ДискретнаяМатематика / Student Solutions Manual / chapter 7

5.How many ways can ve identical advertisements be placed in three mailboxes if each mailbox receives at least one advertisement? How many ways if a mailbox may receive none? (The order in which a messenger delivers a message is immaterial.)

5. The number of solutions of the system x1 + x2 + x3 = 2 with x1 0; x2 0; x3 0 gives the answer to the rst part of the question. The number of solutions of the equation x1 + x2 + x3 = 5 gives the answer to the second part.

7.How many ways can you choose eight letters from

aaaaabbbbbb cccccccc

with at least one a, one b, and two c's?

7.After removing the letters that must be chosen, let xa be the number of a's chosen, xb be the number of b's chosen, and let xc be the number of c's chosen. We must solve x1 + xb + xc = 4 where x1 0, xb 0, and xc 0: The answer is C(6; 2):

9.How many ways can two booksellers divide between themselves 300 copies of one book, 200 copies of another, and 100 copies of a third if neither bookseller is to get all the copies of any one of the books?

9.Each bookseller gets between 1 and 299 copies of the rst book, 1 and

199copies of the second, and 1 and 99 copies of the third. The answer is

299199 99:

11.How many ways can six candy bars be distributed among three children if every child is to receive at least one candy bar?

11. Use Theorem 1 with n = 3 and r = 3: C(3 + 3 1; 3 1)

13.A king is placed on the bottom left hand square of an 8 8 chess board and is to move to the top right hand corner square. If the piece can move only up or to the right, how many possible paths does it have?

13. C(14; 7) { of the sixteen moves eight will be up. When the up moves are identi ed, the remaining moves are simply right.

15.Three rst-year, three second-year, and three third-year students are to be seated in a row. The students in each class are indistinguishable. How many ways can they be seated so that no three students of the same class sit together.

15. The total number of permutations of the three classes of students is

9!

3!3!3!

Let A1 be the number of permutations with three rst-year students in a row; A2 be the number of permutations with three second-year students in a row; and A3 be the number of permutations with three third-year students in a row. The answer is

jA1 \ A2 \ A3j = T otal jA1j jA2j jA3j

+jA1 \ A2j + jA1 \ A3j + jA2 \ A3j jA1 \ A2 \ A3j

=1680 3 (140) + 3 (20) 6

=1314

17.A shop sells six avors of ice cream. Each ice cream cone holds one, two, or three scoops of ice cream. How many ways can four ice cream cones be made such that:

(a)All the cones have a di erent avor, and each cone has a single avor for each of its scoops.

(b)Not necessarily all the cones have a di erent avor.

(c)The cones contain only two or three avors of ice cream.

(d)The cones contain three di erent avors.

17.(a) C(6; 4)

17.(b) C(9; 3)

17.(c) C(9; 3) C(6; 4) C(6; 1)

17.(d) C(6; 3)C(3; 1)

19.How many ways can an eight-person committee be chosen from a group of 10 new members and 15 old members if the committee is composed of

(a)Four members from each group

(b)More new members than old members

(c)At least two new members

19.(a) C(10; 4) C(15; 4)

19. (b) Five new + six new + seven new + eight new

8

X

=C(10; k)C(15; 8 k)

k=5

19. (c) 2 new + 3 new + 4 new + 5 new + 6 new + 7 new + 8 new =

8

k=2 C(10; k)C(15; 8 k)

21.A domino is made of two squares, each of which is marked with one, two, three, four, ve, or six spots or is left blank. A set of dominos consists of dominos with all possible pairs showing in the two squares. How many dominos are there in a set.

21. No. doubles + No. di erent pairs = 7 + C(7; 2)

23. Prove Pn C(n; k)2 = C(2n; n):

k=0

(a)Prove the identity using the fact that (1 + x)2n = (1 + x)n(1 + x)n.

(b)Give a combinatorial proof of the identity in part (a).

(c)Find the number of 14-digit binary sequences for which the number of 1's in the rst seven digits is the same as the number of 0's in the last seven digits of the sequence. Enumerate all such sequences of length six.

23. (a)

(1 + x)2n = (1 + x)n(1 + x)n

The expansion of these two polynomials is given by

The coe cient of xn for the lhs polynomial is C(2n; n): The coe cient for xn for the rhs polynomial is

n

X

C(n; k)C(n; n k)

k=0

By Pascal's Identity, C(n; n k) = C(n; k) so

23. (b) The right-hand side counts the number of n element subsets of a set with 2n elements. For the left-hand side divide the 2n element set into two sets of size n: Choose k elements to form the rst set and n k elements to form the second set where k = 0; 1; : : : ; n:

23. (c) There are k ones in the rst seven locations and 7 k ones in the second seven elements where k = 0; 1; : : : ; 7: The answer is C(14; 7) = 3; 432:

25. Sum 13 + 23 + ::: + m3 using the fact that m3 can be represented as

m3 = aC(m; 3) + bC(m; 2) + cC(m; 1)

where a; b, and c are rational numbers. The problem should not be solved using a proof by induction.

