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Transcript of 1 Relations and Functions Chapter 5 1 to many 1 to 1 many to many.
1
Relations and FunctionsChapter 5
1 to many
1 to 1
many to many
2
5.1 Cartesian Products and Relations
Def. 5.1 For sets , , the Cartesian product, orcross product, of and is denoted by and equals {( , )| , }.
A B UA B A B
a b a A b B
•the elements of A B are ordered pairs•|A B|=|A| |B|=|B A|
But, in general AndA B B AA A A a a a a A i nn n i i
.{( , , , )| , }.1 2 1 2 1
3
5.1 Cartesian Products and Relations
Ex. 5.3 The sample space by rolling a die first then flipping a coin
Tree structure
1 2 3 4 5 6
1,H 1,T 2,H 2,T 3,H 3,T 4,H 4,T 5,H 5,T 6,H 6,T
={1,2,3,4,5,6} {H,T}
4
5.1 Cartesian Products and Relations
Trees are convenient tools for enumeration.
Ex. 5.4 At the Wimbledon Tennis Championships, women playat most 3 sets in a match. The winner is the first to win two sets.In how many ways can a match be won?
N
E
N
E
N
E
N
EN
E
Therefore, 6 ways.
(
, )
permutations of NNE and EEN:3!
2! for men:
5!
3! 2!
2 6 2 20
5
5.1 Cartesian Products and Relations
.on
a called is ofsubset Any . to from
a called is ofsubset any ,, setsFor 5.2 Def.
Arelation
binaryAABArelation
BAUBA
In general, for finite sets A,B with |A|=m and |B|=n, thereare 2mn relations from A to B, including the empty relationas well as the relation A B itself.
Ex. 5.7 A=Z+, a binary relation, R, on A, {(x,y)|x<y}
(1,2), (7,11) is in R, but (2,2), (3,2) is not in Ror 1R2, 7R11 (infix notation)
6
5.1 Cartesian Products and Relations
Theorem 5.1 For any set , , :(a) b)
(c) (d) ( ) = ( ) ( )Proof of (a): For any , , ( , ) ( )
A B C UA B C A B A CA B C A B A CA B C A C B CA B C A C B C
a b U a b A B C a Ab B C a A b B b C a b A Ba b A C a b A B A
( ) ( ) ( )( ( ) ( ) ( )
( ) ( ) ( )
( ) , , ( , ) ( ),( , ) ( ) ( , ) ( ) ( C)
7
5.2 Functions: Plain and One-to-One
Def. 5.3 For nonempty sets, A,B, a function, or mapping, f fromA to B, denoted f:A B, is a relation from A to B in which everyelement of A appears exactly once as the first component of anordered pair in the relation.
set A set B
not allowed
8
5.2 Functions: Plain and One-to-One
Def 5.4 Domain, Codomain, Range
abf(a)=b
domain codomain
range
A B
9
5.2 Functions: Plain and One-to-One
Ex. 5.10 the greatest integer less than or equal to
the least integer greater than or equal to
the integer part of
storage of matrix, is an matrix.row major storage:
( )( ). , . ,
( )( ). , . ,
( )( )( . ) , ( . ) , ( )
( ) ( ), , , , , ,
a floor functionf x x x
b ceiling functionf x x x
c trunc functiontrunc x xtrunc trunc truncd A a n n
a a a aij n n
n
3 8 3 3 8 4 3 3
3 8 4 3 8 3 3 3
3 8 3 3 8 3 3 3
11 12 1 21 a a af a initial i n j
a a a a a a af a initial j n i
n nnij
n n nnij
2 31
11 21 1 12 2 13
1 1
1 1
, , ,( ) ( ) ( )
, , , , , , , , ,( ) ( ) ( )
column major storage:
10
5.2 Functions: Plain and One-to-One
If |A|=m, |B|=n, then the number of possible functions from Ato B is nm.
Def 5.5 A function f:A B is called one-to-one, or injective, ifeach element of B appears at most once as the image of an element of A.
If : is one to one, with , finite, we must have For arbitrary sets , , if : is one to one,
then for a1
f A B A BA B A B f A B
a A f a f a a a
| | | |.
, , ( ) ( ) .2 1 2 1 2
For example, is one - to - one.
