The document contains mathematical formulas and sets. It defines several sets including sets A and B and discusses their properties and relationships. It considers concepts like subsets, membership, cardinality, and power sets. Several examples are provided to illustrate set operations and properties.

1. F F F
F .4
F F F F F
F 6 F 5
F (ก) () ก ก F {1, 4, 9, 16, 25, 36} Fก A
F () () ก ก F {0, 1, 4}
F ก (ก) - ( ) ก F2 F
F 5 F 7 3
ก. 1 ∈ A
. F x∈A F 1 ∈A
x
. F x∈A ก F 2x ∈ A ≡ x ∈ A ก F 2x ∉ A
1 ∈A ↔ 2 ∉ A
2 ∉A ↔ 4 ∈ A FF 1
4
∈A
4 ∈A ↔ 8 ∉ A
8 ∉ A ↔ 16 ∈ A FF 1
16
∈A
16 ∈ A ↔ 32 ∉ A
32 ∉ A ↔ 64 ∈ A FF 1
64
∈A
F 6 F 10
ก F S = {x / x 3 = 1} = {−1, 1}
ก 1 : {x / x 3 = 1} = {1}
ก 2 : {x / x 2 = 1} = {−1, 1}
ก 3 : {x / x 3 = − 1} = {−1}
ก 4 : {x / x 4 = x} = {0, 1}
x4 − x = 0
x(x 3 − 1) = 0
x(x − 1)(x 2 + x + 1) = 0
x = 0, 1
1

2. F F F
F 2 F 11
A = {x/x = 1
n n ก} = {1, 1 , 1 , 1 , .....}
2 3 4
A F
F 6 F 12 4
x = ±3 x=2
A = {x / x 2 = 9 2x = 4} = {} , B = {x / x ∈ R x ≠ x} = {}
C = {x / x ∈ R x 2 + 4 = 0} = {}, D = {x / x ∈ R x > x} = R
/
E = {x / x ก } = {}
F 2 F 18 1
ก F
C
3 2 F F B F F F A F Fก F
B 3 F C F F Cก F
A
1 2 4 F F C F
∴ ก 1
4
F 2 F 20 2
1. F
2. ก F ก
3. F F F F
4.
F 5 F 20 4
1. ก F ก
2. ก F F F
3. ก F F F
4. F F F F
F 9 F 21 3
ก F A = {1, 2, {3}}
B = {x / x = A} = {A} = {{1, 2, {3}}}
C = {x / x ∈ A} = {1, 2, {3}}
ก กF กF A=C
2

3. F F F
F 12 F 22
a) F ก F
F ∅ ก F F ก
b) ก
c) F ก ก F
F
d) ก
e) ∅⊆∅
/
f) A⊂B F B⊂A A=B F ก
g) A⊂B F B⊂A F
/ ก A=B
h) A⊂B a∈B F a F F ก A
i) ก F A⊂B ก ก A F F B
F 14 F 22
ก1 : F A ⊂ B, B ⊂ C F A⊂C F 3∉C FF 3∉A
F
ก2 :C C = {u, v, w, x, z}
ก3 :C C = {u, v, w, x, z} ก4 กF
F 16 F 23
A ก3 6!
= 6 C 3 = 3!3! = 20
F 17 F 23 กF F 1 F " ก " ก
ก B⊂A B≠A
B ก F3 = A
− A ก0 − A ก1
− A ก2 − 1
= 2 6 − 6C0− 6 C1− 6 C2 − 1
= 64 − 6! − 6! − 6! − 1 = 41
6!0! 5!1! 4!2!
F 3 F 24
B = {∅} n(B) = 1
21 2
n(P(P(P(B)))) = 2 2 = 22 = 2 4 = 16
3

