ДискретнаяМатематика / Student Solutions Manual / chapter 1

11.There are 76 students enrolled in Anth229, Intermediate Anthropology. Each of these students is also required to enroll in either one or both of Biol113, Physiology and Engl218, Victorian Poets. Of these 76 students there are 35 in Biol313 and 49 in Engl218. How many students are enrolled in all three classes?

11. Let A = fAnth229 students in Biol113g and B = fAnth229 students in Engl218g.

j A [ B j = j A j + j B j j A \ B j

76 = 35 + 49 j A \ B j

j fAnth229 students taking Biol113 and Engl218g j = 8

13. How many numbers between 1 and 1000 are not divisible by 3, 7, or 9?

13. Let D3 = fIntegers in this range divisible by 3g; D7 = fIntegers in this range divisible by 7g; D9 = fIntegers in this range divisible by 9g; D21 = fIntegers in this range divisible by 21g; and D63 = fIntegers in this range divisible by 63g. j D3 j = 333: j D7 j = 142: j D9 j = 111: j D3 \D7 j = j D21 j = 47: j D3 \D9 j = j D9 j = 111: j D7 \ D9 j = j D63 j =

j D3 [ D7 [ D9 j = 1000 (j D3 j + j D7 j + j D9 j

j D21 j j D9 j j D63 j + j D63 j)

=1000 (333 + 142 + 111 47 111 15 + 15)

=572

Easier: since D9 D3; we can just compute jD3 [ D7j = 1000 333 142 + 47 = 572:

15.(a) How many numbers between 1 and 70,000,000, including 1 and 70,000,000, are divisible by 2, 5, or 7?

(b) How many numbers between 1 and 6,000,000, including 1 and 6,000,000, are divisible by 4, 5, or 6?

15. (a) Let D2 = fIntegers in this range divisible by 2g; D5 = fIntegers in this range divisible by 5g; D7 = fIntegers in this range divisible by 7g; D10 = fIntegers in this range divisible by 10g; D14 = fIntegers in this range divisible by 14g; and D35 = fIntegers in this range divisible by 35g. j D2 j = 35; 000; 000: j D5 j = 14; 000; 000: j D7 j = 10; 000; 000: j D2 \ D5 j = j D10 j = 7; 000; 000: j D2 \ D7 j = j D14 j = 5; 000; 000: j D5 \ D7 j = j D35 j = 2; 000; 000: j D2 \ D5 \ D7 j = j D70 j = 1; 000; 000: The number of integers in this range that are divisible by 2, 5, or 7 is:

j D2 [ D5 [ D7 j = j D2 j + j D5 j + j D7 j j D10 j

j D14 j j D35 j + j D70 j

21.(a) Extend Example 9 to cover four Victorian gentlemen and four top hats. With four gentlemen there are 4 3 2 1 = 24 ways to give the hats back.

(b)Modify part (a) to ask the number of ways, with four gentlemen and four hats, that at least two gentlemen can get their own hats back.

(c)Solve Example 9 using an alternate proof that counts the number of ways no gentleman gets his own hat back and subtracts that value from the total number of ways for the hats to be given back.

(d)Challenge: Solve part (b) using the same method as for part (c).

21.(a) Let H1 be the number of times person 1 receives the right hat; H2 be the number of times person 2 receives the right hat; H3 be the number of times person 3 receives the right hat; and H4 be the number of times person 4 receives the right hat.

The answer is:

jH1 [ H2 [ H3 [ H4j = jH1j + jH2j + jH3j + jH4j

jH1 \ H2j jH1 \ H3j jH1 \ H4j

jH2 \ H3j jH2 \ H4j jH3 \ H4j + jH1 \ H2 \ H3j + jH1 \ H2 \ H4j

+ jH2 \ H3 \ H4j jH1 \ H2 \ H3 \ H4j

=6 + 6 + 6 + 6 2 2 2 2 2 2

+1 + 1 + 1 + 1 1

=15

21.(b) Calculate the number of ways at least one gentleman gets the right hat back and subtract the number of ways exactly one gentleman gets the right hat back. The answer is 7.

21. (c) Suppose the gentlemen and their respective hats are numbered 1, 2, 3, and 4. Each way of returning the hats gives rise to an ordered 4-tuple, where the number in position i for 1 i 4 tells which hat was returned to gentleman i. We will count the number of 4-tuples such that the number of position i is not equal to i for any i such that 1 i 4. There are three choices for the rst position. If the rst position is lled with 2, then there are three choices for position 2. Whatever choice is made for position 2, there is only one way to ll the remaining two positions because at least one of the remaining hats belongs to either 3 or 4. Therefore there are three possibilities if the rst position receives 2.