But, is not. (
f x x
g x x x g g
( )
( ) ( ) ( ) )
3 7
0 1 04
11
5.2 Functions: Plain and One-to-One
If |A|=m, |B|=n, and , then the number of one-to-one functions from A to B is P(n,m)=n!/(n-m)!.
m n
Def. 5.6 If : and then( for some and( is called the image of under .
1
1
f A B A Af A b B b f a a Af A A f
,) { | ( ), })
1 11
Ex. 5.16 Let :R R be given by ( ) = Then(R) = the range of = [0,+ ). The image of Z under is(Z) = {0, 1, 4, 9, 16, } and for we get(
g g x xg g gg Ag A
2
11
2 10 4
.
[ , ]) [ , ].
12
5.2 Functions: Plain and One-to-One
Theorem 5.2 Let : , with Then(a) ((b) ((c) ( when is injective.Proof: for part (b). For any , ( for some
= ( ) ( = ( ) and or ( = ( ) and or ( (
1
11
1 12 2
f A B A A Af A A f A f Af A A f A f Af A A f A f A f
b B b f A Aa A A b f a b f a a A b f aa A b f A b f A b f A
, .) ( ) ( )) ( ) ( )) ( ) ( )
), )
) ( ( )) )) ( )
21 2 1 2
2 1 22 1 2
1 22
1 1 f Ab f A f A b f A b f A
a A b f a a A b f aa A A b f a b f A A
( )) ( ) ( ) )
( , ,, ).
22 1
1 1 1 2 2 21 2 1 2
Conversely, ( or (for some = ( )) or (for some = ( ))
for some = ( ) (
1 2
1 2 3
a b
13
5.3 Onto Functions: Stirling Number of the Second Kind
Def 5.9 A function : is called , or , if( ) = , that is, if for all there is at least one
with ( ) = .
f A B onto surjectivef A B b B a A
f a b
Ex 5.19 f:R R defined by f(x)=x3 is onto. But f(x)=x2 is not.
Ex. 5.20 f:Z Z where f(x)=3x+1 is not onto. g:Q Q where g(x)=3x+1 is onto. h:R R where h(x)=3x+1 is onto.
If A,B are finite sets, then for any onto function f:A B topossibly exist we must have |A| |B|. But how many ontofunctions are there?
14
5.3 Onto Functions: Stirling Number of the Second Kind
Ex. 5.22 If = { , , } and = {1,2} then all functions: are onto except and
So there are | |onto functions from to .
In general, if | |= 2 and | |= 2, then there are 2 ontofunctions from to .
2| |
m
A x y z Bf A B f x y z
f x y z BA B
A m BA B
A
13
1 1 1
2 2 2 2 2 2 6
2
{( , ), ( , ), ( , )}
{( , ), ( , ), ( , )}.
15
5.3 Onto Functions: Stirling Number of the Second Kind
Ex. 5.23 For A={w,x,y,z} and B={1,2,3}, there are 34 functionsfrom A to B. Among all functions:(1) {1} is not mapped: 24 functions from A to {2,3}(2) {2} is not mapped: 24 functions from A to {1,3}(1) {3} is not mapped: 24 functions from A to {1,2}But the functions A to {1} or {2} or {3} are all counted twice.Therefore, number of onto functions from A to B is
33
22
3
11 36
33
22
3
11
4 4 4
. In general, if | |= 3
and | |= 3, there are 3
3 onto
functions from to .
A m
B
A B
m m m
(What about m=1 or m=2?)
16
5.3 Onto Functions: Stirling Number of the Second Kind
General formulaFor finite sets , with | |= and | |= , there areA B A m B nn
nn
n
nn
n
nn
n n n
n kn k
n
n kn k
m m m
n m n m k m
k
n
k m
k
n
11
22
12
2 11
1 1
1
2 1
0
1
0
( ) ( )
( ) ( ) ( ) ( )
( ) ( )
17
5.3 Onto Functions: Stirling Number of the Second Kind
Examples at the beginning of this chapter (P217)
(1) seven contracts to be awarded to 4 companies such that every company is involved?
4
44
4
33
4
22
4
11 84007 7 7 7
ways
(2) How many seven-symbol quaternary (0,1,2,3) sequences have at least one occurrence of each of the symbols 0,1,2, and 3?(3) How many 7 by 4 zero-one matrices have exactly one 1 in each row and at least one 1 in each column?
