Chapter 1: combinatorial analysis
elements
" "" " "
in order / ordered
arrangements ( basically order of elements matter )
,
0 e.g. ABC ≠ BAC ↓
considered
n
distinguishable objects
as diff
.
sets
Total
☐
same elements in diff
sets are the same
same
"
connects to ABC = BAC =
BCA
the Binomial
theorem > distinct objects
• Proof of the basic principle of counting ,
.
m outcomes
.
µ
2 experiments are
performed Together
•
: mxn
→
•
Experiment 1 has M possible outcomes % i
2)
,
n outcomes
possible
3
Experiment 2 has n outcomes
•
a
• If
exp .
I can result in
any 1 Of m
possible outcomes for each possible outcome of
exp . I and there are n
possible outcomes, then together there are mn possible outcomes of the 2 experiments
, ( 1,1 ) ( 1,2 ) . . .
( 1 n )
, , ,
( 2 , 1) 12 2) . .
(2 n )
,
.
, , ,
:
Cm
,
1) ,
( M 2)
, ,
- . .
(M
,
N ) ] Total : mxn
i. set of possible outcomes consists of m rows
, containing n elements
•
Permutations :
Number of permutations of n distinct objects :
n ( n -
1) ( n -
2) . . .
2- 1 .
= n !
With different groups of groups of arrangements / permutations :
If the groups can be switched around : XM !
,
where m is the no .
Of places
4 !
e. g. 4-1 . .
3-1 .
✗ 2! .
1 ! ✗
Number of of objects where identical
permutations n
,
n, ,
nz are :
formula of
If we
just use normal
,
we are
going to
get repetitions arrangements .
e.
g. BOB
: we need the number of distinguishable / different permutations :
Pn .
Pn
n.im?!...nr,. number of permutations of n
objects where hi nz nr are identical
=
.
. .
, ,
e. g. • PEPPER problem
• 10
competitors where
just nationalities are listed
•
Different arrangements of objects of identical colours
•
Handshakes
•
Grid problem
6!
• A BEFORE B with A B C D E F }
2 !
^
A B
, combinations
choosing groups ( combinations ) Of size r
,
from a collection of n objects Groups : not
ordered !
-
not ordered : ABC =
BAC
;
because groups cannot be counted as
permutations
Number of objects from collection of n objects where order doesn't matter:
ways to
pick r a
,
n ( n -
1) ( n -
r!
2) . . .
(n -
r -11 ) =
r
n
! (n
!
-
r) !
=
(Y) ) n choose r
examples :
•
choose a Comm of 3 from 20 ¥
•
men and women comm : (E) (F) = total outcomes
> 2 men refuse to serve together (E) (( E ) ✗
-
-
(mm : )(I÷mf ) )
" "
normal where
men group
group
2
feuding men
serve together
Linear configurations
of which defective
Example of n anntenas
,
m are .
How many linear configurations (
permutations) where no 2 defects are next to each other ?
,
① up then Basically find Of
Line
working antennae : n m no
-
.
,
ni na
ni
defective antennas
}
places for
in
^
T
^ th tht out )
( throw working ones
spaces for
② defective
There are In -
m -11 )
possible m antennae .
③ of these
At most 1 defective anntenae can
go in each spaces
( )
n - m + '
④
NB
SO we choose M leg .
2) of the ( n
-
m -11 ) ( e. g. s )
spaces : : .
m possibilities
e. g. (E)
Of to
no . ways
< ↳ from 5
this is the no .
Of possible choose 2 places
orderings where there is at least
I
working antenna between 2 defective ones
