,h}

B={c,d,e,f}

and

C={a,b,c,f}.

We are to verify that

n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(A∩C) + n(A∩B∩C).

We have

n(A) = 6,  n(B) = 4  and  n(C) = 4.

Now,

A∪B∪C = {a,b,c,e,f,h} ∪ {c,d,e,f} ∪ {a,b,c,f} = {a, b, c, d, e, f, h}.

So, L.H.S. = n(A∪B∪C) = 7.

A∩B = {a,b,c,e,f,h} ∩ {c,d,e,f} = {c, e, f}   ⇒  n(A∩B) = 3,

B∩C = {c,d,e,f} ∩ {a,b,c,f} = {c, f}  ⇒  n(B∩C) = 2,

A∩C =  {a,b,c,e,f,h} ∩ {a,b,c,f} = {a, b, c, f}   ⇒ n(A∩C) =4,

A∩B∩C = {a,b,c,e,f,h} ∩ {c,d,e,f} ∩ {a,b,c,f} = {c, f}  ⇒ n(A∩B∩C) = 2.

So,

R.H.S.

= n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(A∩C) + n(A∩B∩C)

= 6 + 4 + 4 - 3 - 2 - 4 + 2

= 16 - 9

= 7.

Therefore, L.H.S. = R.H.S.

Hence verified.