18
5.3 Onto Functions: Stirling Number of the Second Kind
Examples at the beginning of this chapter (P217)
(4) Seven unrelated people enter the lobby of a building which has four additional floors, and they all get on an elevator. What is the probability that the elevator must stop at every floor in order to let passengers off? 8400/47=8400/16384>0.5(5) For positive integers m,n with m<n, prove that
( ) ( )
1 00
k m
k
n n
n kn k
(6) For every positive integer n, verify that
nn
n kn kk m
k
n! ( ) ( )
10
19
5.3 Onto Functions: Stirling Number of the Second Kind
Ex. 5.25 7 jobs to be distributed to 4 people, each one gets at leastone job and job 1 is assigned to person 1.Ans: case 1: person 1 gets only job 1 onto functions from 6 elements to 3 elements (persons)
( ) ( )
13
33 540
0
3 6k
k kk
case 2: person1 gets more than one job onto functions from 6 elements to 4 elements (persons)
( ) ( )
14
44 1560
0
4 6k
k kk 540+1560
20
5.3 Onto Functions: Stirling Number of the Second Kind
The number of ways to distribute m distinct objects into n different containers with no container left empty is
( ) ( )
10
k m
k
n n
n kn k
If the containers are identical:1
10n
n
n kn kk m
k
n
!( ) ( )
S(m,n): Stirling number of the second kindn!S(m,n) onto functions
Ex. 5.27 distribute m distinct objects into n identical containerswith empty containers allowed
n
i)i,m(S
1
21
5.3 Onto Functions: Stirling Number of the Second Kind
Theorem 5.3 Let , be positive integers with < .Then ( + , ) = ( , - ) + ( , ).
m n n mS m n S m n nS m n
11 1
A a a a am m= { 1 , , , , }2 1 Proof:
n identical containers
S(m+1,n)
am+1 is alone in one container am+1 is not alone in one container
S(m,n-1)Distribute other n objects firstinto n containers. Then am+1 canbe put into one of them.
nS(m,n)
22
5.3 Onto Functions: Stirling Number of the Second Kind
Ex. 5.28 How many ways to factorize 30030 into at least twofactors (greater than 1) where order is not relevant?Ans: 30030 2 3 5 7 11 13
There are at most 6 factors.Therefore, the answer is S(6,2)+ S(6,3)+S(6,4)+S(6,5)+S(6,6)=202
Ex. Prove that for all m n mm
ii S n in
i
m, , ( !) ( , )
Z +
1Proof: mn: ways to distribute n distinct objects into m distinct containers
i!S(n,i): ways to distribute n distinct objects into i distinct containers with no empty containers
23
5.4 Special Functions
Def 5.10 For any nonempty subset , , any function ::is called a on . If , then the binary operationis said to be closed on .
Def 5.11 A function : is called a , or ,operation on .
Ex. 5.29 (a) :Z Z Z, defined by ( , ) = - , is closed binaryoperation on Z.
(b) If :Z Z Z with ( , ) = - , then is not closed.
(c) :R R with ( ) = / is a unary operation on R
+ +
+ + +
A B f A A Bbinary operation A B A
A
g A A unary monaryA
f f a b a b
g g a b a b g
h h a a
1 .
24
5.4 Special Functions
Def 5.12 Let : ; that is, is a binary operation on .(a) is said to be commutative if ( , ) = ( , ), ( , ) .(b) When (that is, when is closed), is said to be associative if for , , , ( ( , ), ) = ( , ( , )).
Ex. 5.32 (a) :Z Z Z with ( , ) = + - . Then is bothcommutative and associative.(b) :Z Z Z with ( , ) = | | . Then (3,-2) = 6 (-2,3). is not commutative. But (
f A A B f Af f a b f b a a b A A
B A f fa b c A f f a b c f a f b c
f f a b a b ab f
h h a b a b h h hh
3
h a b c h a b c a b ch a h b c a h b c a b c a b c a b c h
g g a b a b g
( , ), ) = ( , )| |= | | | | and( , ( , )) = | ( , )|= | | | |= | | | | | |= | | | | . So is
associative.(c) :R R Z with ( , ) = + then is commutative butnot associatve. g(g(3.2,4.7),6.4) = 15 but g(3.2,g(4.7,6.4)) = 16
,
25
5.4 Special Functions
Ex 5.33 If = { , , , }, then | |= 16. Consequently,
there are 4 functions : , that is, 4 closed binary operations on . How many of them are commutative?