27.Prove that 1 2 3+2 3 4 + ::: + n (n+1) (n+2) = n(n + 1)(n + 2)(n + 3)=4. The problem should not be solved using a proof by induction.

27. The general term k(k 1)(k 2) in the sum is equal to

or P (k; 3) = 3! C(k; 3): The sum can be rewritten

3!(C(3; 3) + C(4; 3) + + C(n; 3) + C(n + 2; 3)) = 3!C(n + 3; 4)

29.Construct the rst 10 rows of Pascal's triangle. 29.

31.Expand (1 + x)8 using the Binomial Theorem.

31. 1+C(8; 1)x+C(8; 2)x2 +C(8; 3)x3 +C(8; 4)x4 +C(8; 5)x5 +C(8; 6)x6 + C(8; 7)x7 +C(8; 8)x8 = 1+8x+28x2 +56x3 +70x4 +56x5 +28x6 +8x7 +x8

33.In the expansion of (3x 2y)18 what are the coe cients of:

(a)x5 y13

(b)x3 y15

33. (a) -17,055,940,608

33.(b) -721,944,576

35.Use the Binomial Theorem to prove that

n

X

3n = C(n; k)2n k

k=0

Write out what the identity says for n = 4:

81 = C(4; 0) + 2C(4; 1) + 4C(4; 2) + 8C(4; 3) + 16C(4; 4)

Consider the number of sequences of length n that can be constructed from the letters a, b, and c. Then, consider the number of times the letter a occurs in such a sequence.

If a occurs an even number of times, prove the identity

3n + 1 = 2n + C(n; 2)2n 2 + + C(n; q)2n q 2

where q = n for n even and q = n 1 for n odd.

37. Theorem 4 in Section 7.9 proves that C(n; m)C(m; k) = C(n; k)C(n k; m k): Use this result to prove that C(n; k)C(n k; m) = C(n; m)C(n m; k):

37.The left-hand side counts the number of ways of selecting two sets:rst a set A of m objects and then from A; a set B of k objects. The right-hand side counts the number of ways to select k elements from an n-element set and then m k elements from the remaining n k elements.

39.Prove that C(n; r) > C(n; r 1) if r < n=2 for each row of Pascal's triangle.

39.Let 1 k n.

n

X

(1 + x)n = C(n; k)xk

k=0

Take the derivative of each side giving:

n

X

n(1 + x)n 1 = k C(n; k)xk 1

k=1

Now let x equal to 1.

n

X

n2n 1 = kC(n; k)

k=1

P5

43. Evaluate (1=4) k=1 k (6 k) and deduce the number of points of intersection for the diagonals of an octagon if no three diagonals meet at a point.

43. Number the vertices of an octagon 1 - 8 in clockwise order. We will do the analysis for vertex 1 and then multiply by 8 to get the total number of intersections{each counted more than once. We divide this number by 4 to get the real number of intersections as we count a given intersection for each of the vertices at the four ends of the two edges that determine an intersection.

For any diagonal there are k vertices on one side of the diagonal and 6 k vertices on the other side of the diagonal. Therefore, there are k(6 k) intersections where 1 k 5: The total number of intersec-

tions is 8 P5 (k(6 k)): If we now divide by 4, the number of distinct

k=1

intersections is determined.

45.Show that there are (3n + 1)=2 strings of length n consisting of the letters a, b, and x in which a occurs an even number of times.

45. There are 3n + 1 possible strings counting the empty string. By Theorem 8, there are as many words with an even number of a's as there are words with an odd number of a's.

47.For a positive integer t, de ne [x]t = x(x 1) (x t + 1): We can represent xn as a linear combination of [x]t, where n = 1; 2; 3; : : : and t = 0; 1; 2; : : : ; n: The coe cients for this expansion are denoted as S(n; t) and are known as the Stirling numbers of the second kind. Thus, for

any n, we can write

n

X

xn = S(n; t)[x]t

t=0

The numbers S(n; t) can be de ned for n = 1; 2; 3; : : : as S(n; 0) = 0; S(n; n) = 1; and

S(n; t) = tS(n 1; t) + S(n 1; t 1)

for 1 t n 1: Make a table of the Stirling numbers of the second kind for n = 1; 2; 3; 4; 5; 6:

47.

49. Still another application of the Principle of Inclusion-Exclusion: Let

x1 x2 xn 1 xn

be a permutation of f1; 2; 3; : : : ; n 1; ng such that for 1 i n there is no xi = i. A permutation with this property is called a derangement. We denote the set of derangements 0n n elements as Dn. Find formula for the number of derangements on 1; 2; : : : ; n 1; n for any n 2 N. Evaluate this formula for n = 2; 3; 4; 5; 6. Construct all derangements for n = 2; 3; and 4. (Hint: See Example 12 in Section 7.5.7.)

