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Chapter 6
Teaching Notes, Chapter 6: The material in Sections 6.1 and 6.2 will be more familiar to most
students than the later sections. In the early sections, some students may have difficulty
distinguishing when to use the Multiplication Principle and when to use the Addition Principle.
Others may have difficulty distinguishing when to use permutations and when to use combinations.
You may want to spend some time in class with Exploratory Exercises that focus on these
distinctions. (See Exploratory Exercise 1 from 6.1 and Exploratory Exercises 1 and 2 from 6.2.)
For some classes it may be appropriate to cover only sections 6.1 and 6.2. This minimal coverage
would still allow students sufficient background to cover the first two sections of Chapter 7.
However, we would encourage instructors to consider including Section 6.3 in a minimal approach to
this chapter, in order to further students’ understanding of the connections among combination
numbers, binomial coefficients, and Pascal’s Triangle.
Of the later sections, Section 6.5 is particularly proof-intensive. If the instructor is not
emphasizing proofs, a more comprehensive coverage of this chapter could include Sections 6.4 and
6.6, in which strategies for solving more complex counting problems are developed by recasting them
in terms of the basic problems outlined in the earlier sections. This will allow for a more
comprehensive coverage of probability in Chapter 7.
Exploratory Exercise Set 6.1
1. (a) 3  2  2 = 12
CW
G
WS
CM
WP
CM
WP
CM
CW
(b)
P
WS
CW
T
WP
CM
WP
CM
WP
CM
Abbreviations:
G: Geoboards
P: Pentominoes
T: Tangrams
CW: Crossword
WS: Word Search
CM: Clay Map
WP: Wood Puzzle
WS
WP
(c) 3 + 2 + 2 = 7
(d) The Multiplication Principle is used in part (a) since students must pick one activity from each
subject area. The Addition Principle is used in (b) since students must pick one activity from
one of the subject areas.
2. (a) P(10, 4) = 10  9  8  7 = 5040
(b) 10  10  10  10 = 10,000
(c) 10  9  9  9 = 7290
3. (a) ACE, AEC, CAE, CEA, EAC, ECA
(b) two: CEA and EAC
(c) No; Y is in its original position.
(d) There are 9: EBTS, ESTB, ETBS, SBTE, STBE, STEB, TBES, TSBE, and TSEB.
4. (a) ATE, EAT, MAT, MET, SAT, SEA, SET, TAM, and TEA
(b) P(5, 3) = 5  4  3 = 60
(c) You should get 10: {A, E, M}, {A, E, S}, {A, E, T}, {A, M, S}, {A, M, T}, {A, S, T},
{E, M, S}, {E, M, T}, {E, S, T}, {M, S, T}
111
112
Chapter 6
(d) {A, E, M}
AEM, AME, EAM, EMA, MAE, MEA
{A, E, S}
AES, ASE, EAS, ESA, SAE, SEA
{A, E, T}
AET, ATE, EAT, ETA, TAE, TEA
{A, M, S}
AMS, ASM, MAS, MSA, SAM, SMA
{A, M, T}
AMT, ATM, MAT, MTA, TAM, TMA
{A, S, T}
AST, ATS, SAT, STA, TAS, TSA
{E, M, S}
EMS, ESM, MES, MSE, SEM, SME
{E, M, T}
EMT, ETM, MET, MTE, TEM, TME
{E, S, T}
EST, ETS, SET, STE, TES, TSE
{M, S, T}
MST, MTS, SMT, STM, TMS, TSM
(e) There are 6 permutations in each row, 10 rows, and a total of 6  10  60 permutations.
5. (a) 10: HHTTT, HTHTT, HTTHT, HTTTH, THHTT, THTHT, THTTH, TTHHT, TTHTH,
TTTHH
(b) 6: HTHTT, HTTHT, HTTTH, THTHT, THTTH, TTHTH
(c) 5! = 120
(d) There are 6 arrangements of 2 G’s and 3 B’s with no 2 consecutive G’s, as in part (b):
GBGBB, GBBGB, GBBBG, BGBGB, BGBBG, BBGBG. For each of these arrangements
there are 2! = 2 ways to put Ann and Betty in the two spots for G, and 3! = 6 ways to put
Cole, Don, and Ed in the 3 spots for B. Thus there are 6  2  6 = 72 ways.
(e) In parts (a) and (b) the 2 H’s are indistinguishable, as are the 3 T’s. In parts (c) and (d), Ann
and Betty are distinguishable, as are the three boys. There are 2! = 2 ways to arrange the girls
and 3! = 6 ways to arrange the boys. Thus the answer to (c) is 2  6 = 12 times the answer to
(a), and the answer to (d) is 12 times the answer to (b).
6. Answers will vary.
Exercise Set 6.1
1. (a) 6: (bus, ship), (bus, plane), (train, ship), (train, plane), (plane, ship), (plane, plane)
a
(b)
S
b
a
c
B
P
d
a
S
b
T
P
b
c
d
P
S
T
a
c
b
c
P
d
2. (a) 32
a
d
(b)
a
b
F
c
3. (a) 6  4 = 24
(b) 6  4  4  6 = 576
d
b
c
d
a
b
c
d
a
b
c
d
a
b
c
d
a
b
c
d
Combinatorics
4.
113
2  2  4 = 16
5. (a) 9  10  10 = 900
(b) 9  10  10  10 = 9000
(c) 9  10  10  10  10 = 90,000
(d) 9  10 n 1
6. (a) 9 107  90,000,000
(b) 4  57  312,500
(c) 90,000,000 – 312,500 = 89,687,500
7. 4  4  3 = 48
8. (a) 4  3  2  1 = 24
(b) 7  6  5  4  3  2  1 = 5040
(c) 1
(d) 1
9. (a) 3! = 6
(b) PQR, PRQ, QPR, QRP, RPQ, RQP
10. (a) 12
(b) WX, WY, WZ, XW, XY, XZ, YW, YX, YZ, ZW, ZX, ZY
11. (a) 2! = 2
(b) 3! = 6
(c) 4! = 24
12. (a) 5  4  3 = 60
(b) 6  5  4  3  2 = 720
(c) 8
(d) 9  8 = 72
(e) 7  6 = 42
(f) 8  7  6  5  4  3  2 = 40,320
1
2
13. 6  3 = 18
3
4
5
6
1
2
3
1
2
3
1
2
3
1
2
13
2
31
2
3
14. (a) 9  10  10 = 900
(b) 9  10  3 = 270
r!
k!
r!
r!
r
 = r!
15. (a) P(r, 1) =
(b)
= k(k – 1)
(c)
( r  1)!
( k  2)!
( r  ( r  1))! 1!
k!
k!
k!

(d)
= k!/2
(e)
= k(k – 1)(k – 2)
( k  ( k  2))! 2!
( k  3)!
k!
k!