16 16A a b c d A A
f A A AA
a b c d
abcd
44
46
6 entries
Therefore, 4 46 4
26
5.4 Special Functions
Def. 5.13 Let : be a binary operation on . Anelement is called an for if ( , ) = ( , )= , for all .
Ex. 5.34 (a) :Z Z Z with ( , ) = + has identityelement 0 since ( , ) = ( , ) = for any integer .(b) :Z Z Z with ( , ) = - has no identity element.(c) Let = {1,2,3,4,5,6,7}, and : be the closed binary operation defined by ( , ) = { , }. Then ( ,7) =
( , ) = . So 7 is t
f A A B Ax A identity element f f a x f x a
a a A
f f a b a bf a f a a a
f f a b a bA g A A A
g a b a b g ag a a
0 0
7min
he identity element of .g
27
5.4 Special Functions
Theorem 5.4 Let : be a binary operation. If has anidentity, then that identity is unique.Proof: If are identities, then ( and
( Therefore,
f A A B f
x x f x x xf x x x x x
1 2 1 2 1
1 2 2 1 2
, , ), ) . .
identity identity
28
5.4 Special Functions
A a a a f A A An { , , , }, :1 2
a1 a2 a3 . . . an
a1
a2
a3
.
.
.an
n2 entries, each has n choices
nn2 closed binary operations
29
5.4 Special Functions
A a a a f A A An { , , , }, :1 2
n entries
n n2
2
entries
n nnn n2
2 commutative closed binary operations
a1 a2 a3 . . . an
a1
a2
a3
.
.
.an
30
5.4 Special Functions
A a a a f A A An { , , , }, :1 2 If a1 is the identity
a1 a2 a3 . . . an
a2 a3 . . . ana1
a2
a3
.
.
.an
a1
a2
a3
.
.
.an
n n2 2 1 ( ) entries
nn nn n
11 1 12 2( ) ( ) = closed binary operations
with an identity
31
5.4 Special Functions
A a a a f A A An { , , , }, :1 2 If a1 is the identitya1 a2 a3 . . . an
a2 a3 . . . ana1
a2
a3
.
.
.an
a1
a2
a3
.
.
.an n-1 entries
( ) ( )n n 1 1
2
2 entries
n
n n nnn n n n
11
1 1
2
2
2
2 2( ) ( )
commutative
closed binary operations with an identity
32
5.4 Special Functions
Def. 5.14 For sets and , if , then defined by is called the on the firstcoordinate. is defined similarly. (used in relational databases)
Ex. 5.36 For = = Z, let = {( , )| =| |}, Z andN.
A B D A B D Aa b a projection
A B D x y y x DD
AAB
AB
: ,( , ) ,
( )( ) { , , , , . . .}0 1 2 3
33
5.5 The Pigeonhole Principle
The Pigeonhole Principle: If m pigeons occupy n pigeonholes andm>n, then at least one pigeonhole has two or more pigeons roosting in it.
For example, of 3 people, two are of the same sex. Of 13 people,two are born in the same month.
Ex. 5.42 A tape contains 500,000 words of four or fewer lowerlowercase letters. Can it be that they are all different?
The number of different possible words is
26At least one word must be repeated.
4
26 26 26 475254 5000003 2
34
5.5 The Pigeonhole Principle
Ex. 5.43 Let Z where | |= 37. Then contains twoelements that have the same remainder upon division by 36. 36 pigeonholes (r) and 37 pigeons ( )
Ex. 5.44 Prove that if 101 integers are selected from the set= {1,2,3, ,200}, then there are two integers such that one
divides the other.
Proof: For each , we may write = , with 0, and{1,3,5, ,199}. Since | |= 100 and 101 integers are
selected, there are two distinct integers of the form
= Either | or | .
+S S S
n q r rS
S
x S x y ky T T
a y b y a b b a
k
m n
,
,
, .
36 0 36
2
2 2
35
5.5 The Pigeonhole Principle
Ex. 5.44 Any subset of size six from the set S={1,2,3,...,9} mustcontain two elements whose sum is 10. pigeonholes: {1,9},{2,8},{3,7},{4,6},(5} pigeons: six of themTherefore, at two elements must be from the same subset.
