Download Spring 2007 Math 510 Hints for practice problems

Spring 2007 Math 510
Hints for practice problems
Section 1.8.
3. Imagine a prison consisting of 64 cells arranged like the squares of an 8 × 8-chessboard. There
are doors between all adjacent cells. A prisoner in one the corner cells is told that he will be
released, provided he can get into the diagonally opposite cell after passing through every other
cell exactly once. Can the prisoner obtain his freedom?
Solution. Count the number of the walls the prisoner has to cross to reach to the corner cell
diagonally opposite to the beginning corner cell after passing through every other cells exactly
once. Then count number for walls separating those two cells. (EVEN or ODD?)
13. Read the algorithm of de la Loubère’s method in the book.
Section 2.4.
2.
Proof. Suppose that 100 numbers are chosen. We write each number a = 2k a0 with a0 odd,
which is called the odd part of a. If two numbers from the selection have the same odd part,
then one of the two will divide the other one and we are done. Otherwise, all 100 numbers have
distinct odd parts, in particular all odd numbers from 1 up to 199 are odd parts of the 100 selected
numbers. Let a be one of 100 selected number such that a ≤ 15. Write a = 2k a0 with a0 odd.
Thus 0 ≤ k ≤ 3.
If k = 0, then a = a0 is odd and, 3a, 5a, 7a, 9a, 11a, 13a are odd numbers no larger than 199.
So they are odd parts of some selected numbers. Hence, a divides at least 6 selected numbers.
If 2 ≥ k > 0, then a0 < 8. Thus, a0 ≤ 7. In this case, 3a0 , 9a0 and 27a0 are odd and less than
200 (27 × 7 = 189), thus they are the odd parts of some of the 100 selected numbers, say that
they are b1 = 2e1 3a0 , b2 = 2e2 9a0 , and b3 = 2e3 27a0 . If one of e1 , e2 , e3 is at least 2, then a divides
one of numbers b1 , b2 b3 . Otherwise two of e1 , e2 , e3 equal by the Pigeonhole principle. Then two of
b1 , b2 , b3 with equal powers of 2 will have one divide the other.
If k = 3, then a0 < 2, thus a0 = 1 and 3a0 , 9a0 , 27a0 , 81a0 are odd and less than 200, thus
are odd parts of some of the 100 selected numbers, say that they are b1 = 2e1 3a0 , b2 = 2e2 9a0 ,
b3 = 2e3 27a0 , and b4 = 2e4 81a0 . If one of the four powers of 2 is at least 3, then a divides the
corresponding number. Otherwise, 0 ≤ ei ≤ 2 for i = 1, 2, 3, 4. Pigeonhole principle implies
that two of the powers of 2 in the four numbers b1 , b2 , b3 , b4 have to be the same. That gives two
numbers such that one divides the other.
A in application 5, if the pigeonhole fails, then 100 numbers have 100 distinct odd parts. Thus
the numbers 2e3 3a0 , 2e5 5a0 , 2e7 7a0 , 2e9 9a0 , 2e11 11a0 , 2e13 13a0 are in the selection in addition to a for
some e3 , e5 , e7 , e9 , e11 , e13 . If any one of them is at least 2, then a divides the corresponding number,
otherwise they are all at most 1 and two of them have to equal. Hence those corresponding number
will have one divides the other.
3. Choose n + 1 numbers then follow the argument almost identically as in Application 5.
5. Line up the n + 1 number from small to the largest then add up the consecutive differences.
7. Each integer n can be written as n = 100q + r with |r| ≤ 50.
1
9. For each group of people, the group age is the sum of the ages of people in that group. There
are totally 210 = 1024 different groups, while the group ages have to maximal of 10 × 60 = 600.
Two groups must have the same group age. But the two group could share a common member.
Through away the people from both groups if they are in both groups.
10. Follow the application 4.
15. Prove that, for any n + 1 integers a1 , a2 , . . . , an , an+1 , there exist two of the integers ai and
aj with i 6= j such that ai − aj is divisible by n.