(f)
= k!/6
( k  ( k  3))! 3!
16. P(26, 3)P(10, 3) = 26  25  24  10  9  8 = 11,232,000
17. (a) 10  10  10  10 = 10,000
(b) P(10,4) = 10  9  8  7 = 5040
18. (a) 5  5  5 = 125
(b) P(5,3) = 5  4  3 = 60
(c) The digit 9 will appear in the hundreds place in 4  3 = 12 of the 60 numbers in (b). Similarly,
each of the digits 3, 4, 5, 6, and 9 will appear 12 times in the hundreds place, 12 times in the
tens place, and 12 times in the ones place. Thus the sum of all 60 numbers will be
(3  4  5  6  9) 12  (100  10  1)  27 12 111  35,964 .
19. (a) Gluing the Q and the U together as QU, there are 5! = 120 arrangements.
(b) We can either glue RA together (120 ways), glue RE together (120 ways) or glue RU together
(120 ways). By the Addition Principle, there are 120 + 120 + 120 = 360 arrangements.
20. (a) We can glue BC together or CB together. Each of these can be done in 9! ways, so there are
2  9! 2  362,880 = 725,760 ways that they can sit together.
(b) There are a total of 10! ways to arrange the students, so there are 10! – 2  9! = 3,628,800 –
725,760 = 2,903,040 arrangements in which Bob and Calvin do not sit together.
114
Chapter 6
Exploratory Exercise Set 6.2
1. (a) (1) There are 24: ABC, ABD, ACB, ACD, ADB, ADC, BAC, BAD, BCA, BCD, BDA, BDC,
CAB, CAD, CBA, CBD, CDA, CDB, DAB, DAC, DBA, DBC, DCA, DCB.
P(4, 3) = 4  3  2 = 24.
4!
(2) There are 4: {A, B, C}, {A, B, D}, {A, C, D}, {B, C, D}. C(4, 3) =
= 4.
1! 3!
(3) {A, B, C}
{A, B, D}
{A, C, D}
{B, C, D}
ABC
ABD
ACD
BCD
ACB
ADB
ADC
BDC
BAC
BAD
CAD
CBD
BCA
BDA
CDA
CDB
CAB
DAB
DAC
DBC
CBA
DBA
DCA
DCB
There should be 6 permutations below each combination (Don’t forget the first one.)
(4) P(4, 3) = 6  C (4, 3) or 3  2  1  C (4, 3)
(b) (1) There are 12: AB, AC, AD, BA, BC, BD, CA, CB, CD, DA, DB, DC.
P(4, 2) = 4  3 = 12.
4!
(2) There are 6: {A, B}, {A, C}, {A, D}, {B, C}, {B, D}, {C, D}. C(4, 2) =
= 6.
2! 2!
(3) {A, B}
{A, C}
{A, D}
{B, C}
{B, D}
{C, D}
AB
AC
AD
BC
BD
CD
BA
CA
DA
CB
DB
DC
There should be 2 permutations below each combination.
(4) P(4, 2) = 2  C (4, 2) or 2 1  C (4, 2)
2. (a) Since the order in which the books are chosen is not important, we use combinations.
14! 14 13 12 11
C(14, 4) =

 7  13  11 = 1001.
4! 10!
4  3  2 1
(b) Since the books are displayed, the order is important; we are counting arrangements or
12!
permutations. P(12, 5) =
 12 11 10  9  8 = 95,040.
7!
(c) Since each office is distinct, the order is important. (For example, Ann as president and Bob
as VP is different from Bob as president and Ann as VP.) We use permutations.
P(100, 4) = 100  99  98  97 = 94,109,400.
(d) Since the order in which the committee members are chosen is not important, we use
15! 15 14 13 12
combinations. C(15, 4) =

 1365 .
4! 11!
4  3  2 1
3. (a) {A, B}, {A, C}, {A, D}, {A, E}, {A, F}, {B, C}, {B, D}, {B, E}, {B, F}, {C, D}, {C, E},
{C, F}, {D, E}, {D, F}, {E, F}
(b) {A, B, C, D}, {A, B, C, E}, {A, B, C, F}, {A, B, D, E}, {A, B, D, F}, {A, B, E, F},
{A, C, D, E}, {A, C, D, F}, {A, C, E, F}, {A, D, E, F}, {B, C, D, E}, {B, C, D, F},
{B, C, E, F}, {B, D, E, F}, {C, D, E, F}
(c) They should be equal.
(d) {A, B}, {C, D, E, F}
{A, C}, {B, D, E, F}
{A, D}, {B, C, E, F}
{A, E}, {B, C, D, F}
{A, F}, {B, C, D, E}
{B, C}, {A, D, E, F}
{B, D}, {A, C, E, F}
{B, E}, {A, C, D, F}
{B, F}, {A, C, D, E}
{C, D}, {A, B, E, F}
{C, E}, {A, B, D, F}
{C, F}, {A, B, D, E}
{D, E}, {A, B, C, F}
{D, F}, {A, B, C, E}
{E, F}, {A, B, C, D}
Combinatorics
115
4. (a) HHTTTT, HTHTTT, HTTHTT, HTTTHT, HTTTTH, THHTTT, THTHTT, THTTHT, THTTTH,
TTHHTT, TTHTHT, TTHTTH, TTTHHT, TTTHTH, TTTTHH
(b) {1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {2, 3}, {2, 4}, {2, 5}, {2, 6}, {3, 4}, {3, 5}, {3, 6},
{4, 5}, {4, 6}, {5, 6}
(c) HHTTTT, {1, 2}
HTHTTT, {1, 3}
HTTHTT, {1, 4}
HTTTHT, {1, 5}
HTTTTH, {1, 6}
THHTTT, {2, 3}
THTHTT, {2, 4}
THTTHT, {2, 5}
THTTTH, {2, 6}
TTHHTT, {3, 4}
TTHTHT, {3, 5}
TTHTTH, {3, 6}
TTTHHT, {4, 5}
TTTHTH, {4, 6}
TTTTHH, {5, 6}
(d) Since each sequence of 2 H’s and 4 T’s corresponds to a 2-element subset of {1, 2, 3, 4, 5, 6},
and vice versa, the number of such sequences is equal to the number of such subsets, which is
the combination number C(6, 2). Alternatively, to find a sequence of 6 H’s and T’s, with
exactly 2 positions held by H’s, perform the following two steps. First, choose exactly 2 of
the 6 positions to fill with H’s. Then fill all 4 of the remaining 4 positions with T’s. By the
Multiplication Principle this can be done in C (6, 2)  C (4, 4) = C(6, 2) ways, so there are C(6,
2) sequences of 2 H’s and 4 T’s.
(e) There is a one-to-one correspondence between the sequences of r H’s and n – r T’s and the relement subsets of {1, 2, 3, …, n}: the positions of the tosses that came up heads correspond
to the elements of the subset. Alternatively, to find a sequence of n H’s and T’s, with exactly
r positions held by H’s, perform the following two steps. First, choose exactly r of the n
positions to fill with H’s. Then fill all of the remaining n – r positions with T’s. By the
Multiplication Principle this can be done in C(n, r )  C( r, r )  C(n, r ) ways, so there are
C(n, r) sequences of r H’s and n – r T’s.
5. Answers will vary.
Exercise Set 6.2
8!
876
= 56

3! 5! 3  2  1
8!
8765
(c) P(8, 4) = 8  7  6  5 = 1680
(d) C(8, 4) =
= 70

4! 4! 4  3  2  1
7!
76
4!
43
2. (a) C(4, 2) =
=6
(b) C(7, 5) =
= 21


5! 2! 2  1
2! 2! 2  1
7!
5!
(c) C(5, 5) =
=1
(d) C(7, 0) =
=1
5! 0!
0! 7!
7!
7!
(e) C(7, 1) =
=7
(f) C(7, 7) =
=1
1! 6!
7! 0!
3. (a) There is only one subset of A with no elements: the empty set.
(b) There are n “singleton” sets, each containing a single element of A.
(c) The subsets of A with n – 1 elements are the subsets with exactly one element of A missing.
These are the complements of the singleton sets in (b), so there are exactly n of them.
(d) There is only one subset of A with n elements: the whole set A.
4. (a) C(n, 0) is the number of 0-element subsets of a set with n elements, so by 3(a), C(n, 0) = 1.
(b) C(n, 1) is the number of 1-element subsets of a set with n elements, so by 3(b), C(n, 1) = n.
(c) C(n, n – 1) is the number of (n – 1)-element subsets of a set with n elements, so by 3(c),
C(n, n – 1) = n.
(d) C(n, n) is the number of n-element subsets of a set with n elements, so by 3(d), C(n, n) = 1.
1. (a) P(8, 3) = 8  7  6 = 336
(b) C(8, 3) =
116
Chapter 6
5. (a) C(10, 4) =
10! 10  9  8  7
= 210

4! 6! 4  3  2  1
(c) C(10, 8) + C(10, 9) + C(10, 10) =
6.
7.
8.
9.
10.
11.
(b) C(10, 7) =
10! 10  9  8
= 120

7! 3! 3  2  1
10! 10!
10!
= 45 + 10 + 1 = 56


8! 2! 9! 1! 10! 0!
(d) C(10, 2) + C(10, 3)    C(10, 10) = 210 – (C(10, 0) + C(10, 1)) = 1024 – 11 = 1013
4! 48!
(a) C(4, 4)  C (48, 1) =

 1 48 = 48
4! 0! 1! 47!
13! 13  12  11  10  9
(b) C(13, 5) =
= 1287

5! 8!
5  4  3  2 1
4! 4!
(c) C(4, 3)  C (4, 2) =

 4  6 = 24
3! 1! 2! 2!
39!
39  38  37  36  35
(d) C(39, 5) =
= 575,757

5! 34!
5  4  3  2 1
48!
48  47  46  45  44  43
C(48, 6) =
= 12, 271, 512

6! 42!
6  5  4  3  2 1
50!
50  49  48  47  46
C(50, 5)  C (36, 1) =
 36 
 36  2,118,760  36 = 76,275,360
5! 45!
5  4  3  2 1
20! 20  19  18 17 16  15  14  13
(a) C(20, 8) =
= 125,970

8! 12!
8  7  6  5  4  3  2 1
(b) C(9, 3)  C (11, 5) + C(9, 4)  C (11, 4) =
9! 11!
9! 11!