Ex. 5.45 Triangle ACE is equilateral with AC=1. If five points areselected from the interior of the triangle, there are at least two whosedistance apart is less than 1/2.
region 1 region 2 region 3 region 4 4 pigeonholes
5 pigeons
36
5.5 The Pigeonhole Principle
Ex. 5.46 Let S be a set of six positive integers whose maximumis at most 14. Show that the sums of the elements in all thenonempty subsets of S cannot all be distinct.
For any nonempty subset A of S, the sum of the elements in A,denoted SA, satifies 1 9 10 14 69 SA , and there are
26-1=63 nonempty subsets of S. (two many pigeonholes!)
Consider the subset of less than 6 elements.pigeonholes=10+11+...+14=60pigeons=26-1-1=62
37
5.5 The Pigeonhole Principle
.m.)mm
).(m)qq(q,q
,rmqrmq
mts
ts.
,,,,m
.mn
therem
sts
stsst
ts
ts
m
m
n
12| Hence 12,gcd( odd, is
Since 12222 So N.for
12 and 12 Hence m.by division upon
remainder same thehave 12 and 12 where1,+<1
withZ, exists thereprinciple, pigeonhole By the 12
121212 integers positive 1+ heConsider t :Proof
12 divides such that integer
positive a exists that Prove odd. m with ZLet 5.47 Ex.
-
1221
21
+1
21
+
38
5.5 The Pigeonhole Principle
Ex. 5.48 28 days to play at most 40 sets of tennis and at least 1 play per day. Prove there is a consecutive span of days duringwhich exactly 15 sets are played.
For 1 28, let be the total number of sets played from the start to the end of ith day. Then 1
Of the 56 integers, since their maximum is 55, two of them must bethe same. Hence there exist 1 < 28 with From day + 1 to the end of day , exactly 15 sets are played.
i xx x x
x x x
j i x xj i
i
i j
1 2 281 2 28
4015 15 15 40 15 55
15
,.
.
39
5.6 Function Composition and Inverse Functions
Inverse of addition: and -Inverse of multiplication: and 1 /
u uu u
Def 5.15 If f:A B, then f is said to be bijective, or to be aone-to-one correspondence, if f is both one-to-one and onto.
1234
wxyz
A B
must be |A|=|B|(if )but could be
Ex. 5.50
A B B A or
40
5.6 Function Composition and Inverse Functions
Def 5.16 The function 1 defined by for all , is called the identity function for .
Def. 5.17 If , : , we say that and are equal and write= , if ( ) = ( ) for all .
Ex. 5.51 Let :Z Z, :Z Q where ( ) = = ( ) Z.Yet, ! is a one - to - one correspondence, whereas is one - to - one but not onto.
Ex. 5.52 , :R Z defined as follows:
( ) =,
- ( ) =
A AA A a aa A A
f g A B f gf g f a g a a A
f g f x x g x xf g f g
f g
f xx if x Z
x if x R Zg x x
: , ( )
,
1
1, for all R, then = .x f g
41
5.6 Function Composition and Inverse Functions
Def. 5.18 If : and : , we define the compositefunction, which is denoted : , by ( )( ) = ( ( )),for each .
f A B g B Cg f A C g f a g f a
a A
Ex. 5.53
1234
AB
C
abc
wxyz
fg
( )( ) ( ( )) ( ) , ( ) , ( ) ,g f g f g a x gf x gf y 1 1 2 3 gf(4) = z
42
5.6 Function Composition and Inverse Functions
Ex. 5.54 Let , :R R be defined by ( ) =
Then ( ) = ( , whereas ( ) = ( + ) =
( + ) Therefore, the composition is not commutative.
Theorem 5.5 Let : and :(a) If , are one - to - one, then is one to one.(b) If , are onto, then is onto.
f g f x x g x x
g f x g x x f g x f x
x
f A B g B Cf g g ff g g f
2
2 2
2
5
5 5
5
, ( ) .
)
.