4. F F F
F 3 F 25
F F 4 {∅}
A = {∅ , {∅} , { P(∅) }}
P(A) = {∅ , {∅} , {{∅}} , {{{∅}}} , {∅ , {∅}} , {∅ , {{∅}}}
, {{∅} , {{∅}}} , {∅ , {∅} , {{∅}}}
F 5 F 25 1
ก P(A) Fก F F A = {0, 1 , 2}
2
ก 1 : 2x 3 − 5x 2 + 2x = 0 → x(2x 2 − 5x + 2 = 0)
→ x(2x − 1)(x − 2) = 0 → x = 0, 1 , 2
2
∴ {x/2x 3 − 5x 2 + 2x = 0} = {0, 1 , 2} = A
2
F 6 F 26
F F g) ก F A F P(A) FF
FF A ก P(A) ก F
F F h) F Fก B F n(B) ≤ n(A)
F F F F A⊂B
F 7 F 26 F ก2 F 2 3
F 8 F 27
A = {x / x ∈ N x < 100} = {1, 2, 3, 4, ....., 99}
B = {x / x ∈ A 5 x } = {5, 10, 15, ....., 95}
n(B) = 19 n(P(B)) = 2 n(B) = 2 19
F 9 F 27 3
กF F F E = {a, b} F ก F E⊂A
{a, b} ⊂ A ก3 F ก Fก A
F 10 F 27 3
ก1 ก: F S = ∅, T = {∅}
ก2 ก: F S = ∅ T = {1} U = {∅}
ก3 ก4 ก
4

5. F F F
F 11 F 28
a) P(B) ⊂ P(A) ก ก A B Fก F
B⊂A FF P(B) ⊂ P(A)
b) C F P(C) ≠ ∅ ก FF C
F 12 F 28
n(P(A)) − n(P(C)) = 63 → 2 n(A) − 2 n(C) = 63 (1)
n(P(A)) + n(P(B)) = 96 → 2 n(A) + 2 n(B) = 96 (2)
ก (1) ก F ก F F
ก 2 n(C) = 1 → n(C) = 0
F 2 n(C) = 1 (1) F 2 n(A) − 1 = 63 → 2 n(A) = 64
n(A) = 6
F 2 n(A) = 64 (2) F 64 + 2 n(B) = 96
2 n(B) = 32 → n(B) = 5
∴ n(A) + n(B) + n(C) = 6 + 5 + 0 = 11
F 2 F 29
ก1 ก {∅} ⊂ P(A) → ∅ ⊂ A ก
ก2 ก {∅} ∈ P(A) → ∅ ∈ A ก
ก3 {{a}, {a, b}} ⊂ P(A) → {a}, {a, b} ⊂ A
กF F F F
ก4 ก {{a, b}} ∈ P(A) → {a, b} ∈ A ก
F 3 F 29
ก A = {1, {1}, 2, {1, 2}, 3, {1, 2, 3}}
1. ก : {1, 2, 3} ∈ P(A) ก P F 1, 2, 3 ∈A ก
2. ก : ∅ ∈ P(A) P F ∅⊂A ก F F
∅ ∈ P(A) ก F ก ก F
∅ ⊂ P(A) ก P F ∅⊂A ก
5

6. F F F
3. : A ⊂ P(A)
ก A = {1, {1}, 2, {1, 2}, 3, {1, 2, 3}} ⊂ P(A) ก P F
1 , {1} , 2 , {1, 2} , 3 , {1, 2, 3} ⊂ A
F A F F ก
F
{A} ∈ P(A) ก P F A∈A
F ก
4. ก : {{{1, 2}, 3}} ⊂ P(A) ก P F {{1, 2}, 3} ⊂ A
ก F {1, 2}, 3 ∈ A ก
F 6 F 30
1. ก : ∅ ∈ P(P(B)) P F ∅ ⊂ P(B) ก
2. ก : ∅ ⊂ P(P(B)) ก F ก
3. : {∅} ⊂ P(B)
/ ก P F ∅⊂B
/
4. ก : F F x=∅ F ∅ ∈ {∅} ∈ B
F 7 F 30
(ก) ก {∅} ⊂ P(A) → ∅ ⊂ A ก
() A ⊂ P(A) ก ก A=∅ F
() ก {∅, A} ⊂ P(A) → ∅, A ⊂ A ก
() {A} ∈ P(A) → ∅ ∈ ∅
F 7 F 34
A B F ก 8 9
4 { }, {8}, {9}, {8, 9}
F 8 F 35 4
ก F A = {1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10}
↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓
1 × 1 × 1 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2 7 = 128
ก Fก X F 1, 2 F F 3
F X F 128
6

