《组合数学》第五版 第6章答案.pdf

set size
justification
S
17 3
17 = 14 + 4 − 1
Ai
8 3
17 − 9 = 8
Ai ∩ Aj 0 17 − 9 − 9 = −1 < 3
By inclusion/exclusion
17
8
|A1 ∩ A2 ∩ A3 ∩ A4| =
3
− 4 = 456. 3
8. Let S denote the set of positive integral solutions for x1 + x2 + x3 + x4 + x5 = 14. For 1 ≤ i ≤ 5 let Ai denote the set of elements in S with xi ≥ 6. We seek |A1 ∩A2 ∩A3 ∩A4 ∩A5|. We have
set size
justification
S 104
A 100
1002 = 104
B 21 213 = 9261 and 223 = 10648
A ∩ B 4 46 = 4096 and 56 = 15625
By inclusion/exclusion |A ∩ B| = 104 − 100 − 21 + 4 = 9883.
3
By inclusion/exclusion
14
7
7
10
3
3
|X ∩ Y ∩ Z| =
−−−
+ + = 10.
2
2
2
2
2
2
7. Let S denote the set of nonnegative integral solutions for x1 + x2 + x3 + x4 = 14. For 1 ≤ i ≤ 4 let Ai denote the set of elements in S with xi ≥ 9. We seek |A1 ∩ A2 ∩ A3 ∩ A4|. We have
By inclusion/exclusion
|A ∩ B ∩ C| = 104 − 2500 − 2000 − 1666 + 500 + 333 + 833 − 166 = 5334.
2. Define the set S = {1, 2, . . . , 104}. Let A (resp. B) (resp. C) (resp. D) denote the set of integers in S that are divisible by 4 (resp. 6) (resp. 7) (resp. 10). We seek |A ∩ B ∩ C ∩ D|. We have
Y ∩ Z ∩ W 0 13 − 5 − 6 − 8 = −6 < 3
By inclusion/exclusion
13
8
7
5
|Y ∩ Z ∩ W | =
− − − = 185.
3
3
3
3
6. We seek the number of integral solutions for
x + y + z = 12, 0 ≤ x ≤ 6, 0 ≤ y ≤ 6, 0 ≤ z ≤ 3.
set
size
justification
S
13 3
13 = 10 + 4 − 1
Y
8 3
13 − 5 = 8
Z
7 3
13 − 6 = 7
W
5 3
13 − 8 = 5
Y ∩Z
0
13 − 5 − 6 = 2 < 3
Y ∩W
0
13 − 5 − 8 = 0 < 3
Z ∩W
0 13 − 6 − 8 = −1 < 3
4
set
size
justification
S
16 3
16 = 13 + 4 − 1
A1
10 3
16 − 6 = 10
A2
8 3
16 − 8 = 8
A3
11 3
16 − 5 = 11
A4
11 3
16 − 5 = 11
A1 ∩ A2
0 16 − 6 − 8 = 2 < 3
A1 ∩ A3
5 3
A1 ∩ A4
Let S denote the set of nonnegative integral solutions to y1 + y2 + y3 + y4 = 13. Let A1 (resp. A2) (resp. A3) (resp. A4) denote the set of elements in S such that y1 ≥ 6 (resp. y2 ≥ 8) (resp. y3 ≥ 5) (resp. y4 ≥ 5). We seek |A1 ∩ A2 ∩ A3 ∩ A4|. We have
set
S A B C A∩B B∩C A∩C A∩B∩C
size 104 2500 2000 1666 500 333 833 166
justification
104 = 2500 × 4 104 = 2000 × 5 104 = 1666 × 6 + 4 104 = 500 × 20 104 = 333 × 30 + 10 104 = 833 × 12 + 4 104 = 166 × 60 + 40
set
size
justification
S
15 3
15 = 12 + 4 − 1
X
10 3
Y
11 3
Z
10 3
W
9 3
15 − 5 = 10 15 − 4 = 11 15 − 5 = 10 15 − 6 = 9
X ∩Y
6 3
X ∩Z
5 3
X ∩W
4 3
Y ∩Z
6 3
Y ∩W
5 3
Z ∩W
4 3
15 − 5 − 4 = 6 15 − 5 − 5 = 5 15 − 5 − 6 = 4 15 − 4 − 5 = 6 15 − 4 − 6 = 5 15 − 5 − 6 = 4
set
size
justification
S
14 2
14 = 12 + 3 − 1
X
7 2
14 − 7 = 7
Y
7 2
14 − 7 = 7
Z
10 2
14 − 4 = 10
X ∩Y
0
14 − 7 − 7 = 0 < 2
X ∩Z
3 2
Y ∩Z
3 2
14 − 7 − 4 = 3 14 − 7 − 4 = 3
X ∩ Y ∩ Z 0 14 − 7 − 7 − 4 = −4 < 2
Math 475
Text: Brualdi, Introductory Combinatorics 5th Ed.