Proof. Hint: Consider the n boxes labeled by 0, 1, . . . , n − 1. Place the integers ai in the box
labeled r is the remainder of ai is r when divided by n. Pigeonhole principle implies that there is
a box with two difference objects ai and gj with i 6 j, i.e., ai and aj have the same remainder when
divided by n. Hence the difference of these two numbers have the remainder 0 when divided, i.e.,
ai − aj is divisible by n.
17. Write ai = 2mi to be the number of acquaintances of the ith person. Then 0 ≤ mi ≤ 49. If
each number appears exactly twice, then there will be two people who don’t know any one while
there is one person who knows 98 people not including him/herself. That is impossible.
19. (a). Divide the triangle into four (1 + 3) equilateral triangles with sides length 1/2.
(b) Divide the triangle into nine (1 + 3 + 5) equilateral triangles with sides length 1/3.
(c) Divide the triangle into (1 + 3 + 5 + · · · + (2n − 1)) equilateral triangles with sides length
1/n.
Section 3.6
1. easy.
5. Count the number of the factors 2 and 5.
8. In how many ways can six men and six ladies be seated at a round table if the men and ladies
are to sit in alternate seats?
Solution. The arrangement can be achieved through two stages.
9. Count the number of the situations that A and B sit next to each other by considering AB
and BA as single person. For the send part, consider the case when AB occurs.
10. A committee of 5 is to be chosen from a club that boast a membership of 10 men and 12
women. How many ways can the committee be formed if it has to have at least 2 women? How
many ways if, in addition, one particular man and one particular woman who are members of the
club, refuse to serve together on the committee.?
22
Solution. The total number of possible committees can be formed
without
any
restrict
is
.
5
10
Among them, the number of committees without
any women is 5 and the number of the com
10
mittee with exactly one woman
is 12 · 4 . Thus the number of committees with at least two
22
10
10
women is 5 − 5 − 12 · 4 .
If A is the particular man and B is the particular woman who can not serve
together. Then
the total number of committees with both A and B on the committee is 20
without
any other
3 9
− 93
restrictions. But the number of such committees with exactly one woman is 3 . Thus 20
3
is the number of committees with at least two women and A and B are both on. This number
should be excluded from the earlier answer. Thus the number
of committees
with9at least two
10
10
20
women and not A and B both on the committee is 22
−
−
12
·
−
(
− 3 ).
5
5
4
3
2
11. How many sets of 3 numbers each can be formed from the numbers {1, 2, 3, . . . , 20} if no 2
consecutive numbers are to be in a set.
Solution. First count the total number of 3 number subsets regardless two consecutive numbers
being in a set or
not. This number is the number of the 3-combinations of the set of 20 elements,
20
which is
. Then count the number of 3-element subsets with exactly one pair of consecutive
3
numbers in it. For each pair {i, i + 1} for 2 ≤ i ≤ 18, there 20 − 4 = 16 ways to add the third
number that won’t make new consecutive numbers with two already chosen, i.e., any number other
than {i − 1, i, i + 1, i + 2}. For i = 1 or 1 = 19, there are 20 − 3 = 17 choices. So the total is
17 × 16 + 2 × 17 = 17 × 18. We still need to count the number of 3-number subsets that contain
3-consecutive numbers. There are
18 of them ({i, i + 1, i + 2} for i = 1, . . . , 18. Finally the total
20
number of 3-number subsets is
− 17 × 18 − 18 = 1140 − 324 = 816.
3
12. A football team of 11 players is to be selected from a set of players, 5 of whom can play only
in the backfield, 8 of them can only play on the line, 2 of them can play either in the backfield, or
on the line. Assuming a football team has 7 men on the line and 4 men in the backfield, determine
the number of football teams possible.
Solution. We divide the formation of the team into the following situations:
(1) None of those two players (A
and
B) who can player in the backfield or on the line is on
8
5
the team. In this case, there are 7 · 4 = 40.
(2) Exactly one of A and B is on the team. There two choices. For
each choices, there are two
5 8
situations: (a) This person plays in the backfield
and
there
are
= 160 many possible. (b)
3 7
5 8
This person is plays on the line, there are 4 6 = 5 · · · 28 = 140 possible teams. Thus the total
number of teams when only one of A and B plays is 2(160 + 140) = 600.