A B Cf g
43
5.6 Function Composition and Inverse Functions
Ex. 5.55 Let , , :R R where ( ) =
Then (( ) )( ) = (
Theorem 5.6 If : , : , : , then ( ) =( ).
f g h f x x g x x h x
x h g f x hg x h x
x h g f x h g f x
f A B g B C h C D h g fh g f
2
2 2 2
2 2
5
2 5
5 2
, ( ) , ( )
. ) ( )
( ) ( ( ))( ) ( ( ( )))
A B Cf g
Dh
gf
hg
h(gf)
(hg)f
44
5.6 Function Composition and Inverse Functions
Def. 5.19 If : , we define and for Z
Ex. 5.56 = {1,2,3,4}, : defined by = {(1,2), (2,2),
(3,1), (4,3)}. Then
Def. 5.21 If : , then is said to be if there is afunction : such that = and =
1 +
.
f A A f f n
f f f
A f A A f
f f f
f f n
f A B f invertibleg B A g f f g
n n
n
A B
, ,
( ).
{( , ), ( , ), ( , ), ( , )},
{( , ), ( , ), ( , ), ( , )} ,
1
2
31 2 2 2 3 2 4 1
1 2 2 2 3 2 4 2 3
1 1
A B
f
g
45
5.6 Function Composition and Inverse Functions
Ex. 5.58 , :R R, ( ) = + , ( ) = Then
( ) = ( + ) = and ( ) = (
and are both invertible functions
Theorem 5.7. invertible function of : is unique.Proof: If : is also inverse of , then = and
= Consequently, =
f g f x x g x x
gf x g x x x fg x f x
x x f g
f A Bh B A f h f
f h h h h f g h f gg g
A
B B
A
2 51
25
2 51
22 5 5
1
25
21
25 5
11 1
1
( ).
( ) ( ))
[ ( )] .
. ( ) ( )
.
(
Since it is unique, we use to represent the inverse of . )-1f f
46
5.6 Function Composition and Inverse Functions
Theorem 5.8 A function : is invertible if and only ifit is one - to - one and onto.
Theorem 5.9 If : , : are invertible functions, then
: is invertible and (g f)
How to find the inverse of a function?Ex. 5.60 :R R = {( , )| = + }
-1
-1
f A B
f A B g B C
g f A C f g
f x y y mx b
f x y y mx b y x y mx b x y x my b
x y ym
x b f xm
x b
c
1 1
11 1
.
{( , )| } {( , )| } {( , )| }
{( , )| ( )}. ( ) ( )
47
5.6 Function Composition and Inverse Functions
Ex. 5.61 f:R R , f(x) = e f is - - .
f
Theorem 5.11 Let : for finite sets and , where| |=| | . Then the following statements are equivalent:(a) is one - to - one(b) is onto(c) is invertible.
+ x
-1
.
{( , )| } {( , )| } {( , )| }
{( , )| ln }. ( ) ln .
one to one and onto
x y y e y x y e x y x e
x y y x f x x
f A B A BA B
fff
x c x y
1
48
5.7 Computational Complexity
problem algorithm 1 algorithm 2
algorithm k
Which one is best?We need measures.
time-complexity or space-complexitya function f(n) where n is the size of the input
lower bounds, best cases, average cases, worst cases
49
5.7 Computational Complexity
Def. 5.23 Let , :Z R. We say that dominates (or is
dominated by ) if there exit constants R and Z
such that | ( )| | ( )| for all Z where .(write as ( ), order or big - Oh of )
+
+ +
+
f g g f f
g m k
f n m g n n n kf O g g g
,
Big-Oh Form NameO(1) constantO(log2n) LogarithmicO(n) LinearO(nlog2n) nlog2nO(n2) QuadraticO(n3) CubicO(nm),m=0,1,2,3,... PolynomialO(cn),c>1 ExponentialO(n!) Factorial
50
5.7 Computational Complexity
18 1019. microseconds 2.14 10 days 5845 centuries8
Order of Complexity
problem size n log2n n nlog2n n2 2n n!
2
16
64
1 2 2 4 4 2
4 16 64 256 6.5 104 2.1 1013
6 64 384 4096 1.84 1019 >1089
51
Summaries (m objects, n containers)
Objects Containers Some Number Ar Are Containers ofDistinct Distinct May Be Empty Distributions Yes Yes Yes Yes Yes No Yes No Yes Yes No No No Yes Yes No Yes No
nn S m nS m S m S m nS m nn m
mn m n
m n
m
m n
m
n
m
! ( , )( , ) ( , ) ( , )( , )
( )
1 2
1
1 1 1
1
Put one object in each container first.
52
Exercise. P252: 7,12 P258: 6,14,20,27 P256: 7,9,10,12 P272: 5,6,8,14 P277: 6,7,10,13,20 P305: 12,25, 27