 84  462  126  330 = 80,388
3! 6! 5! 6! 4! 5! 4! 7!
15! 11! 15  14 11 10  9
(a) C(15, 2)  C (11, 3) =



 105  165 = 17,325
2! 13! 3! 8!
2 1
3  2 1
(b) C(15, 2)  C (11, 3) + C(15, 1)  C (11, 4) + C(15, 0)  C (11, 5)
= 105  165  15  330  1  462 = 17,325 + 4950 + 462 = 22,737
(a) C(2, 0)  C (18, 4) = 3060
(b) C(2, 1)  C (18, 3) = 2  816 = 1632
(c) C(2, 2)  C (18, 2) = 153
(d) 153 + 1632 + 3060 = 4845
Teaching Notes 6.3: We have found Exploratory Exercise 1 to be a good introduction to the main
theme of this section: the interconnections among Pascal’s Triangle, combination numbers, and
binomial coefficients.
Related Website: For an elegant discussion of combinatorial proofs check out
http://faculty.oxy.edu/jquinn/boulder/Default.htm and the related text, Proofs that Really Count, by
Art Benjamin and Jennifer Quinn.
Exploratory Exercise Set 6.3
1. (a)
Type of Subset
0-combinations
1-combinations
2-combinations
3-combinations
List of Subsets
{}
{a}, {b}, {c}
{a, b}, {a, c}, {b, c}
{a, b, c}
Number of r-combinations
C(3, r)
1
3
3
1
Combinatorics
117
(b) ( x  y ) 3  ( x  y )( x  y )( x  y )  x 3  3x 2 y  3xy 2  y 3 ; the coefficients are 1, 3, 3, and 1
(c) The number of combinations in (a) are the same as the coefficients in (b).
(d) Row 3: 1 3 3 1
14! 14 13 12 11
2. (a) C(14, 4) =

 1001
4! 10!
4  3  2 1
(b)
1 10 45 120 210 252 210 120 45 10 1
1 11 55 165 330 462 462 330 165 55 11 1
1 12 66 220 495 792 924 792 495 220 66 12 1
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1
1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1
so C(14, 4) = 1001.
(c) The answers should be the same.
3. (b) C(11, 4) = 330 = 2  165 = 2  C (11, 3)
(c) In Rows 2, 5, and 8
(d) Row 14
(e) Yes; C(14, 5) = 2002 = 2  1001 = 2  C (14, 4)
4. (a) The coefficients of xy 9 , x 3 y 7 , and x 4 y 6 are C(10, 1) = 10, C(10, 3) = 120, and
C(10, 4) = 210, respectively.
(b) The coefficients of x 6 y 4 , x 7 y 3 , and x 9 y are C(10, 6) = 210, C(10, 7) = 120, and
C(10, 9) = 10, respectively.
(c) In the expansion of ( x  y ) n , the coefficient of x r y n r is equal to the coefficient of x n r y r .
(d) By Theorem 3, the coefficient of x r y n r in the expansion of ( x  y ) n is C(n, r). By the
Leftover Principle, C(n, r) = C(n, n – r). Using Theorem 3 again, C(n, n – r) is the coefficient
of x n r y r . Therefore the coefficient of x r y n r is equal to the coefficient of x n r y r .
5. (a) xxxx, xxxy, xxyx, xxyy, xyxx, xyxy, xyyx, xyyy, yxxx, yxxy, yxyx, yxyy, yyxx, yyxy, yyyx, yyyy.
(b) Four sequences have exactly three x’s: xxxy, xxyx, xyxx, and yxxx.
(c) The coefficient of x 3 y in the expansion of ( x  y ) 4 is C(4, 3) = 4.
(d) The expansion of ( x  y ) 4 is the sum of the 16 products of x’s and y’s given in part (a). Of
these 16 products, the four in part (b) are equal to x 3 y , so the coefficient in (c) is equal to the
number of sequences in (b).
6. (a) Row 1: 1 – 1 = 0
(b) Row 2: 1 – 2 + 1 = 0
Row 3: 1 – 3 + 3 – 1 = 0
Row 4: 1 – 4 + 6 – 4 + 1 = 0
Row 5: 1 – 5 + 10 – 10 + 5 – 1 = 0
Row 6: 1 – 6 + 15 – 20 + 15 – 6 + 1 = 0
etc.
(c) The alternating sum of the numbers in Row n of Pascal’s Triangle is 0 .
(d) It does not hold for Row 0, the top row. The alternating sum of the numbers in Row n of
Pascal’s Triangle, where n > 0, is 0 .
(e) If n is odd, then in the alternating sum C(n, r) and C(n, n – r) will have opposite signs and will
cancel, giving an alternating sum of 0.
(f) Using Theorem 3, substituting –1 for x and 1 for y, then ( 1  1) n 
C (n, n)  ( 1) n  C (n, n  1)  ( 1) n1 11  C (n, n  2)  ( 1) n2 12    C (n, 1)  ( 1) 1n 1  C (n,0) 1n =
C(n, 0) – C(n, 1) + C(n, 2)    ( 1) n 1 C(n, n – 1) + (1) n C(n, n),
which is the alternating sum. If n > 0, then ( 1  1) n  0 n  0 , so the alternating sum is equal
to 0 as well.
7. Answers will vary.
118
Chapter 6
Exercise Set 6.3
1. (a) C(8, 2) = 28
(b) C(8, 3) = 56
(c) C(8, 4) = 70
8
(d) C(8, 5) = 56
(e) 2  256
2. (a) C(6, 2) = 15
(b) C(6, 3) = 20
(c) C(6, 5) + C(6, 6) = 6 + 1 = 7
(d) C(6, 4) + C(6, 5) + C(6, 6) = 15 + 6 + 1 = 22
(e) 2 6  64
3. There are C(10, 4) = 210 ways.
4. You could have 0, 1, 2, 3, or 4 toppings, so the number of possible pizzas is
C(10, 0) + C(10, 1) + C(10, 2) + C(10, 3) + C(10, 4) = 1 + 10 + 45 + 120 + 210 = 386.
5. (a) C(14, 4) = 1001
(b) C(14, 5) = 2002
(c) C(14, 6) = 3003
6. (a) C(14, 4) = 1001
(b) C(14, 5) = 2002
(c) C(14, 6) = 3003
7. (a) C(14, 4) = 1001
(b) C(14, 5) = 2002
(c) C(14, 6) = 3003
8. (a) C(9, 2) = 36
(b) C(9, 4) = 126
(c) C(9, 7) = 36
(d) C(9, 8) = 9
(e) C(9, 9) = 1
9. (a) ( x  y ) 6  x 6  6 x 5 y  15 x 4 y 2  20 x 3 y 3  15 x 2 y 4  6 xy 5  y 6
(b) ( x  y ) 6  x 6  6 x 5 y  15 x 4 y 2  20 x 3 y 3  15 x 2 y 4  6 xy 5  y 6
(c) ( x  y ) 7  x 7  7 x 6 y  21x 5 y 2  35 x 4 y 3  35 x 3 y 4  21x 2 y 5  7 xy 6  y 7
(d) ( x  y ) 7  x 7  7 x 6 y  21x 5 y 2  35 x 4 y 3  35 x 3 y 4  21x 2 y 5  7 xy 6  y 7
10. (a) C(9, 4) = 126
(b) Substituting 2s for x and –t for y into Theorem 3 gives the term
C(9, 4) (2s) 4 ( t ) 5 = 126  2 4  s 4  ( 1) 5  t 5  126 16  ( 1)  s 4 t 5  2016 s 4 t 5 ,
so the coefficient of s 4 t 5 is –2016.
(c) Substituting s for x and 3t for y into Theorem 3 gives the term
C(9, 4) s 4 (3t ) 5 = 126  35  s 4t 5  30,618s 4t 5 , so the coefficient is 30,618.
(d) Substituting –s for x and 2t for y into Theorem 3 gives the term
C(9, 4) (  s) 4 (2t ) 5 = 126  s 4  2 5  t 5  4032 s 4 t 5 ,
so the coefficient of s 4 t 5 is 4032.
11. (a) (2a  b) 5  1  (2a ) 5  5  (2a ) 4 b  10  (2a ) 3 b 2  10  (2a ) 2 b 3  5  (2a )  b 4  1  b5
 32 a 5  80a 4 b  80a 3b 2  40a 2 b 3  10ab 4  b 5
(b) (a  3b)5  1  a 5  5a 4 ( 3b)  10a 3 ( 3b) 2  10a 2 ( 3b) 3  5a( 3b) 4  1  ( 3b)5
 a 5  15a 4 b  90a 3b 2  270 a 2 b 3  405ab 4  243b 5
(c) ( x  2) 5  x 5  5x 4 ( 2)  10 x 3 ( 2) 2  10 x 2 ( 2) 3  5x( 2) 4  ( 2) 5
 x 5  10 x 4  40 x 3  80 x 2  80 x  32
(d) (3s  2t ) 5  (3s) 5  5(3s) 4 ( 2t )  10(3s) 3 ( 2t ) 2  10(3s) 2 ( 2t ) 3  5(3s)(2t ) 4  ( 2t ) 5
 243 s 5  5  81  2 s 4 t  10  27  4 s 3t 2  10  9  8s 2 t 3  5  3  16 st 4  32t 5
 243 s 5  810 s 4 t  1080 s 3t 2  720 s 2 t 3  240 st 4  32t 5
12. (a) C(9, 0) + C(9, 1) + C(9, 2) + C(9, 3) = 1 + 9 + 36 + 84 = 130
(b) First choose the flavor; then the toppings.
39  (C(10, 0) + C(10, 1) + C(10, 2) + C(10, 3)) = 39  (1 + 10 + 45 + 120)
= 39  176 = 6864
13. There are C(4, 1) = 4 ways to choose the lettuce; C(11, 0) + C(11, 1) + C(11, 2) + C(11, 3) +
C(11, 4) = 1 + 11 + 55 + 165 + 330 = 562 ways to choose the toppings; and C(6, 0) + C(6, 1)
= 1 + 6 = 7 ways to choose the dressing, for a total of 4  562  7 = 15,736 different salad
plates.
Combinatorics
119
14. (a) 2 9  512
(b) C(9, 6) = 84
(c) C(9, 7) + C(9, 8) + C(9, 9) = 36 + 9 + 1 = 46
(d) 2 9  (C(9, 0) + C(9, 1)) = 512 – 10 = 502
n!
n!
15. C ( n, r )  C ( n, r  1) 