7. F F F
F 9 F 35
กS = {X ∈ P(A) / 1 ∈ X n(X) ≥ 5}
X⊂A F F X กA
ก n(X) ≥ 5 F3ก
n(X) = 5 n(X) = 6 n(X) = 7
ก n(X) = 5 F X F 6!
1⋅ 6 C 4 = 4!2! = 15
ก n(X) = 6 F X F 1⋅ 6 C 5 = 6! = 6
5!1!
ก n(X) = 7 F X F 1⋅ 6 C 6 = 6! = 1
6!0!
X Fก กก X ก
ก ก
∴ X F = 15 + 6 + 1 = 22
F 10 F 36 3
ก F {1, 2} ⊂ A F A F 1, 2
ก {1, 2, 4, 8, 16}
↓ ↓ ↓ ↓ ↓
1 × 1 × 2 × 2 × 2 = 23 = 8
= 8
A F ∴ F = 8−1 = 7
F 11 F 36 1
1 A ก3 ∴ n(P(A)) = 2 3 = 8
F P(A) ก F F 1
1+ 2+ 3+ 4+ C5 +8 C6 +8 C7
8C 8C 8C 8C 8
= 8! + 8! + 8! + 8! + 8! + 8! + 8!
7!1! 6!2! 5!3! 4!4! 3!5! 2!6! 1!7!
= 8 + 28 + 56 + 70 + 56 + 28 + 8 = 254
2 A ก3
F P(A) ก F F 1
= P(A) − F ก − 1
= 2 8 − 1 − 1 = 254
7

8. F F F
F 12 F 36 4
n(P(A)) = 32 → 2 n(A) = 2 5 → n(A) = 5
F B⊆A F ก B F F ก4
n(P(B)) ก = 2 4
F 13 F 37 1
ก F P(B) = {∅, {0}, {{0, 1}}, { 0 {0, 1}}}
ก x ∈ P(A) Fx⊂A F F x กA
x ⊂ P(B) F ก x F F P(B)
ก F A = {∅, {∅}, 0, 1}
↓ ↓ ↓ ↓
2 × 1× 1 × 1 = 2 ∴ n(C) = 2
F 14 F 37
a) A⊂X⊂B FF A⊂B F
F F X F ก กF
X = 0
b) A⊂X
/ X⊂B F ก X F ก กB F
∴ X = 2 5 = 32
F 15 F 38
B⊂E F 1, 2, 3 F F E F
E⊂A F 4, 5, 6, 7 F F F Eก F
{1, 2, 3, 4, 5, 6, 7}
↓ ↓ ↓ ↓ ↓ ↓ ↓
1 × 1 × 1 × 2 × 2 × 2 × 2 = 24
∴ E F = 2 4 = 16
8

9. F F F
F 16 F 38
X = {B ∈ P(A) / 1 ∈ B 2∈B 3 ∈ B}
B⊂A
F F B F 1 2 3
B ก A F F 1
A = {1, 2, 3, 4, 5, 6, 7}
↓ ↓ ↓ ↓ ↓
(2 3 − 1) × 2 × 2 × 2 × 2 = 7 × 2 4 = 112
ก F 1, 2, 3 F B
ก 1, 2, 3
F B
∴ F X = 2 112 − 1
F 17 F 38
X = {B ∈ P(A) / 1 ∈ B 2 ∉ B}
B⊂A B F 1 F
F 2 ก
ก 1, 2 F
B F B
F 1 F 2 (ก F 1)
A = {1, 2, 3, 4, 5, 6, 7, 8, 9}
↓ ↓ ↓ ↓ ↓ ↓ ↓
(2 2 − 1) × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 3 × 2 7
B F 3 × 10 7 ∴ n(X) = 3 × 10 7
F 3 F 39
U = {0 , 2} A = {x / x 2 = 4} F A = {2}
x = −2, 2
F x = −2 F F F F กU
9