If the rst position receives either 3 or 4, there are two possibilities for position 2 cannot be put in position 2 but either one 1 or x can be chosen for position 2 where x is either 3 or 4 depending on which hat was put in position 2. In the case 3 or 4 is put in the rst two positions there are two possibilities for position 3 since either hat 1 or 2 can be put in position 3. This leaves a single remaining possibility for position 4. If one of 3 or 4 is not put in position 1 or 2, there is only one way to ll positions 3 and 4. Altogether there are three ways to distribute the hats so that no owner received the right hat if the rst gentleman received hat 3. The same statement is true if the rst gentleman received hat 4. Since putting hat 3 in the rst position is a di erent case from putting hat 4 in the rst position, this reasoning leads to six possibilities for passing out the hats so that no one receives the right one.

Therefore, there are 9 ways to pass out the hats so that no gentleman receives the right one. Since there are 24 ways to pass out the hats, there are 15 ways to pass out the hats so that at least one owner receives the right hat.

21. (d) The solution comes in two parts. The rst part is to count the number of ways no one receives the right hat. This is 9 from (c). The second part is to count the number of ways only one person receives the right hat.

The number of assignments for which exactly one hat is given to the right person can be counted as follows. Since one hat is given to the right person, the idea is to solve a new problem and count how many ways three hats can be assigned so that each hat goes to someone other than the owner.

For the new problem, suppose there are three hats numbered 1, 2, and 3 that must be put in positions 1, 2, and 3 so that no hat is put in the position of its number. Suppose 2 is put in position 1. There is only one way to complete the assignment so that 1 and 3 are put in positions 2 and 3 and 3 is not put in position 3. There is also one possibility if 3 is put in position 1. Thus for each assignment with exactly one hat in the right place there are two ways to complete the assignment. Therefore there are 8 ways to assign four hats so that exactly one hat is given to its owner.

There are 9 ways to assign hats so that no one gets the right hat and 8 ways to assign hats so that only one person gets the right hat. Therefore there are 17 ways to assign the hats so that at most one person gets the right hat. The answer is 24 - 17 = 7.

1.9 Exercises

Assume all variables not given an explicit domain are elements of N:

1. Show that for n = 0; 1; 2 the following is true

12 + 22 + 32 + + n2 = n(n + 1)(2n + 1)=6

1. When n = 0, a sum of the form F (1) + f (2) + + f (n) is understood to consist of just the f (n) term, which is f (0).

1:1 1 = 1(2)(3)/6

2:5 5 = 2 (3) (5) / 6

3.Write out the information that describes what the inductive step assumes and what the step must prove for proving

12 + 22 + 32 + + n2 = n(n + 1)(2n + 1)=6

with n0 given.

3. The inductive hypothesis is that n 2 T , that is: n n0 and 12 + 22 + 32 + + n2 = n(n + 1)(2n + 1)=6

What must be proven is that n + 1 2 T ; that is, n + 1 n0 and that

12 + 22 + 32 + + n2 + (n + 1)2 = (n + 1)((n + 1) + 1)(2(n + 1) + 1)=6

5.Write out the information that describes what the inductive step assumes and what the step must prove for proving that 6 divides n3 + 5n with n0 given.

5. The inductive hypothesis is that n 2 T , that is: n n0 and that n 2 T ; that is, 6 j (n3 + 5n): What must now be proven is that n + 1 2 T ; that is: n + 1 n0 and 6 j ((n + 1)3 + 5(n + 1)):

7. Show for n = 0; 1; 2 that

(n + 1)(2n + 1)(2n + 3)=3 + (2n + 3)2 = (n + 2)(2n + 3)(2n + 5)=3

9. Show that

n2 + n + 2(n + 1) = (n + 1)2 + (n + 1)

9.

n2 + n + 2(n + 1) = n2 + n + 2n + 2 = n2 + 3n + 2

= n2 + 2n + 1 + (n + 1) = (n + 1)2 + (n + 1)

11.For which elements n 2 f0; 1; 2; 3; 4; 5g does 6 divide n3 + 5n? 11.

13.Find the smallest n 2 N such that 2n2 + 3n + 1 < n3:

13.

The answer is 4.