( n  r )! r! ( n  r  1)! ( r  1)!
n!
r 1
n!
nr



( n  r )! r! r  1 ( n  r  1)! ( r  1)! n  r
n! ( r  1)
n! ( n  r )


( n  r )! ( r  1)! ( n  r )! ( r  1)!
n! ( r  1  n  r )

( n  r )! ( r  1)!


n! ( n  1)
( n  r )! ( r  1)!
( n  1)!
( n  1  ( r  1))! ( r  1)!
 C ( n  1, r  1)
16. (a) 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28; This is the seventh triangular number, t 7 .
(b) 12 + 11 + 10 + … + 3 + 2 + 1 = 78; this is the twelfth triangular number, t12 .
n(n  1)
(c) The nth triangular number: t n 
2
(d) t1  t 2  t3    t12  1  3  6    78 = 364
(e) 364 = C(14, 3)

Teaching Notes 6.4: This section on combinations and permutations with repetition will be new to
most students, but can be developed naturally from the material in the preceding sections. There are a
number of elegant “tricks” that turn problems that look quite hard at first glance into more familiar
problems. Exploratory Exercises 3 and 4 introduce such tricks for problems with nonadjacency
conditions and for combinations with repetitions, respectively.
Related Website: http://www.ping.be/~ping1339/tel.htm has a nice overview of combinatorial
problems with and without repetition.
Exploratory Exercise Set 6.4
1. (a) EEEPT, EEETP, EEPET, EEPTE, EETEP, EETPE, EPEET, EPETE, EPTEE, ETEEP,
ETEPE, ETPEE, PEEET, PEETE, PETEE, PTEEE, TEEEP, TEEPE, TEPEE, TPEEE
There are 20.
(b) E E E__ __, E E __ E __, E E __ __ E, E __ E E __, E __ E __ E,
E __ __ E E, __ E E E __, __ E E __ E, __ E __ E E, __ __ E E E
There are C(5, 3) = 10 ways to fill 3 of 5 blanks with E’s.
(c) There are 2! = 2 ways to fill the 2 blanks with 2 different letters.
(d) Arrange the letters in 2 stages. First position the E’s (C(5, 3) = 10 ways), then the remaining
two letters (2! = 2 ways). The Multiplication Principle says the number of arrangements
should be the product: 10  2  20 .
120
Chapter 6
2. (a) DDDEEE, DDEDEE, DDEEDE, DDEEED, DEDDEE, DEDEDE, DEDEED, DEEDDE,
DEEDED, DEEEDD, EDDDEE, EDDEDE, EDDEED, EDEDDE, EDEDED, EDEEDD,
EEDDDE, EEDDED, EEDEDD, and EEEDDD. There are 20.
(b) DEDEDE, DEDEED, DEEDED, EDEDED
(c) DEDEDE: 1st, 2nd, and 3rd blanks
DEDEED: 1st, 2nd, and 4th blanks
DEEDED: 1st, 3rd, and 4th blanks
EDEDED: 2nd, 3rd, and 4th blanks
(d) DEDEDE, EDDEDE, EDEDDE, EDEDED
(e) DEDEDE: 2nd, 3rd, and 4th blanks
EDDEDE: 1st, 3rd, and 4th blanks
EDEDDE: 1st, 2nd, and 4th blanks
EDEDED: 1st, 2nd, and 3rd blanks
3. (a) AAAAA, AAAAB, AAAAE, AAABB, AAABE, AAAEE, AABBB, AABBE, AABEE,
AAEEE, ABBBB, ABBBE, ABBEE, ABEEE, AEEEE, BBBBB, BBBBE, BBBEE, BBEEE,
BEEEE, and EEEEE. There are 21.
(b) / / / / / | |, / / / / | / |, / / / / | | /, / / / | / / |, / / / | / | /, / / / | | / /, / / | / / / |, / / | / / | /, / / | / | / /,
/ / | | / / /, / | / / / / |, / | / / / | /, / | / / | / /, / | / | / / /, / | | / / / /, | / / / / / |, | / / / / | /, | / / / | / /,
| / / | / / /, | / | / / / /, and | | / / / / /. There are C(7, 2) = 21.
(c) Yes.
(d) AAAAA, / / / / / |
AAAAB, / / / / | / |
AAAAE, / / / / | | /
AAABB, / / / | / / |
AAABE, / / / | / | /
AAAEE, / / / | | / /
AABBB, / / | / / / |
AABBE, / / | / / | /
AABEE, / / | / | / /
AAEEE, / / | | / / /
ABBBB, / | / / / / |
ABBBE, / | / / / | /
ABBEE, / | / / | / /
ABEEE, / | / | / / /
AEEEE, / | | / / / /
BBBBB, | / / / / / |
BBBBE, | / / / / | /
BBBEE, | / / / | / /
BBEEE, | / / | / / /
BEEEE, | / | / / / /
EEEEE, | | / / / / /
4. Answers will vary. Look for a reformulation of the problems in terms of slashes for tally marks
and vertical lines for column dividers.
Exercise Set 6.4
1. (a) Inventory: 2 A’s, 2 R’s, 2 N’s, 1 G, 2 E’s, 1 M, and 1 T (11 letters)
11!
39,916,800
The number of arrangements is
= 2,494,800.