10. F F F
F 3 F 45 3
ก1 A ∩ B = {1, {1}} ≠ A
ก2 A ∪ B = {∅, 1, {1}, {1, {1}}} ≠ A
ก3 ก A ∩ B = {1, {1}} ∈ B
ก4 A ∪ B = {∅, 1, {1}, {1, {1}}} ∈ B
F 5 F 46
1. C ∩ D = {4} ⊂ A
/
2. ก A − D = {3} C − B = {3}
A−D ⊂C−B
3. A ∩ C = {3}
(A ∩ C) ∪ B = {3, 4} ⊂ D
/
4. A − D = {3} {3} ∈ A − D
F 8 F 46
A − B = {1} B − A = {{1}}
∴ (A − B) ∪ (B − A) = {1 , {1}}
F 9 F 47
ก1 A ∩ C = {2} ≠ ∅
ก2 ก A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9}
(A ∪ B) = {}
ก3 B ∩ C = B − C = {1, 9}
(B ∩ C ) = (B − C) = {0, 2, 3, 4, 5, 6, 7, 8} ≠ 0
ก4 A − C = {0, 4, 6, 8}
(A − C) = {1, 2, 3, 5, 7, 9} ≠ ∅
10

11. F F F
F 10 F 47 2
ก 1) B ∩ C = {3, 5, 7}
2) C ∩ B → C = {0, 1, 4, 6, 8, 9} ∴ C ∩ B = {1, 9}
3) C − B = {2}
4) A − B → A = {1, 3, 5, 7, 9} ∴ A −B = ∅
∴ ก 2 = {1, 9}
F 15 F 49
15.1 ก F ก
F F F
A B
7, 8 11 1
9 4,5 6
2, 3 10, 12
C U
15.2 1) A∪C = {2, 3, 4, 5, 7, 8, 9, 11}
(A ∪ C) = {1, 10, 12}
B − (A ∪ C) = {1, 4, 5, 6, 11} − {1, 10, 12} = {4, 5, 6, 11}
2) กF F
[C ∪ (C ∩ A ∩ B)] ∩ [A ∪ (A ∩ B ∩ C )] ∩ [B ∩ (B ∪ C ∪ A )]
=  C∪(A ∩ B ∩ C )  ∩  A ∪(A ∩ B ∩ C )  ∩  B ∪(A ∩ B ∩ C ) 
     
= (C ∩ A ∩ B ) ∪ (A ∩ B ∩ C ) = {2 , 3} ∪ {11} = {2, 3, 11}
11

12. F F F
F 16 F 50
ก) I 6 ∪ I 12 = I 12
(I 6 ∪ I 12 ) ∩ I 8 = I 12 ∩ I 8 = I 8
) I 20 − I 15 = {16, 17, 18, 19, 20}
(I 20 − I 15 ) ∩ I 40 = {16, 17, 18, 19, 20}
) I 5 = {6, 7, 8, ...} I 9 = {10, 11, 12, ...}
I 5 ∩ I 9 = {10, 11, 12, ...} = I 9
) I7 ∪ I4 = I7
(I 7 ∪ I 4 ) ∩ I 10 = I 7 ∩ I 10 = ∅
 (I 7 ∪ I 4 ) ∩ I 10  = ∅ = U
 