15.Prove by induction:

(a)12 + 22 + 32 + + n2 = n (n + 1) (2n + 1)=6 for n 0

(b)13 + 23 + 33 + + n3 = (1 + 2 + 3 + + n)2 for n 0

(c)14 + 24 + 34 + + n4 = n(n + 1)(2n + 1)(3n2 + 3n 1)=30 for n 0

(d)15 + 25 + 35 + + n5 = 16 n6 + 12 n5 + 125 n4 121 n2 for n 0

(Base step) For n = 0 the left-hand side and the right-hand side both equal 0, so n0 2 T .

(Inductive step) Assume n 2 T : Now prove that n + 1 2 T : That is, assume that n n0 and that

12 + 22 + 32 + + n2 = n(n + 1)(2n + 1) 6

is true and show that

12 + 22 + 32 + + n2 + (n + 1)2 = (n + 1)(n + 2)(2n + 3) 6

and that n + 1 n0 are true.

(Base step) For n = 0 the left-hand side and the right-hand side both equal 0, so n0 2 T .

(Inductive step) Assume n 2 T : Now prove that n + 1 2 T : That is, assume that n n0 and that

13 + 23 + 33 + + n3 = (1 + 2 + 3 + + n)2 are true and prove that

13 + 23 + 33 + + n3 + (n + 1)3 = (1 + 2 + 3 + + n + (n + 1))2 and that n + 1 n0 are true.

13 + 23 + n3 + (n + 1)3 = (13 + 23 + n3) + (n + 1)3

Since n + 1 > n, it follows that n + 1 n + 0. Therefore, n + 1 2 T :

By the Principle of Mathematical Induction, T = N.

15. (c) Let n0 = 0: Let

T = fn 2 N : 14 + 24 + 34 + + n4 = n(n + 1)(2n + 1)(3n2 + 3n 1)=30g

(Base step) For n = 0 the left-hand side and the right-hand side both equal 0, so 0 2 T .

(Inductive step) Assume n 2 T : Now prove that n + 1 2 T : That is, assume that n n0 and that

14 + 24 + 34 + + n4 = n(n + 1)(2n + 1)(3n2 + 3n 1)=30

are true and prove that

14+24+34+ +n4+(n+1)4 = (n+1)(n+2)(2n+3)(3n+1)2+3(n+1) 1)=30

is true.

14 + 24 + 34 + + n4 + (n + 1)4 = (14 + 24 + 34 + + n4) + (n + 1)4

=n(n + 1)(2n + 1)(3n2 + 3n 1)=30 + (n + 1)4 (by ind. hyp.)

=(n + 1)(6n4 + 39n3 + 91n2 + 89n + 30)=30

=(n + 1)(n + 2)(2n + 3)(3(n + 1)2 + 3(n + 1) 1)=30

Of course, n + 1 > n n0. Therefore, n + 1 2 T :

By the Principle of Mathematical Induction, T = N.

15. (d) Let n0 = 0: Let

T = fn 2 N : 15 + 25 + 35 + + n5 = 16 n6 + 12 n5 + 125 n4 121 n2g

(Base step) For n = 0 the left-hand side and the right-hand side both equal 0, so 0 2 T .

(Inductive step) Assume n 2 T : Now prove that n + 1 2 T : That is, assume that n n0 and that

15 + 25 + 35 + + n5 = 16 n6 + 12 n5 + 125 n4 121 n2

are true and prove that

15+25+35+ +n5+(n+1)5 = 16 (n+1)6+ 12 (n+1)5+ 125 (n+1)4 121 (n+1)2

and n + 1 n0 are true.

15 + 25 + 35 + + n5 + (n + 1)5 = (15 + 25 + 35 + + n5) + (n + 1)

=16 n6 + 12 n5 + 125 n4 121 n2 + (n + 1)5 (by ind. hyp.)

Of course n + 1 > n n0. Therefore, n + 1 2 T :

By the Principle of Mathematical Induction, T = N.

for n 1

(Base step) For n = 1 the left-hand side and the right-hand side both equal 1/2, so 1 2 T .

(Inductive step) Assume n 2 T : Now prove that n + 1 2 T : That is, assume that n n0 = 1 and that

n1

=n + 1 + (n + 1) (n + 2) (by ind. hyp.)

1(n + 1)2

Of course, n + 1 n0. Therefore, n + 1 2 T :

By the Principle of Mathematical Induction, T = fn 2 N : n 1g.

(Base step) For n = 1 the left-hand side and the right-hand side both equal 1/2, so 1 2 T .