2! 2! 2! 1! 2! 1! 1!
24
(b) Inventory: 3 R’s, 2 E’s, 1 O, 1 D, 1 I, 1 N, and 1 G (10 letters)
10!
3,628,800
The number of arrangements is
= 302,400.

3! 2! 1! 1! 1! 1! 1!
62
(c) Inventory: 3 T’s, 3 I’s, 4 N’s, 2 A’s, 1 B, 1 U, 1 L, 1 O (16 letters)
16!
20,922,789,888,000
The number of arrangements is
= 12,108,096,000.

3! 3! 4! 2! 1! 1! 1! 1!
62  24  2
(d) Inventory: 1 D, 3 E’s, 1 F, 2 N’s, 1 S, 2 T’s, 1 R, 1 A, 1 I, 1 O (14 letters)
14!
87,178,291,200
The number of arrangements is
= 3,632,428,800.

1! 3! 1! 2! 1! 2! 1! 1! 1! 1!
6  22
13!
6,227,020,800
2. There are
= 90,090 ways to assign the students to the rooms.

4! 4! 5!
24 2  120
9!
362,880
3. There are
= 1260 ways to donate the paintings.

3! 4! 2! 6  24  2
Combinatorics
121
23!
= 412,275,623,800 ways to divide his class.
6! 4! 7! 6!
6!
6!
5. (a)
= 60
(b)
= 60
2! 3! 1!
2! 3! 1!
4. There are
(c) The coefficient of x 2 y 3 z in the expansion of ( x  y  z ) 6 is equal to the number of
arrangements of xxyyyz, and hence equal to the number of arrangements of SEEDED. Notice
that there are two D’s, three E’s, and one S in SEEDED; this corresponds to the number of
factors of x, y, and z, respectively, in x 2 y 3 z .
14!
14!
6. (a)
= 2,522,520
(b)
= 2,522,520
2! 5! 3! 4!
2! 5! 3! 4!
(c) The coefficient of w 2 x 5 y 3 z 4 in the expansion of ( w  x  y  z )14 is equal to the number of
arrangements of wwxxxxxyyyzzzz.
7. (a) 2 8  256
(b) C(8, 5) = 56
(c) In the arrangement __ 0 __ 0 __ 0 __ 0 __ 0 __, fill 3 of the 6 blanks with one 1 each. There
are C(6, 3) = 20 ways to do this.
8. (a) C(10, 6) = 210
(b) In the arrangement __ H __ H __ H __ H __ H __ H __, fill 4 of the 7 blanks with one T each.
There are C(7, 4) = 35 ways to do this.
(c) In the arrangement __ T __ T __ T __ T __, it is not possible to fill the 5 blanks with 6 H’s if
no more than one H can go in each blank. There are 0 ways.
9. If we label the disruptive students as D and the remaining students as G, then there are C(10, 5) =
252 ways to fill 5 of the 10 blanks in the arrangement
__ G __ G __ G __ G __ G __ G __ G __ G __ G __
with one D each. For each such arrangement, there are 5! = 120 ways to place the 5
disruptive students in the D positions, and 9! ways to place the other 9 students in the G
positions. By the Multiplication Principle, the total number of ways to line up the students is
C (10,5) 5! 9! = 252 120  362,880 = 10,973,491,200.
10. (a) 54  625
(b) 10 4 = 10,000
(c) 26 4  456,976
11. (a) There are 26 + 10 = 36 choices for each character: 36 8 = 2,821,109,907,456
(b) First choose an arrangement of 3 L’s and 5 D’s (for letters and digits); then choose a letter for
each L position; finally choose a digit for each D position. There are a total of C(8, 3)
 263 105  56 17,576 100,000 = 98,425,600,000
(c) This is similar to part (b), but we must choose an arrangement of L’s and D’s with no two
consecutive L’s. In the arrangement __ D __ D __ D __ D __ D __, fill 3 of the 6 blanks with
one L each. There are C(6, 3) = 20 ways to do this, so the total number of passwords is C(6,
3)  263 105  20 17,576 100,000 = 35,152,000,000
12. With a column for each of the 6 flavors, arrange 9 tally marks ( / ) and 5 vertical column dividers
( | ). There are C(9 + 5, 9) = C(14, 9) = 2002 ways to do this, so there are 2002 possible
combinations of flavors.
13. (a) / / | | / / / | / | / / | /
(b) / | / / | | | / / / / / / |
(c) | | / / | / / / / | | / / /
(d) / | / / | / / / | / / | / |
14. (a) 2 apple, 1 banana, 2 grape, 1 lemon, 3 orange
(b) 5 cherry, 2 lemon, 2 orange
(c) 5 apple, 4 cherry
(d) 2 apple, 1 banana, 2 cherry, 1 grape, 2 lemon, 1 orange
122
Chapter 6
15. With a column for each of the 4 types of gifts, arrange 10 tally marks and 3 vertical column
dividers. There are C(13, 10) = 286 ways to do this.
Teaching Notes 6.5: This is the most proof-intensive section in this chapter. You may want to
spend time in class with Exploratory Exercise 2 or 3. The method of proof developed in these
exercises will be helpful in many of the later exercise. Exercises 11 and 12 give examples of
problems combining geometry with the Pigeonhole Principle.
Related Website: Check out http://www.cut-the-knot.org/do_you_know/pigeon.shtml for an
intriguing array of proofs using the Pigeonhole Principle, including a proof that there are at least two
people in New York City with exactly the same number of hairs on their bodies.
Exploratory Exercise Set 6.5
1. (a) (Answers may vary.) If m coins are put into n drawers, where m > n, then there must be at
least one drawer with more than one coin in it. The drawers are analogous to the
pigeonholes.
(b) (Answers may vary.) If m shoes are put into n shoeboxes, where m > n, then there must be at
least one shoebox with more than one shoe in it. The shoeboxes are analogous to the
pigeonholes.
2. (a) Let g i be the position in line of Girl i for i = 1, 2, 3, …, 10. Since there are 19 children in the
class, each number g i must be an integer between 1 and 19, inclusive. Consider the
lists g1 , g 2 , , g10 and g1  1, g 2  1, , g10  1 . There are a total of 20 numbers in the two
lists, and each number must be between 1 and 19 + 1 = 20, inclusive. With 20 numbers
between 1 and 20, inclusive, the Pigeonhole Principle does not guarantee us that two of the
numbers must be equal. For example, we could have g1  1, g2  3, g3  5,, g10  19 , so
that the numbers in the two lists would be
1, 3, 5, 7, 9, 11, 13, 15, 17, 19 and 2, 4, 6, 8, 10, 12, 14, 16, 18, 20. Thus we cannot prove
that two girls must be next to each other in line.
(b) GBGBGBGBGBGBGBGBGBG is the only arrangement that will work.
3. (a) Let bi be the position of the ith black card for i = 1, 2, 3, …, 16. Since there are a total of 26
cards, each number bi is an integer between 1 and 26, inclusive. Consider the lists
b1 , b2 , , b16 and b1  5, b2  5, , b16  5 . There are a total of 32 numbers in the two lists;
each one is an integer between 1 and 26 + 5 = 31, inclusive. With the numbers in the lists as
the pigeons and the integers between 1 and 31, inclusive, as the pigeonholes, the Pigeonhole
Principle tells us that some two numbers in the two lists must be equal. Since two black cards
cannot occupy the same position, bi  b j for i  j , so it must be true that bi  b j  5 for
some i  j . Therefore, the ith and jth black cards are exactly 5 cards apart. The proof