F 17 F 50 3
A n = [ 1 , 2n)
n A 1, A 2, A 3, A 4 F
A 1 = [1, 2) F
A4
A2 = [ 1 , 4) A3
2
A2
A3 = [ 1 , 6) A1
3
1 1 1 1 2 4 6 8
A 4 = [ 1 , 8) 4 3 2
4
ก F A1 ∪ A4 = [ 1, 1)
4 2
(A1 A4 ) - (A2 3 A 3)
3
A 2 3 A3
A 2 ∩ A 3 = [4, 8)
A 1 A4 3
(A 1 ∪ A 4 ) − (A 2 ∩ A 3 ) = [ 1 , 1 ) ∪ [4, 8) 1 1
4 2 4 8
4 2
F 19 F 51 3
F 20 F 51 2
F 1 F 60
1.1 ก Fก F x F d, e, f, g
x⊂A F F x F กA
A = { a, b, c, d, e, f, g}
↓ ↓ ↓ ↓ ↓ ↓ ↓
2 × 2 × 2 × 1 × 1 × 1 × 1 = 23 ∴ n(S) = 2 3 = 8
12

13. F F F
1.2 a b F 22 − 1 = 3
b = {a, b}
a
− = {a}
b = {b}
−
− = { } ×
∴ A = { a , b, c, d, e, f, g}
↓ ↓ ↓ ↓ ↓ ↓
(2 2 − 1) × 2 × 2 × 2 × 2 × 2 = 3(2) 5
1.3 ก F S F a b c d
n(S) = (2 4 − 1) ⋅ 2 3 = 15 ⋅ 2 3
F 2 F 60
ก F S∩B ≠ ∅ F S F 1 2
F F S A A
A = {1, 2, 3, a, b, c}
↓ ↓ ↓ ↓
(2 2 − 1) × 2 × 2 × 2 × 2 = 3 × 2 4 = 48
∴ A S∩B ≠ ∅ F ก 48
F 3 F 60
ก F A ∩ {1, 2, 3, 5, 7} = {2, 5, 7}
A F 2, 5, 7 A F F 1, 3
ก A⊂ U F F A ก U
ก F
U = {1, 2, 3, 4, 5, 6, 7, ....., 18}
↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓
1 × 1 × 1 × 2 × 1 × 2 × 1 × 2 11 = 2 13
∴ A F ก 2 13 F ก กF
13

14. F F F
F 2 F 61
A∪B = {2, 3}
P(A ∪ B) = {∅, {2}, {3}, {2, 3}}
P(A) = {∅} P(B) = {∅, {2}, {3}, {2, 3}}
P(A) ∪ P(B) = {∅, {2}, {3}, {2, 3}}
P(A ∪ B) = P(A) ∪ P(B)
F 4 F 62 กF F ก n(P(A)) − n(P(A)) = 63 n(P(A)) − n(P(B)) = 63
ก n(P(A)) − n(P(B)) = 63
2 n(A) − 2 n(B) = 63
F F ก 2 n(B) ก
F 2 n(B) = 1 → n(B) = 0
F 2 n(A) − 1 = 64 → 2 n(A) = 64 → n(A) = 6
∴ n(A) + n(B) = 6 + 0 = 6
F 7 F 63
ก F ก P(A) ก B ∅, {∅}, {0, {0, 1}}
n(P(A) ∩ B) = 3
P(A) B ก n(P(A) − B) = n(P(A)) − n(P(A) ∩ B)
= 2 n(A) − 3
P(A) - B = 2 4 − 3 = 13
F 9 F 63
ก F ก P(A) ก A ∅, {∅}, {0} n(P(A) ∩ A) = 3
A P(A) n(P(A) − A) = n(P(A)) − n(P(A) ∩ A) = 2 n(A) − 3 = 2 6 − 3 = 61
A P(A) n(A − P(A)) = n(A) − n(P(A) ∩ A) = 6 − 3 = 3
A P(A) n[(P(A) − A) ∪ (A − P(A))] = 61 + 3 = 64
14

