(i) {1 ,2}
(ii) {4,5,6,7}
(iii) {1,2,5,6}

(i) {5,9}
(ii) {1,2,3,4,5,7,8,9}

22

(a) A = {Jan, March, May, July, Aug, Oct, Dec}
(b) B = {13,26,39,52,65,78,91}
(c) C = {a,e}
(d) D = {6,7,8,9,10}
(e) E = {1,4,9}

(X\(\cup \)Y)'

B' = {1, 4, 6}
(B')' = {0, 1, 2, 3, 4, 5, 6, 7} - {1, 4, 6}
= {0, 2, 3, 5, 7}

n(A) = 25, n(B) = 40, n(AUB) = 50 and n(B') = 25
n(A\(\cap\)B) = n(A) + n(B) - n(AUB)
n(A\(\cap\)B) = 25 + 40 - 50
= 65 - 50
= 15
n(U) = n(B) + n(B)'
= 40 + 25
= 65
\(\therefore\) n(A\(\cap\)B) = 15 and n(U) = 65

Let V be the set of people who had Vanilla ice cream and C be the set of people who had Chocolate ice cream.
Then n(V) = 35, n(C) = 30, n(VUC) = 60,
Let x be the number of people who had both ice creams.
From the Venn diagram

35 – x + x +30 – x = 60
65 – x = 60
x = 5
Hence 5 people had both ice creams.
(i) Number of people who had only Vanilla ice cream = 35 – x
= 35 – 5 = 30
(ii) Number of people who had only Chocolate ice cream = 30 – x
= 30 – 5 = 25.
We have learnt to solve problems involving two sets using the formula n(A U B) = n(A)+n(B)−n(A ∩ B). Suppose we have three sets, we can apply this formula to get a similar formula for three sets.

(A – B)′

A = {20,22, 23, 24} , B = {25, 30, 40, 45}
A ∩ B = {20,22, 23, 24} ∩ {25, 30, 40, 45}
= { }
Since A ∩ B = ∅, A and B are disjoint sets.

A Δ B = (A – B) ∪ (B – A)

U = {c, d, e, f, g, h, i, j}, A = {c, d, g, j}
A′ = {e, f, h, i}

Given X = {a, b, c, x, y, z}.Then, n(X) = 6
The number of subsets   = n[P(X)] = 2⁶ = 64
The number of proper subsets = n[P(X)]-1 = 2⁶ – 1
= 64 – 1 = 63.

PRINCIPAL
B = {P, R, I, N, C, A, L}

(i) AUB

(ii) A∩B

(iii) (A∩B)′

(iv) (B–A)'

(v) A′UB′

(vi) A′∩B′


(vii) From the venn diagram (iii) and (v), We have
\((A \cap B)^{\prime}=A^{\prime} \cup B^{\prime}\)
Now \(\mathrm{B} \cap \mathrm{C}=\left\{\frac{1}{4}, 2, \frac{5}{2}\right\}\)
\(A \cap(B \cap C)=\left\{\frac{1}{4}, 2\right\}\)............(1)
Then \(A \cap B=\left\{0, \frac{1}{4}, \frac{3}{4}, 2\right\}\)
\((A \cap B) \cap C=\left\{\frac{1}{4}, 2\right\}\).............(2)
From (1) and (2), it is verified that
\((A \cap B) \cap C=A \cap(B \cap C)\)