2 nu e Nnn 2 no Y e 1 Combinatorial Theorem xn y 7 Mlb Inn yn - - PDF document
CS 70 July 15 2020 Counting 2 1 Combinatorial Theorem and Exclusion Simple Inclusion 2 and Exclusion Inclusion 3 4 Derangement's Sampling 5 and 6 Star Bars e nu 2 nm no t T XZ 2mg The number of subsets LHS 2 X 2 as of Is g
Counting 2
CS 70 July 15 2020
1 Combinatorial
Theorem
2
Simple
Inclusion
and Exclusion
3
Inclusion
and Exclusion
4
Derangement's
5
Sampling
6 Star
and
Bars
2
nm
e nu
t
no
T
LHS
2
2 X
XZ
2mg The number of subsets
as
Is
g n
RHS
subsets iz
y
I
2
n
b
F
e
nu
2
no
Y
e
Nnn
1 Combinatorial Theorem
Hey
Mlb
xn y 7
a
Inn yn
How
to
distribute
my ball
among I red bins and y blue bins
LHS
City acxey
X
XLxty
w
make subsequent choices
Hey'T possibilities
I20 h
RHS
inred.aerni
i
r.edeuef
fin
u i i
Indy xn
X
no
x
e 7 x
y
e
h yn
City
Eno
2n
2 Simple Inclusion 1 Enclusion
sum
Rule
for disjoint sets A and B
1AABled
to count the number of elements of A V B
we have
IA UBI
Alt 1B I
when A and
B
have
common element
Inclusion Enclusion Rude e
k
1AvB
IAABI
A AB
A um
i'It'mII
in
IAnBl
are
counted twice
subtract IA ABI
Example
Howmany lb digit Phone numbers have
5 as their
first
digit
A numbers with
5
as
first digit
stood
as
9
B numbers
with
5
as
second digit 1131 10
g
lo
A
5
B
5
AAB
numbers
with 5
as
first and second digit
IAAB1
108
IAU Bls IAI11131 IA
Is 18 18
108
Three way inclusion Exclusion
Rule
Set
It
Beg
LAU BU GI
IAI 1431 14
IAABI
land
HAD
1 Brett
Ian Brel
3 Inclusion
Exclusion principle
sets
Ai
Ah
L yn
E
IAah Ain l
2Hiv to
n
4 Derangement
see
Permutations of l
n
n
How many permutations where
no item in its proper
place or
fixed Points
Derangement
Example
number of derangement 123
Derangement
Tooo
231
Yes
we can count the complement.ro count permutations
with
at least
fined Points
Permutation
where
I is
fined Points
2slg2,3
Complement
Permutations
with
at
least
fined Points
1A VA zu Ast
IA UAzUA3l
IA il 11A de IA31 IA in Az I
1AM
Azl
IANAste IAA AznAg
2
2
t
2
I
I
I
I
4
Subtract this from the total Permutations
Derangement
3
4 2
For
n items
h
permutations
Permutations with at least one fined point
YELII Efi of Aintidt
him
nan I
7 Ch y
th cn u
GyntInn
h
7 Cn D
t 4 In 21
e C un
nu
m
h
I
1
eosin
e
un
n xd
h x Eot.fi
n xte
W
n
a
Roughly
0 37
the Permutations
are
5 Sampling
derangements
Assume sample K items
n
as
Without replacement
nx Ch y x n 2
Ah
Kel
ht
a
Ch K
Second rule
divid by number of orders
k
nd
h K
K
with replacement
K
h X h X
x n
n
can
we
use
second rule
doesn't
a
Problem
depends
each
item
we
choose
For
chosen string
ABCD 4
a n n
4z
Different number of
map to
each unordered Anotherexample
How many ways
can Alice and Bob
split
5 For each
Pick Alice
25 and divid out order
A 5 B D CA A A A A
i
1
I
A 4 B
1B Aga A A B gLAgAgAyBgA
I
5
A
I 13 45
Ag BoB B B
g
5
A O
B S B B B B B
t
second
rule
counting
is
no
good
here
Anotherexample How ma
ways
can Alice Bob and Eve split
5
Idea
separat
Alice's dollars
from Bob's
and
then Bob's from Eve's Assume dollars
are
5
stars.o
see
manddbajars
split Alice D
Bob 3 Eve 3
Zeroth Rule Counting
If
there
is
a
to
mapping
between
two
set they have
the
Same
size
So
we
can
ask
How
many different
sequence
5
stars
and
2
bars
are
there µ
00000
A
A
a
a
a
x
a
e
Ig
Positio
7 positions
in which
to
Place
2
bars
7
7 choose
2 e
z
way
to do this
ways
to
split
5 among 3 People
6 Star and Bars
ways
to
split
k dollarsamong n people
K
from
n
with
replacement
where
correspondence n
1
bars
to
split
the K
stars
I
I
I
I
nyk
1
Positions
from which
to
choose
n 1
bar
positions