works.
(b) Part (a) proves that it is impossible to find such a sequence of B’s and R’s.
(d) (a) Let bi be the position of the ith black card for i = 1, 2, 3, …, 15. Since there are a total of
25 cards, each number bi is an integer between 1 and 25, inclusive. Consider the lists
b1 , b2 , , b15 and b1  5, b2  5, , b15  5 . There are a total of 30 numbers in the two
lists; each one is an integer between 1 and 25 + 5 = 30, inclusive. The Pigeonhole
Principle does not guarantee us that two of the numbers must be equal. The proof does
not work.
(b) BBBBBRRRRRBBBBBRRRRRBBBBB is the only arrangement that will work.
4. (a) Answers will vary.
Combinatorics
123
(b), (c) It should always be the case that there are two students who know the same number of
people.
(d) Each person can know k other people in the room, where k is an integer between 0 and
n – 1, inclusive. We consider two cases.
Case 1: If each person knows at least one other person in the room, then each person knows
between 1 and n – 1 people, inclusive. With the n people as the pigeons, and the n – 1
integers between 1 and n – 1, inclusive, as the pigeonholes, the Pigeonhole Principle tells
us that there must be at least two people who know the same number of people, since n >
n – 1.
Case 2: If there is somebody who knows nobody in the room, then nobody else in the room
knows that person. Thus one person knows 0 people, and the remaining n – 1 people
know between 0 and n – 2 people, inclusive. If one of these n – 1 people also knows 0
people, then we are done, since there would be two people who both know 0 people. If
not, then one person knows 0 people, and each of the n – 1 other people knows between 1
and n – 2 people, inclusive. With the n – 1 people as the pigeons, and the n – 2 integers
between 1 and n – 2, inclusive, as the pigeonholes, the Pigeonhole Principle tells us that
there must be at least two people who know the same number of people, since n – 1 > n –
2.
5. (a) (Answers will vary.) One possibility: 3, 6, 1, 2, 30, 21, 5
(b) (Answers will vary.) One possibility: 21 – 1 = 20
(c) Other possibilities:
3, 8, 25, 12, 9, 4, 0:
8 + 12 = 20
5, 2, 9, 3, 4, 7, 1:
9 + 1 = 10
23, –42, 6, 220, 8, 11, 25:
–42 – 8 = –50
(d) Conjecture: If S is any set of 7 integers, then there are two integers in S whose sum or
difference is a multiple of 10.
(e) If n is an integer, define f(n) to be the distance between n and the nearest multiple of 10.
Since every integer is at most 5 units away from a multiple of 10, the range of this function f
is {0, 1, 2, 3, 4, 5}. With the 7 elements of the set S as the pigeons and the 6 elements of the
range of f as the pigeonholes, the Pigeonhole Principle tells us that there must be two
elements of S, say x and y, such that f(x) = f(y) = k, where k is 0, 1, 2, 3, 4, or 5. That means
that x and y are both k units away from a multiple of 10. If x and y are both k units more than
a multiple of 10, then x = 10m + k, and y = 10n + k, where m and n are integers, so x – y =
(10m + k) – (10n + k) = 10(m – n), which is a multiple of 10. Similarly, if x and y are both k
units less than a multiple of 10, then x = 10m – k, and y = 10n – k, where m and n are integers,
so x – y = (10m – k) – (10n – k) = 10(m – n), which is a multiple of 10. If one of x and y, say
x, is k units more than a multiple of 10, and the other, say y, is k units less than a multiple of
10, then x + y = (10m + k) + (10n – k) = 10(m + n), which is a multiple of 10. Thus in each
case, either the sum or the difference of x and y is a multiple of 10.
6. Answers will vary.
Exercise Set 6.5
1. (a) the suits
(b) the cards
(c) 5, since 5 > 4
2. (a) the colors
(b) the socks
(c) 7, since 7 > 6
3. (a) colors
(b) the pencils
(c) 81, since 81 > 4  20
4. From the Multiplication Principle there are 4  3  3 = 36 possible full names. With the 36 names
as the pigeonholes and the 40 people as the pigeons, the Pigeonhole Principle says there must be
at least two people with the same full name, since 40 > 36.
5. (a) Since there are 8 odd numbers in {1, 2, 3, …, 15}, you must choose at least 9 to be sure to
have at least one even number.
(b) Since there are 7 even numbers in {1, 2, 3, …, 15}, you must choose at least 8 to be sure to
have at least one odd number.
124
Chapter 6
(c) If n is an odd positive integer, then there are
you must choose at least
n 1
2
1 
n 3
2
you must choose at least
1 
n 1
2
odd numbers in the set {1, 2, 3, …, n}, so
numbers to be sure to have at least one even number.
(d) If n is an odd positive integer, then there are
n 1
2
n 1
2
n 1
2
even numbers in the set {1, 2, 3, …, n}, so
numbers to be sure to have at least one odd number.
6. In the shuffled deck, let ri be the position of the ith red card. Then 1  r1  r2    r25  44 ,
since there are a total of 25 + 19 = 44 cards in the deck. Consider the two lists r1 , r2 , , r25 and
r1  4, r2  4, , g25  4 . The numbers in each list must be positive integers no greater than 44 +
4 = 48. With the numbers 1, 2, 3, …, 48 as the pigeonholes and the 50 numbers in the two lists as
the pigeons, at least two of the numbers in the two lists must be equal. Since no two of the
numbers ri are equal to each other, it must be the case that ri  rj  4 for two distinct integers i
and j. Thus the ith and jth red cards are exactly 4 cards apart.
7. Number the 44 seats going across each row from left to right so that the front row has seats 1, 2,
3, and 4, the second row has seats 5, 6, 7, and 8, and the number of each seat (except those in the
first row) is 4 more than the number of the seat directly in front of it. Let g i represent the seat
number of the ith girl, where 1  g1  g 2    g 25  44 and consider the two lists:
g1 , g 2 , , g 25 and g1  4, g 2  4, , g 25  4 . The numbers in each list must be positive integers
no greater than 44 + 4 = 48. With the numbers 1, 2, 3, …, 48 as the pigeonholes and the 50
numbers in the two lists as the pigeons, at least two of the numbers in the two lists must be equal.
Since no two of the numbers g i are equal to each other, it must be the case that gi  g j  4 for
two distinct integers i and j. Thus the ith girl sits directly behind the jth girl.
8. Let wi be the position in line of the ith woman, where 1  w1  w2    w65  120 . In the lists
