Let A and B be two sets and U be their Universal set. If n(U)= 120, n(A)= 42, n(B)= 50 and n(AꓵB)= 21 then find,
(i) n (AUB), n(A – B), n(B – A) and n(AˊꓵBˊ)
(ii) n(Bˊ), n(Aˊ), n(AꓴB)ˊ
(iii) n(PUQ) and n(PꓵQ) where P= A – B and Q= AꓵB (iv) How many elements are there in the set U – (AUB)
n(A)= 42
n(B)= 50
n(AꓵB)= 21
i) We know that
n (AUB)= n(A) + n(B) – n(AꓵB)
=42+50 – 21
=71
Again,
n (A ̶ B)= n(A) ̶ n(AꓵB)
= 42 – 21
= 21
n (B ̶ A)= n(B) ̶ n(AꓵB)
= 50 – 21
= 29
And
= 42 – 21
= 21
n (B ̶ A)= n(B) ̶ n(AꓵB)
= 50 – 21
= 29
And
n(AˊꓵBˊ)= n(AꓴB)ˊ
= n(U) – n(AUB)
= 120 – 71
= 49
(ii) n(Bˊ) = n(U) – n(B)
= 120 – 50
= 70
n (Aˊ)= n(U) – n(A)
= 120 – 42
= 78
And
= n(U) – n(AUB)
= 120 – 71
= 49
(ii) n(Bˊ) = n(U) – n(B)
= 120 – 50
= 70
n (Aˊ)= n(U) – n(A)
= 120 – 42
= 78
And
n(AUB)ˊ= n(U) – n(AUB)
= 120 – 71
= 49
(iii) n(PUQ)= n [(A ̶ B) U (AꓵB)]
= n[(AꓵBˊ) U (AꓵB)]
= n[Aꓵ(BˊUB)] (distribution law)
= n [AꓵU]
= n(A)
= 42
And
= 120 – 71
= 49
(iii) n(PUQ)= n [(A ̶ B) U (AꓵB)]
= n[(AꓵBˊ) U (AꓵB)]
= n[Aꓵ(BˊUB)] (distribution law)
= n [AꓵU]
= n(A)
= 42
And
n(PꓵQ)= n [(A ̶ B) ꓵ (AꓵB)]
= n[(AꓵBˊ) ꓵ (AꓵB)]
= n[ A ꓵ (BꓵBˊ)]
= n [A ꓵ ф]
= n(ф)
= 0
(iv) n[U – (AUB)] = n(U) – n(AUB)
= 120 – 71
= 49
= n[(AꓵBˊ) ꓵ (AꓵB)]
= n[ A ꓵ (BꓵBˊ)]
= n [A ꓵ ф]
= n(ф)
= 0
(iv) n[U – (AUB)] = n(U) – n(AUB)
= 120 – 71
= 49
Exercise: 1.1