15. F F F
F 24 F 75
(A∩B) ∪ (A∩B ) ∪ (A ∩ B) = [A ∩ (B ∪ B )] ∪ (A ∩ B)
U
= A ∪ (A ∩ B)
= (A ∪ A ) ∩ (A ∪ B)
= U ∩ (A ∪ B) = A ∪ B
F 25 F 76
ก. F กF B ∩C B ∪C
[B ∪ ((A ∪ B ) ∩ (B ∪ C))] ∩ [B ∪ (B ∩ A)] ∩ (B ∪ A)
= [B ∪ B ∪ (A ∩ C)] ∩ [B ∪ (B ∩ A)] ∩ (B ∩ A )
= U ∩ [(B ∪ B) ∩ (B ∪ A)] ∩ (B ∩ A )
= U ∩ B ∩ (B − A) = B − A
. [(A ∪ B ∪ C) ∩ (A ∩ B) ∩ (A − B) ] ∪ [((A ∩ B) − C) ∩ (C − (A ∩ B ))]
A∪B
= (C ∩ A ∩ ∅ ) ∪ [∅ ∩ (C ∩ (A ∩ B ) )]
= (C ∩ A ∩ U ) ∪ [ U ∩ ∅] = A ∪ ∅ = A
F 26 F 76
(A ∆ B) = [(A − B) ∪ (B − A)]
= [(A ∩ B ) ∪ (B ∩ A )]
= [(A ∩ B ) ∩ (B ∩ A )]
= [(A ∪ B) ∩ (B ∪ A)]
= (A ∪ B) ∩ (A ∪ B ) F . ก
= (A ∩ (A ∪ B )) ∪ (B ∩ (A ∪ B ))
= ((A ∩ A) ∪ (A ∩ B )) ∪ ((B ∩ A) ∪ (B ∩ B ))
= (∅ ∪ (A ∩ B )) ∪ ((B ∩ A) ∪ ∅)
= (A ∩ B ) ∪ (B ∩ A)
= (B ∩ A) ∪ (A ∩ B )
= (A ∩ B) ∪ (A ∩ B ) F ก. ก
15

16. F F F
F 5 F 87 1
กF Fก F F
ก F
FF
3 1
ก
2
2 = 2 + 3 + 1 + 10 + 2 + 4 + 5 = 27
10 4
5
U
F 6 F 87
F Humanities Sciences x
ก F
Humanities Social = 20
8-x 2 6 8 − x + 2 + 6 + x + 0 + 0 + 6 − x = 20
0
x 0 x = 2
6-x ∴ Humanities Sciences 2
Sciences U
F 9 F 89
กF ก
F F F
= 35% = 48% F =x
ก
35 - x x=12 48 - x 35 + (48 − x) + 29 = 100
= 23 = 36
x = 12
29 U = 100
1. = 35 + 36 = 71%
2. = 12%
3. F F = 23%
4. F = 36%
5. F = 35 + 29 = 64%
16

17. F F F
F 16 F 92
กF ก
F F F
F F F ก1 = 65 + 9 + 6 + 7
F A F B
6 = 87
9 6
3
2
2
F F FFF
65 F F ก1 87
= 100 = 0.87
7
F C
U = 100
F 17 F 93
กF Fก F F
ก F
F F F 40
5 x 10 F 5 + x + 10 + 10 = 40 → x = 15
0
y z F 90
30 10
F 5 + y + 30 + 10 = 90 → y = 45
F U
F F 70
F 10 + z + 30 + 10 = 70 → z = 20
∴ 2 F F = x + y + z = 15 + 45 + 20 = 80
F 18 F 93
กF Fก F F
F U = 100%
F F
a + b + d + x ≥ 75 (1)
a b c
x b + c + e + x ≥ 70 (2)
d e
d + e + f + x ≥ 65 (3)
f
กF U = 100
17