49. Let Pi denote the set of permutation that have xi = i for 1 i n:

XX

Dn = n! N (Pi) + N (PiPj ) + ( 1)nN (P1P2 Pn)

N (Pi) = n 1)! since xi = i and the remaining positions can be lled arbitrarily. For the sum P N (Pi Pk ) consisting of l terms, we have

X

N (Pi Pk ) = C(n; l)(n l)!

We get

Dn = n!(1 1=1! + 1=2! + ( 1)n(1=n!))

7.12 End of Chapter Materials

Starting to Review

1.What are the values of C(5; 3) and P (5; 3)?

(a)60, 10

(b)60, 20

(c)60

(d)None of the above

1.a

3.How many ways can you roll two dice and get a total of six appearing on the top faces?

3.Just observe that one die has values 1, 2, 3, 4, 5, 6 as possibilities and to get a sum of 6 you simply complement the values 1, 2, 3, 4, 5. A solution that gives a more general setting to the problem is the following. x1 + x2 = 6 minus the solutions 0, 6 and 6, 0. C(6 + 2 1; 2 1) 2 Alternatively put 1 in each value and distribute the 4 left. This works because there is no 0 on a die.

5.To graduate, Sally needs two courses to complete the general education requirement. Courses in anthropology and economics will satisfy the requirement. If Sally has satis ed the prerequisites for 18 anthropology and

21economics courses, how many possible schedules are possible?

5.741

7.There are 10 geography books, 12 chemistry books, and 18 detective novels. How many ways can you pick two books from each of two di erent groups of books?

7.C(12; 2) C(18; 2) C(10; 2)

9.Find n such that C(n; 0) + C(n; 1) + + C(n; n) = 128:

9. n = 7 works. C(7; 0) + C(7; 1) + C(7; 2) + C(7; 3) + C(7; 4) + C(7; 5) + C(7; 6) + C(7; 7) = (1 + 1)7 = 128

Review Questions

1.Find the number of ways in which nine 3's and six 5's can be placed in a row so that no two 5's are together.

1.Place the nine 3's in positions numbered 1, 2, . . . , 9. Now, create positions a, 1a, 2a, 3a, 4a, . . . , 9a with position a to the left of the rst 3 and positions ia between position i and position i + 1 for 1 i 8. You have created 9 new positions. Now add position 9a to the right of position 9 giving 10 new positions in all. Choose six of these new positions for the 5's. There are C(10; 6) ways to choose positions for the 5's.

3.In the interest of e ciency spelling rules for words have been revised. The word relief can be spelled in the ways described by the following rules:

The number of letters must not exceed 6.

The word must contain at least one l.

The word must begin with an r and end with an f.

There is just one r and one f, and only the letters e, i, and l may occur in the middle positions.

How many ways can relief be spelled?

3. 1 + 2 3 + 3 32 + 4 33. Place l and then ll with the letters e, i, and l in all possible ways.

5.How many solutions are there for x1 +x2 +x3 +x4 = 30 with 10 xi 20 for 1 i 4:

5. Solve x1 + x2 + x3 + x4 = 70 in C(73; 3) ways. Now, subtract the number of solutions with an xi 20, which is done using the Principle of Inclusion-Exclusion. Let Ci = # solutions with xi > 20 for 1 i 4.

N (C1) = N (C2) = N (C3) = N (C4) = C(52; 3)

N (C1C2) = N (C1C3) = N (C1C4) = N (C2C3) =

N (C2C4) = N (C3C4) = C(31; 3)

N (C1C2C3) = N (C1C2C4) = N (C1C3C4) = N (C2C3C4) = C(10; 3) N (C1C2C3C4) = 0

The number of solutions with at least one xi having a value greater than 20 is:

N (C1C2C3C4) = C(73; 3) 4N (C1) + 6N (C1C2) 4N (C1C2C3) +N (C1C2C3C4)

7. How many solutions in integers are there for

x1 + x2 + x3 = 14

-2 as if you could put -2 elements in box 1, and then solve the resulting equation for 16. The upper bounds are handled by the Principle of Inclusion-Exclusion using the sets A1 = f(x1; x2; x3) 2 U : x1 > 5g; A2 = f(x1; x2; x3) 2 U : x2 > 6g; and A3 = f(x1; x2; x3) 2 U : x3 > 7g:)

7. Solve

x1 + x2 + x3 = 14

with x1; x2; x3 0: Then let C1 = # of solutions with x1 > 4; C2 = # solutions with x2 > 5; and C3 = # solutions with x3 > 6: Solve

x1 + x2 + x3 = 9 to nd jC1j

x1 + x2 + x3 = 8 to nd jC2j

x1 + x2 + x3 = 7 to nd jC3j

Now, compute

N (C1C2C3) = C(16; 2) (N (C1) + N (C2) + N (C3)

N (C1C2) N (C1C3) N (C2C3)

+N (C1C2C3))