w1 , w2 , , w65 and w1  9, w2  8, , w65  9 , there are 130 numbers, each of which must be a
positive integer no greater than 120 + 9 = 129. With the numbers 1, 2, 3, …, 129 as the
pigeonholes and the 130 numbers in the two lists as the pigeons, at least two of the numbers in the
two lists must be equal. Since no two of the numbers wi are equal to each other, it must be the
case that wi  w j  9 for two distinct integers i and j. Thus the ith and jth women are exactly 9
people apart in the line.
9. Let hi represent the total number of hits in the first i games of the streak. Since he had at least
one hit in each of the 34 games, 1  h1  h2    h34  52 . Consider the two lists h1 , h2 ,  , h34
and h1  15, h2  15, , h34  15 . The numbers in each list must be positive integers no greater
than 52 + 15 = 67. With the numbers 1, 2, 3, …, 67 as the pigeonholes and the 68 numbers in the
two lists as the pigeons, at least two of the numbers in the two lists must be equal. Since no two
of the numbers hi are equal to each other, it must be the case that hi  h j  15 for two distinct
integers i and j. Thus the number of hits in the first i games is 15 more than the number of hits in
the first j games, so the number of hits in games
j + 1, j + 2, …, i must be 15.
10. Since we want more than 3 batteries of the same type, we use the Extended Pigeonhole Principle
with k = 3. Let the 2 types of batteries represent the n = 2 pigeonholes. The smallest integer
larger than k  n  3 2 = 6 is m = 7. You must pull out at least 7 batteries.
11. Since we want more than 10 screws of the same type, we use the Extended Pigeonhole Principle
with k = 10. Let the 4 types of screws represent the n = 4 pigeonholes. The smallest integer
greater than k  n  10  4 = 40 is m = 41. You must remove at least 41 screws.
Combinatorics
125
12. Divide the plot of land into twelve 2 miles  2 miles squares, as
shown at right. With the 13 farmers as the pigeons and the 12
squares as the pigeonholes, there must be some square that
contains at least 2 farmers. The farthest apart those two farmers
can be is the length of the diagonal of the square, which is
13.
14.
15.
16.
22  22  8  2 2 miles.
Divide the hexagon into 6 equilateral triangles, as shown at right,
each with edges of length 5 cm. With the 7 points as pigeons and
the 6 triangles as pigeonholes, at least two points must lie on or
inside the same triangle. The maximum distance between these
two points is 5 cm.
Each member has played between 0 and 31 other members, inclusive. We consider two cases.
Case 1: If every member has played at least one other member, then each of the 32 members has
played between 1 and 31 other members, inclusive. With the 32 members as the pigeons and
the integers between 1 and 31, inclusive, as the pigeonholes, some two members must have
played the same number of members.
Case 2: If there is a member who has played no other members, then nobody has played that
member, so each of the other 31 members has played at most 30 other members. If a second
member has played no other members, then there are two members who have played 0 other
members. Otherwise, there is one member who has played 0 other members and the
remaining 31 members have played between 1 and 30 members, inclusive. With the 31 other
members as the pigeons and the integers between 1 and 30, inclusive, as the pigeonholes,
some two members must have played the same number of members.
(a) Answers will vary. One possibility is S = {1, 3, 5, 7, 8}
(b) Answers will vary. With S = {1, 3, 5, 7, 8}, one possibility is {5, 7} and {1, 3, 8}.
(c) Answers will vary. Other possibilities are S = {1, 3, 4, 6, 7}, {1, 3} and {4};
S = {1, 2, 4, 6, 7}, {1, 2, 4} and {7}; and S = {4, 5, 6, 7, 8}, {4, 7} and {5, 6}.
(d) There are at least two disjoint subsets of S for which the sums of their elements are equal.
(e) Recall that S has a total of 2 5  32 subsets, so it has 31 non-empty subsets. The sum of the
elements of any non-empty subset of S must be a positive integer no greater than 4 + 5 + 6 +
7 + 8 = 30. With the 31 non-empty subsets of S as the pigeons and the numbers 1, 2, 3, …,
30 as the pigeonholes, there must be at least two distinct subsets whose elements add up to
the same number. If these two subsets are disjoint, then we are done. If not, remove any
common elements from both subsets. Since the original subsets were distinct and one was
not a subset of the other, the new subsets will be non-empty, they will be distinct, and the
sums of their elements will still be equal.
(a) There are 26 3 = 17,576 possible monograms, so there would have to be at least 17,577
students to be sure that at least two students have the same monogram.
(b) Since we want more than 4 students with the same monogram, we use the Extended
Pigeonhole Principle with k = 4. With the n = 17,576 monograms as the pigeonholes, the
smallest integer greater than k  n  417,576 = 70,304 is m = 70,305. There must be at least
70,305 students to be sure that at least five students have the same monogram.
Exploratory Exercise Set 6.6
1. (a) (Answers will vary.) One possibility: X 1 = {1, 3, 4, 6, 7, 8, 10}; n ( X 1 )  7
(b) One possibility: X 2 = {2, 3, 4, 6, 7, 9}; n( X 2 )  6
(c) X 1  X 2 = {1, 2, 3, 4, 6, 7, 8, 9, 10}; n( X 1  X 2 ) = 9
(d) No; n( X 1 )  n( X 2 ) = 7 + 6 = 13, while n( X 1  X 2 ) = 9. n( X 1 )  n( X 2 ) is greater by 4.
126
Chapter 6
(e) X 1  X 2 = {3, 4, 6, 7}; n( X 1  X 2 ) = 4.
(f) n( X 1 )  n( X 2 ) is greater than n( X 1  X 2 ) by n( X 1  X 2 ) . Indeed, Theorem 11 says that
n( X 1  X 2 ) = n( X 1 )  n( X 2 ) – n( X 1  X 2 ) .
2. (a) This is equivalent to saying that d is a divisor (or factor) of m.
(b) There is an integer k such that m = dk .
(c) Since m and d are positive, k must be a positive integer as well.
n
(d) That means that d  k  n and k  .
d
n
(e) Therefore k must be a positive integer  .
d
n
(f)
k
n
(g)
k
3. (a) 12, 24, 36, 48, 60, 72, 84, 96 (Notice these are multiples of 12.)
(b) 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96 (These are multiples of 6.)
(c) 12, 24, 36, 48, 60, 72, 84, 96 (These are multiples of 12.)
(d) The integers that are multiples of both a and b are multiples of the least common multiple of a
and b, or lcm(a, b). Thus the number of positive integers  n that are multiples of both a and