(3) Both A and B in the team. Now thereare four
different cases.
5
8
(a) Both A and B are on line, there are 4 · 5 ways.
(b) Both A and B are in the back field, there are 52 · 87 ways.
(c) One of the A and B is on the line and the other is in the backfield. There are 2 · 53 · 86
ways.
Finally the total number of ways is the sum of the numbers above.
14. A classroom has two rows of 8 seats each. There are 14 students, 5 of whom always sit in
the front row and 4 of whom always sit in the back row. In how many ways can the students be
seated.
Solution. We will regard the eight seats in each row different. The arrangement process can be
divided into three stages.
1. Let the 5 students who always sit in the font row choose their seats. There are P (8, 5)
possible outcomes.
2. Let the 4 students who always sit in the back row choose their seats. There are P (8, 4)
possible outcomes.
3. Let the remaining five students to fill choose their seats from the remaining 16 − 9 seats.
There are P (7, 5) different ways.
By the multiplication principle, the total number of ways to seat the students is P (8, 5)·P (8, 4)·
P (7, 5).
15. (a) Select 15 women out of the 20 then distribute them to 15 men.
3
(b) First choose 10 men then choose 10 women to match.
18. In how many ways can 2 red and 4 blue rooks be placed on an 8-by-8 board so that no two
rooks can attach on another.
Solution. Note that there cannot be more than one rook in any column or row in order to avoid
any rook attacking any other rooks. The placement can be done to two stages. 8
Stage 1. Place 2 red rooks. Ch oose 2 rows out of 8 rows and there are
ways. Then
2
for
each
place the two rooks in the two rows chosen and there are 8 · 7 of choices. Thus there are
8
· 8 · 7 ways to place 2 red rooks.
2
Stage 2. For each placement of the two
red
rooks, choose 4 rows out the six remaining 6 rows
6
to place the four blue rooks. There are
ways. Then place the four rooks in the remaining
4
6 columns
for each choice of the four rows. There 6 · 5 · 4 · 3 ways to do so. Thus there are
6
· 6 · 5 · 4 · 3 ways to place the four blue rooks in the remaining board.
4
(8 · 7)2 (6 · 5 · 4 · 3)2
By multiplication principle, there are
·
ways to place the six rooks on the
2!
4!
board so that no two rooks attack one another.
Another way: First place 6 indistinguishable rooks on an 8 × 8-board. Then choose 2 of them
to be painted red.
19. (a) Choose 5 rows for the red rooks others are for blue rooks. Decide the column positions
8! ways of doing so.
(b) Choose 5 rows for red rooks, then choose from the remain rows (7 of them) choose three
for blue rooks. Then chose 8 columns to place the rooks standing in 8 different rows.
32. Determine the number of 11-permutations of the multiset
{3 · a, 4 · b, 5 · c}.
Solution. There are three types of objects in the multiset S which has size 12. To get 11permutation, one of the objects has be left behind. There are three different ways to leave one
object behind, that is, either an a, or a b, or a c.
11! ;
The number of 11-permutations with 2 a’s is
2! · 4! · 5!
11! ;
The number of 11-permutations with 3 b’s is
3! · 3! · 5!
11! ;
The number of 11-permutations with 4 c’s is
3! · 4! · 4!
11!
11!
11! .
Thus the total number of 11-permutations is
+
+
2! · 4! · 5! 3! · 3! · 5! 3! · 4! · 4!
There twelve objects in the multiset. decide which one of the object to be unselected.
37. A bakery sells 6 different kinds of pastry. If the bakery has at least a dozen of each kind, how
many different options for a box of dozen pastry are there? What if each box contains at least one
of each kind?
Solution. Let x1 , x2 , x3 , x4 , x5 , x6 be, respectively, the numbers of the first, second, third, fourth,
fifth, and sixth pastry in a dozen. Then each xi is a non-negative integer and x1 + x2 + x3 + x4 +
x5 + x6 = 12. This becomes a question of 12-combinations
of a multiset
with six types of objects
12 + 6 − 1
and infinite repetition numbers. Then total number is
.