Prof: Paul Terwilliger
Selected solutions for Chapter 6
1. Define the set S = {1, 2, . . . , 104}. Let A (resp. B) (resp. C) denote the set of integers in S that are divisible by 4 (resp. 5) (resp. 6). We seek |A ∩ B ∩ C|. We have
5 3
A2 ∩ A3
3 3
A2 ∩ A4
3 3
A3 ∩ A4
6 3
16 − 6 − 5 = 5 16 − 6 − 5 = 5 16 − 8 − 5 = 3 16 − 8 − 5 = 3 16 − 5 − 5 = 6
Let S denote the set of nonnegative integral solutions to x + y + z = 12. Let X (resp. Y ) (resp. Z) denote thΒιβλιοθήκη Baidu set of elements in S such that x ≥ 7 (resp. y ≥ 7) (resp. z ≥ 4). We seek |X ∩ Y ∩ Z|. We have
set size
justification
S
13 4
13 = 14 − 5 + 5 − 1
Ai
8 4
13 − 5 = 8
Ai ∩ Aj 0 13 − 5 − 5 = 3 < 4
By inclusion/exclusion
13
8
|A1 ∩ A2 ∩ A3 ∩ A4 ∩ A5| =
4
− 5 = 365. 4
1
By inclusion/exclusion
|A ∩ B ∩ C ∩ D| = 104 − 2500 − 1666 − 1428 − 1000 + 833 + 357 + 500 + 238 + 333 + 142 − 119 − 166 − 71 − 47 + 23
= 5429.
3. Define a set S = {1, 2, . . . , 104}. Let A (resp. B) denote the set of integers in S that are perfect squares (resp. perfect cubes). We seek |A ∩ B|. We have
9. We make a change of variables
y1 = x1 − 1, y2 = x2, y3 = x3 − 4, We seek the number of integral solutions to
y4 = x4 − 2.
y1 + y2 + y3 + y4 = 13, 0 ≤ y1 ≤ 5, 0 ≤ y2 ≤ 7, 0 ≤ y3 ≤ 4, 0 ≤ y4 ≤ 4.
Let S denote the set of nonnegative integral solutions to x + y + z + w = 10. Let Y (resp. Z) (resp. W ) denote the set of elements in S such that y ≥ 5 (resp. z ≥ 6) (resp. w ≥ 8). We seek |Y ∩ Z ∩ W |. We have
2
By inclusion/exclusion
15
10
11
10
9
|X ∩ Y ∩ Z ∩ W | =
−
−
−
−
3
3
3
3
3
6
5
4
6
5
4
++++++
3
3
3
3
3
3
= 34.
5. These 10-combinations correspond to the integral solutions for
x + y + z + w = 10, 0 ≤ x, 0 ≤ y ≤ 4, 0 ≤ z ≤ 5, 0 ≤ w ≤ 7.
4. These 12-combinations correspond to the integral solutions for
x + y + z + w = 12, 0 ≤ x ≤ 4, 0 ≤ y ≤ 3, 0 ≤ z ≤ 4, 0 ≤ w ≤ 5.
Let S denote the set of nonnegative integral solutions to x + y + z + w = 12. Let X (resp. Y ) (resp. Z) (resp. W ) denote the set of elements in S such that x ≥ 5 (resp. y ≥ 4) (resp. z ≥ 5) (resp. w ≥ 6). We seek |X ∩ Y ∩ Z ∩ W |. We have
set
S A B C D A∩B A∩C A∩D B∩C B∩D C ∩D A∩B∩C A∩B∩D A∩C ∩D B ∩C ∩D A∩B ∩C ∩D
size 104 2500 1666 1428 1000 833 357 500 238 333 142 119 166 71 47 23
justification
104 = 2500 × 4 106 = 1666 × 6 + 4 104 = 1428 × 7 + 4
104 = 1000 × 10 104 = 833 × 12 + 4 104 = 357 × 28 + 4
104 = 500 × 20 104 = 238 × 42 + 4 104 = 333 × 30 + 10 104 = 142 × 70 + 60 104 = 119 × 84 + 4 104 = 166 × 60 + 40 104 = 71 × 140 + 60 104 = 47 × 210 + 130 104 = 23 × 420 + 340
X∩Y ∩Z
0
15 − 5 − 4 − 5 = 1 < 3
X∩Y ∩W 0
15 − 5 − 4 − 6 = 0 < 3
X ∩Z ∩W
0 15 − 5 − 5 − 6 = −1 < 3
Y ∩Z∩W
0
15 − 4 − 5 − 6 = 0 < 3
X ∩ Y ∩ Z ∩ W 0 15 − 5 − 4 − 5 − 6 = −5 < 3