n
n
b is the greatest integer less than or equal to
, or INT 
 .
lcm ( a , b)
 lcm ( a , b) 
4. Answers will vary; we give one possibility.
Question: This semester at Alpha U., 290 students took art, 260 took biology, 240 took
chemistry, 220 took dentistry, 105 took art and biology, 95 took art and chemistry, 85 took art
and dentistry, 80 took biology and chemistry, 70 took biology and dentistry, 60 took
chemistry and dentistry, 30 took art, biology, and chemistry, 25 took art, biology, and
dentistry, 20 took art, chemistry, and dentistry, 15 took biology, chemistry and dentistry, and
5 took art, biology, chemistry, and dentistry. How many students took at least one of art,
biology, chemistry, or dentistry?
Solution using the Inclusion-Exclusion Principle: We label as A, B, C, and D the sets of
students who took art, biology, chemistry, and dentistry, respectively.
A1  n( A)  n( B )  n(C )  n( D ) = 290 + 260 + 240 + 220 = 1010;
A2  n( A  B )  n( A  C )  n( A  D )  n( B  C )  n( B  D )  n(C  D )
= 105 + 95 + 85 + 80 + 70 + 60 = 495;
A3  n( A  B  C )  n( A  B  D )  n( A  C  D )  n( B  C  D )
= 30 + 25 + 20 + 15 = 90;
A4  n( A  B  C  D ) = 5;
and n( A  B  C  D )  A1  A2  A3  A4 = 1010 – 495 + 90 – 5 = 600.
Solution using a Venn diagram: We begin by placing a 5 in the square region representing the
intersection of all four sets. (See the diagram that follows.) We proceed outward based on
the number of students who took three courses. For example, since n( A  B  C )  30 and
n( A  B  C  D)  5 , then n( A  B  C  D )  25 , so we place a 25 in the square region
above the one with a 5 in it. Then we continue outward based on the number of students who
took two courses and finally one course. We then add all of the numbers in the disjoint
regions to get a total of 600.
Combinatorics
127
A
B
75
55
70
C
50
25
40
65
15
5
10
30
45
20
35
60
D
Exercise Set 6.6
1. (a) INT(100/6) = INT(16.66…) = 16
(b) INT(100/8) = INT(12.5) = 12
(c) INT(100/13) = INT(7.69…) = 7
(d) INT(100/19) = INT(5.26…) = 5
(e) INT(100/20) = INT(5) = 5
2. (a) These are the multiples of 5  8  40 that are less than or equal to 100; there are 2.
(b) If A is the set of multiples of 5  100 and B is the set of multiples of 8  100 , then the
Inclusion-Exclusion Principle says that there are n( A)  n( B)  n( A  B) = INT(100/5) +
INT(100/8) – INT(100/40) = 20 + 12 – 2 = 30 positive integers  100 that are multiples of
either 5 or 8.
(c) From Exploratory Exercise 3, the common multiples of 6 and 8 are the multiples of the least
common multiple of 6 and 8, denoted lcm(6, 8), which is 24. Thus the number of positive
integers  100 that are multiples of both 6 and 8 is the number of positive integers  100 that
are multiples of lcm(6, 8) = 24, which is INT(100/24) = 4.
(d) If A is the set of multiples of 6  100 and B is the set of multiples of 8  100 , then the
Inclusion-Exclusion Principle says that there are n( A)  n( B)  n( A  B) = INT(100/6) +
INT(100/8) – INT(100/24) = 16 + 12 – 4 = 24 positive integers  100 that are multiples of
either 6 or 8.
3. (a) Let A, B, C, and D represent the sets of positive integers  100 that are multiples of 2, 3, 5
and 7, respectively. Using the Inclusion-Exclusion Principle,
A1  n( A)  n( B )  n(C )  n( D ) = INT(100/2) + INT(100/3) + INT(100/5) + INT(100/7)
= 50 + 33 + 20 + 14 = 117;
A2  n( A  B )  n( A  C )  n( A  D )  n( B  C )  n( B  D )  n(C  D )
= INT(100/6) + INT(100/10) + INT(100/14) + INT(100/15) + INT(100/21)
+ INT(100/35) = 16 + 10 + 7 + 6 + 4 + 2 = 45;
A3  n( A  B  C )  n( A  B  D )  n( A  C  D )  n( B  C  D )
= INT(100/30) + INT(100/42) + INT(100/70) + INT(100/105)
= 3 + 2 + 1 + 0 = 6;
A4  n( A  B  C  D ) = INT(100/210) = 0;
and n( A  B  C  D )  A1  A2  A3  A4 = 117 – 45 + 6 – 0 = 78.
(b) From part (a), there are 78 positive integers less than or equal to 100 that are multiples of 2, 3,
5, or 7. Of these 78 integers, 2, 3, 5, and 7 are prime and the rest are composite. That gives
78 – 4 = 74 composite integers less than or equal to 100. Recall that 1 is neither prime nor
composite; that leaves 100 – (74 + 1) = 25 prime numbers less than or equal to 100.
128
Chapter 6
4. The total number of three-digit positive integers is 9  10  10 = 900, since the first digit cannot be
a 0. The number of three-digit positive integers in which no digit is a 5 or a 7 is 7  8  8 = 448, so
the number that have either a 5 or a 7 as one of the digits is 900 – 448 = 452.
5. (a) If M and H represent the sets of students in the cohort taking math and history, respectively,
then n( M  H )  n( M )  n( H )  n( M  H ) = 90 + 75 – 40 = 125.
(b)
M
H
50
40
35
25
(c) 25
6. (a) Let B, C, and D be the sets of students taking band, chorus, and drama, respectively. Then
n( B  C  D)  n( B)  n(C )  n( D)  (n( B  C )  n( B  D)  n(C  D))  n( B  C  D)
= 150 + 180 + 165 – (70 + 85 + 90) + 40 = 495 – 245 + 40 = 290.
(b)
B
C
30
35
60
40
45
50
30
D
110
(c) From the Venn diagram, n( B  C  D) = 45.
(d) From the Venn diagram, n( B  C  D ) = 110.
7. We use a Venn diagram as in Exercise 6.
A
B
10
30
20
15
5
25
40
C
5
(a) From the Venn diagram, n( A  B  C ) = 30.
(b) From the Venn diagram, n( A  B  C ) = 5.
8. Let A be the set of rearrangements of 123456 in which 1 is followed by 3;
let B be the set of rearrangements of 123456 in which 2 is followed by 4;
let C be the set of rearrangements of 123456 in which 3 is followed by 5; and
let D be the set of rearrangements of 123456 in which 4 is followed by 6.
Combinatorics
129
Thinking of gluing together the 1 and the 3 to make 13, we see that that n(A) = 5! = 120, and
similarly n(B) = n(C) = n(D) = 5! = 120. Thus
A1  n( A)  n( B )  n(C )  n( D )  C ( 4, 1)  5!  4  120  480 .
An element of A B corresponds to an arrangement in which 13 and 24 appear; the number of
ways of permuting 13, 24, 5, and 6 is 4! = 24. Similarly, there are 4! = 24 arrangements in the
intersection of any two of the four sets, 3! = 6 arrangements in the intersection of three of the sets,
and only 2! = 2 arrangements in the intersection of all four sets, namely 135246 and 246135, so
A2  C ( 4, 2)  4!  6  24  144 , A3  C ( 4, 3)  3!  4  6  24 , A4  C ( 4, 4)  2!  1  2  2 , and
n( A  B  C  D )  A1  A2  A3  A4 = 480 – 144 + 24 – 2 = 358.
Since the total number of arrangements is 6! = 720, the number of arrangements in which no digit
is 2 more than the preceding one is 720 – 358 = 362.
9. Let A be the set of rearrangements of 123456 in which 1 is followed by 2;
let B be the set of rearrangements of 123456 in which 2 is followed by 4; and
let C be the set of rearrangements of 123456 in which 3 is followed by 6. As in the previous
exercise, we can think of gluing together the 1 and the 2, the 2 and the 4, and/or the 3 and the 6,
so that
A1  n( A)  n( B )  n(C )  C (3, 1)  5!  3  120  360 ;
A2  n( A  B )  n( A  C )  n( B  C )  C (3,2)  4!  3  24  72 ; and
A3  n( A  B  C )  C (3, 3)  3!  6 ; so
n( A  B  C )  A1  A2  A3  360 – 72 + 6 = 294; and
n( A  B  C )  720 – 294 = 426.
10. Let A be the set of combinations with at least 4 apple jellybeans; similarly, let C, G, L, and O
represent the sets of combinations with at least 4 of the cherry, grape, lemon, and orange
jellybeans, respectively. Then the number of elements in A is the number of ways to choose the
four remaining jellybeans from among any of the five flavors. In Section 6.4 we learned that the
number of such combinations with repetitions allowed is C(4 + 5 – 1, 4) = C(8, 4) = 70. (Think
of arranging 4 tally marks ( / ) and 5 – 1 = 4 column dividers ( | ).) Thus,
A1  n( A)  n(C )  n(G )  n( L)  n(O )  C (5, 1)  C (8, 4)  5  70  350 .
If a combination is in two of the five sets, then it must have four of one flavor and four of a
second flavor. As this accounts for all 8 of the jellybeans in the combination, then there is only
one possible combination for each pair of flavors, and A2  C (5, 2)  1  10 . Since it is not
possible to have at least four jellybeans of three or more flavors, A3  A4  A5  0 , and
n( A  B  C  D  E )  A1  A2  350  10  340 .
Since the total number of combinations of jellybeans is C(8 + 5 – 1, 8) = C(12, 8) = 495, then the
number of combinations with no more than three of the same flavor is 495 – 340 = 155.
11. The ten combinations are: CCGGLL, CCGGLO, CCGGOO, CCGLLO, CCGLOO, CCLLOO,
CGGLLO, CGGLOO, CGLLOO, and GGLLOO.
12. Let A, B, C, and D be the sets of arrangements in which the letters A, B, C, and D, respectively,
appear three times in a row. Then the number of elements in A is the number of arrangements of
AAA (glued together), 3 B’s, 3 C’s, and 3 D’s, which is the multinomial coefficient
10!
3,628,800

 16,800 . (See the MISSISSIPPI problem from Section 6.4.) Since n(A) =
3
216
1! (3!)
n(B) = n(C) = n(D), then
10!
A1  n( A)  n( B)  n(C )  n( D)  C (4, 1) 
 4 16,800  67,200 .
1! (3!) 3
130
Chapter 6
In a similar manner, to get an arrangement in A B , we glue together AAA and BBB, and then
permute these along with the three C’s and the three D’s (which do not have to be together).
8!
40,320
Thus n( A  B) 

 1120 , and by symmetry,
2
2
36
(1!) (3!)
8!
A2  C (4, 2) 
 6  1120  6720 .
2
(1!) (3!) 2
Gluing together three of the four groups of identical letters tells us that
6!
A3  C (4, 3) 
 4  120  480 .
(1!) 3 3!
If we glue together all four of the groups of identical letters, then there are only 4! = 24
arrangements, so A4  24 , and
n( A  B  C  D )  A1  A2  A3  A4 = 67,200 – 6720 + 480 – 24 = 60,936.
12!
Since the total number of arrangements is
 369,600 , the number of arrangements in which
(3!) 4
no letter appears three times in a row is 369,600 – 60,936 = 308,664.