12
4
If there is at least one of each kind in the dozen, then each
xi ≥ 1. Set yi = xi − 1. Then
6+6−1
y1 + y2 + y3 + y4 + y5 + y6 = 6. Thus there are
.
6
41. In how many ways can 12 indistinguishable apples and 1 orange be distributed among three
children in such a way that each child gets at least one piece of fruit?
Solution. Since each child should get at least one piece of fruit and all fruits have to be distributed,
do the distribution in two stages.
Stage 1. Choose one of the three children to get the orange and the other two for one apply
each. There are 3 ways.
Stage 2. For each outcome of the stage 1, distribute the remaining 10 apples to the three
children. Each child has the ability to take as many as he/she is given. This becomes problem
of 10-combinations of a multiset with three types of objects (apples received by each
child) with
10 + 3 − 1
infinite repetition numbers (each child can take as many as given). Thus there are
10
ways.
10 + 3 − 1
Finally, multiply the ways from two stages 3 ·
to get the total number of ways
10
distributing the fruits to three children such that each child gets at least one piece of fruit.
42. Distribute the lemon and the lime drinks first then give each of the remaining two children
an orange drink. Then distribute the rest of orange drinks without any restrictions. (combinations
of multiset of the number of the integral solutions of an equation with conditions.
Section 5.8
6. Expand the binomial with 3x and −2y as two variables first.
7. Consider x = 1 and y = 2 in the binomial theorem.
9. Set x = 1 and y = −10 binomial theorem.
11. The left hand side is the number of the k-combinations that has at least one of a, b, c in it.
For the right have side,
n−1
is the number k-combinations of S with a in them;
k−1 n−2
is the number of k-combinations without a but has b in them;
k−1 n−3
is the number of k-combinations without a and b but has c in them;
k−1
15. Consider the binomial expansion of the function f (x) = (1 + x)n and compute f 0 (x) on both
sides. Then evaluate f 0 (−1) with n − 1 > 0.
(−r)(−r−1)···(−r−k+1)
and then reverse the order of the
=
21. Take all negative signs out from −r
k
k!
product of the numerator.
23. (a). Thepaths are determined by when to walk in an east-west direction street. The number
of paths is 24
.
10
9
ways from her friends home
(b). 4 ways to walk to her friends home. There are 15
5
to school.
9 15
Total number ways from
her
home to school passing through her friend’s home is 4 5 .
9 9 6
(c). There are 4 3 3 ways.
25. Consider a group people with m1 women and m2 men. To choose n-people from the group,
consider the the number k of women chosen.
5
31. Note that the clutter of size at least con not have 0-combination nor 4-combination. Then
show that there can not be any 3-combinations in it. Otherwise, say {1, 2, 3} is in the clutter of
size 6. Among the total 24 = 16 combinations, only 14 are usable and seven of them are subsets
of {1, 2, 3}. Seven of the remaining 9 subsets, all contain 4 in them and {4} cannot in the clutter
nor can any three combination with 4 in them (which will have four subsets with 4 in them) leave
not enough subset for 6 subsets to form a clutter. By duality, there cannot be any 1-combination
in the clutter.
40. Use multinomial coefficients.
41. Use multinomial expansion.
43. Consider the n-permutations of the multiset S = {n1 ·a1 , · · · , nt ·at }. Counting those starting
with a1 , those starting with a2 , etc.
Section 6.7
1. Use inclusion-exclusion principle by considering the numbers that are divisible by 4, 5, or
6 respectively and their intersections. A number is divisible by both 4 and 6 if and only if it is
divisible by LCM (4, 6) = 12.
3. Use inclusion-exclusion principle. Let A1√be the set of all perfect squares from 1 to 10000, and
A2 be the set of perfect cubes. Then |A1 | = [ 10000] and |A2 | = [100001/3 ], |A1 ∩ A2 | = [100001/6 ],
where [x] is the largest integer that is less than or equal to x for any given real number.
5. Do similarly as in Prob. 4.
8. Change into a question of combinations of multiset similar to problem 4. Or consider the set
of all solutions with xi , . . . at least 9.