18. F F F
(1) + (2) + (3) : (a + b + c + d + e + f + x) +b + d + e + 2x ≥ 210%
= 100%
(b + d + e + x) + x ≥ 110%
F 100%
x ≥ 10%
∴ F F 10% 3
F 19 F 94 3
กF Fก F F F
ก F FF
ก
F ก F Fก
F F F F F 3
ก F F
ก F
1. Fก F
ก F
2. ก F Fก
F 3. F Fก
4. Fก ก
F 20 F 95 1
กF Fก F F
Fก y F ≠ 0 z = 5y
MATH ENG F y = 2 F z = 10 2x + 6 + y + z = 28
x x 0 2x + 6 + 2 + 10 = 28
y
6 z x = 5
0
F y = 4 F z = 20 2x + 6 + y + z = 28
HIS U = 100
2x + 6 + 4 + 20 = 28
x = -1
F F
∴ ก ก F F 5
18

19. F F F
F 21 F 95
F n(A ∩ B ∩ C) = x
กF ก
F F F
n(A ∪ B ∪ C) = 80
A B(42)
20 + 2 + (37 − x) + 5 + 3 + x + (20 − x) = 80
20 2 42-2-3-x
3 = 37-x กF ก F x = 7
5 x
28-5-3-x
∴ n(A ∩ B ∩ C) = 7
= 20-x
C(28)
F 25 F 97
n(P(A) ∪ P(B)) = n(P(A)) + n(P(B)) − n(P(A) ∩ P(B))
= n(P(A)) + n(P(B)) − n(P(A ∩ B))
= 16 + 8 − 4 = 20
F 26 F 98 2
กF Fก F F
ก
= 100
U
n(B ∪ A ) = 45 + 30 + 10 = 85 F 1
n(A − B ) = n(A ∩ B) = 30 F 2 ก
15 45 30
n(A ∪ B ) = n((A ∩ B) ) = 15 + 30 + 10 = 55
10
F 3
n(B ∩ A) = n(A ∩ B ) = 15 F 4
F 27 F 98
n(P(A ∪ B)) = 2 n(A∪B) = 2 n(A) + n(B) − n(A∩B)
n(P(A)) ⋅ n(P(B))
= 2 n(A) ⋅ 2 n(B)
2 n(A ∩ B)
= n(P(A ∩ B))
(1)
ก n(P(A) ∪ P(B)) = n(P(A)) + n(P(B)) − n(P(A) ∩ P(B))
z = x + y − n(P(A ∩ B))
n(P(A ∩ B)) = x+y−z
F n(P(A)) , n(P(B)) n(P(A ∩ B)) (1)
F n(P(A ∪ B)) = x + y − z
xy
19

20. F F F
F 28 F 98 2
n(x ∪ y ∪ z)= n(x) + n(y) + n(z) − n(x ∩ y) − n(x ∩ z) − n(y ∩ z) + n(x ∩ y ∩ z)
70 = 30 + 44 + 28 − n(x ∩ y) − 18 − 11 + 3
n(x ∩ y) = 6
F 29 F 99
F A = F F F F
B = F F F F
C = F F F F
กF Fก F F
A( F ) B( F )
a y b
3
x z
c
C( F ) U = 40
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(A ∩ C) − n(B ∩ C) + n(A ∩ B ∩ C)
40 = 21 + 16 + 19 − (y + 3) − (x + 3) − (z + 3) + 3
x + y + z = 10
ก (a + b + c) + (x + y + z) + 3 = 40
a + b + c + 10 + 3 = 40
a + b + c = 27
∴ F ก ก F = a + b + c = 27
F 31 F 99
ก n[(A ∪ C) ∪ (B ∪ C)] = n(A ∪ C) + n(B ∪ C) − n[(A ∪ C) ∩ (B ∪ C)]
n(A ∪ B ∪ C) = 32 + 29 − n[(A ∩ B) ∪ C]
n(A ∪ B ∪ C) = 32 + 29 − 28 = 33
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