10. There is an r-combination means that the equation x1 + x2 + · · · + xk = r has a non-negative
integer solution with x1 ≤ n1 , x2 ≤ n2 , . . . , xk ≤ nk . In particular n1 + n2 + · · · + nk ≥ r. But the
fact that A1 ∩ A2 ∩ · · · ∩ Ak is non-empty means that there integers x1 > n1 , x2 > n2 , . . . , xk > nk
such that x1 + x2 + · · · + xk = r > n1 + n2 + · · · nk .
12. One has to consider which
four integers are in their natural positions, while all other are not
in their natural positions. 84 ways to choose the
four positions. D4 ways to arrange the other
8
four with none in its natural position. Hence 4 D4 .
14. 84 D4 . See # 12 for argument.
(−1)n
n ). Then use induction on n.
24(b). The forbidden
positions are divided into 3 clusters. r1 = 3 · 4.
r2 = 3 × 2 + 32 · 4 · 4.(The two positions are in the same “cluster” or in separate clusters.)
r3 = P (3,
2)2 · 4 +4 · 4 · 4. (The three positions are in two clusters or in three clusters.)
r4 = 32 · 2 · 2 + 31 2 · 4 · 4. (Four positions are in two clusters or in three clusters with one of then
have two
positions.)
3
r5 = 1 · 4 · 2 · 2. (one of the cluster has two positions and the other two has two positions.)
r6 = 2 · 2 · 2.
20. Note that Dn = n(Dn−1 +
26. Convert the problem into a 6 × 6 chess-board with forbidden positions.
6
27. Use circular permutations and inclusion-exclusion principle. A1 is all circular permutations
with 12 appears, ... A7 all circular permutations with 78 appears, and A8 all circular permutations
with 81 appears.
30. How many circular permutations are there of the multiset {2 · a, 3 · b, 4 · b, 5 · d} where, for
each type of letter, all letters of that type do not appear consecutively, (i.e., not all letters of that
type appear consecutively).
Solution. Let S be the set of all circular permutations of the multiset. Let Aa be the set of
all circular permutations with all a’s appear consecutively. Similarly, let Ab , Ac , and Ad be the
set of all circular permutations with all b’s, c’s, and d’s, respectively, appear consecutively. To
compute |Aa |, we change into the question of linear permutations with all a’s appear at the very
12!
beginning. Thus there are 3+4+5
many such linear permutations, i.e., |Aa | = 3!·4!·5!
. Similarly,
3,4,5
10!
9!
10!
11!
|Ab | = 2!·4!·5! , |Ac | = 2!·3!·5! , and |Aa | = 2!·3!·4! . |Aa ∩ Ab | = 1!·4!·5! (treating all a’s as one a and
at the beginning to change into a linear permutation question and all b’s as just one b. Similarly,
8!
8!
7!
6!
9!
, |Aa ∩ Ad | = 3!·4!·1!
, |Ab ∩ Ac | = 2!·1!·5!
, |Ab ∩ Ad | = 2!·4!·1!
, |Ac ∩ Ad | = 2!·3!·1!
,
|Aa ∩ Ac | = 3!·1!·5!
7!
6!
5!
7!
|Aa ∩ Ab ∩ Ac | = 1!·1!·5!
, |Aa ∩ Ab ∩ Ad | = 1!·4!·1!
, |Aa ∩ Ac ∩ Ad | = 3!·1!·1!
, |Ab ∩ Ac ∩ Ad | = 2!·1!·1!
,
3!
and |Aa ∩ Ab ∩ Ac ∩ Ad | = 1!·1!·1! .
One needs to calculate |S|. Note that since the greatest common divisor of 2, 3, 4, 5 is 1, each
circular permutation of the multiset {2 · a, 3 · b, 4 · c, 5 · d} corresponds to exactly 2 + 3 + 4 + 5
different linear permutations of the multiset (why?) Then |S| = (2+3+4+5)!
/(2 + 3 + 4 + 5). Now
2!·3!·4!·5!
use the inclusion and exclusion principle to compute |Āa ∩ Āb ∩ Āc ∩ Ād |.
Note that the circular permutation of {3 · a, 6 · b, 3 · c} determined by the linear permutation
abcbabcbabcb corresponds to only 4 different linear permutations.
Section 7.8
16. The characteristic equation is x3 − 3x + 2 = 0. Observe that x = 1 is a solution and factorize
to get (x1 )2 (x − 2). Then general solution is hn = c1 + c2 n + c3 (−2)n by Theorem 7.2.2.
P∞
1
n n
28(b).
n=0 (−1) x = 1 + x .
P∞ 1 n
x
28(d).
n=0 n! x = e .
20(c). g(x) − xg(x) − 9x2 g(x) + 9x3 g(x) = h0 + (h1 − h0 )x + (h2 − h1 − 9h0 )x2 = x + x2 .
1
1
1
−
−
x + x2
x + x2
3 +
12 +
8 .
=
=
g(x) =
1 − x − 9x2 − 9x3
(1 + 3x)(1 − 3x)(1 − x)
1 − 3x 1 + 3x 1 − x
36. g(x) =
1 · 1 · 1 · 1 .
1 − x2 1 − x5 1 − x 1 − x7
Section 9.5
1.
7
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X
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HXX
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e
HH X
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\
Z
@ZXXX
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Xe
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HHX\
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4.
Any bipartite graph with at least one vertex meeting five or more edges will not be a
domino-bipartite graph of any board.
7. Count the number of edges through x-vertices and then through Y -vertices.
9. Follow the matching algorithm starting a with matching (try to find the matching with as
many edges as you can to save time).
10. First select a matching M (with as many edges as you could find). The follow the matching
algorithm. If you get a break-through, find the m-alternating chain and a new matching with more
edges. Then apply the matching algorithm with new matching and repeat till a nonbreak-through
happens. In that case the matching is a max matching.
12.
Check the Marriage Condition. |A1 ∩ A3 ∩ A4 ∩ A5 | = 3 < 4. Having A5 removed the
remaining the sets has an SDR by checking the Marriage Condition.
14. First take an SDR {e1 , e2 , . . . , en } for the collection A. If x is one of the member, then it is
an SDR we are looking for. Otherwise, replace e1 by x to get a new SDR {x, e2 , . . . , en }.
Example: A1 = {a, x} and A2 = {x} and A = (A1 , A2 ). Then e1 = a, e2 = x give an SDR,
but A1 will never be represented by x as x has to represent A2 .
16. A sequence of numbers e1 , e2 , . . . , en from {1, 2, . . . , n} for an SDR for A with ei in Ai if
any only if that ei 6= i and all are distinct. Since there are only n different numbers, thus the
sequence give forms a permutation of the set {1, 2, ,̇n}, and thus a derangement. Conversely, each
derangement gives an SDR. Dn > 0 if n > 1.
1, 2 2, 1
19.
.
2, 1 1, 2
8
21. Consider the preferential ranking

w1
1, n

w2  n, 1
w3 
n − 1, 2
..  ..
. 
 .
wn−1 3, n − 2
wn 2, n − 1
matrix
2, n − 1 3, n − 2
1, n
2, n − 1
n, 1
1, n
..
..
.
.
4, n − 3 5, n − 4
3, n − 2 4, n − 3

···
n, 1
· · · n − 1, 2

· · · n − 2, 3

.. 
..
.
. 

· · · 2, n − 1
···
1, n
Prove that, for each k = 1, 2, . . . , n, the complete marriage in which each woman gets her kth
choice is stable.
Proof. If each woman gets her kth choice, then w1 → mk , w2 → mk+1 , · · · , wn−k+1 → mn , wn−k+2 →
m1 , wn → wk−1 . Note that mi is his wifes’ n − k + 1th choice. Note that for each pair (wi , mj )
the ranking p of mj by wi and the ranking q of wi by mj satisfy the equation p + q = n + 1. If wi
ranks mj higher than her husband, then p < k then q > n − k + 1. Thus mj ranks wi lower than
his wife. Thus the marriage is stable.
9