CS301 Algorithmic Combinatorics I Lecture 3

Computer Science Division: CS301
Course CS301, Part:
“ALGORITHMIC COMBINATORICS I”
( 3rd Year Students )
Prepared by:
http://www.researchgate.net/profile/Ahmed_Abdel-Fattah
Lecture #3
N.B. The Mid-Term Exam will be held on 08-NOV-2014
(and it includes this lecture)
Dr. Ahmed Ashry “Algorithmic Combinatorics I” for 3rd Year Students — Lecture #3 1 (of 16)

Selections with REPETITIONS:
When an object can be chosen more than once.
Example: Unordered Pairs of Numbers
A domino is a rectangular piece whose top face is divided into 2 squares.
Each of the squares contains between 0 (blank) and 6 dots.
How many different domino faces are possible?

Theorem (Selections with Unlimited Repetitions:a)
a([Tucker, 2007, Theorem 2, Page 188])
The number of selections with repetition of r objects chosen from
n (distinct) types of objects is C(r + n − 1, r ).
Proof.
1 We use n-1 separators to separate the n types.
2 The r objects to be selected can be represented by r !’s.
n types, n-1 separators ! "# $
#!$!!"
1st
%%%
#!!$!!!"
2nd
%%%
· · ·
%%%
#!$!!!"
nth
3 The number of sequences with r !’s and n-1 separators is
&
r + (n − 1)
r
'

Example: Unordered Pairs of Numbers
How many different domino faces are possible?
We use the theorem to find the number of ways to construct an unordered
pair of 2 numbers (with doubles allowed) between 0 and 6. This means:
1 We need to select r = 2 numbers,
2 With possible repetition, and
3 From n = 7 types of numbers.
" C(r + n − 1, r) = C(2 + 7 − 1, 2) = 8!
2!×6! = 28

Example: Selling Doughnuts
How many ways are there to fill a box with a dozen doughnuts (i.e. 12
pieces) chosen from 5 different varieties with the requirement that at
least one doughnut of each variety is picked?
Procedure:
Pick one doughnut of each variety
(N.B. this gives a total of five doughnuts in 1 way!!!)
The remaining 7 doughnuts are selected with repetition
Therefore, apply the theorem:
n = 5 (types of doughnuts)
r = 7 (objects needed to be selected)
" C(r + n − 1,
r) =
!
11
7
"
= 330

Example: Constrained Collections
How many ways are there to pick a collection of exactly 10 balls from a
pile of red balls, blue balls, and green balls if there must be:
1 at least 5 red balls?
2 at most 5 green balls?
Solutions:
1 Simply:
Pick 5 red balls (in one way)
Pick the remaining 5 balls arbitrarily in
!
5 +3 − 1
5
"
= 21 ways
2 Count !the complementary set:
10 + 3 − 1
10
"
= 66 ways to pick 10 balls without restriction
!
4 +3 − 1
4
"
= 15 ways to pick a collection with " 6 green balls
Answer: 66 − 15 = 51 ways (to get 10 balls with # 5 green balls)

DISTRIBUTIONS:
The “Distribution” Process
Assigning (distinct/identical) objects to (distinct/identical) cells:
(ordered/unordered) collections in containers?
other constraints?
Generally equivalent to selection or arrangement (with repetition).
Usually broken into subcases.
distribution of distinct objects corresponds to arrangements.
distribution of identical objects corresponds to selections.

Distinct Objects " Distinct Containers
We have r distinct objects and n distinct cells
" Each object can go in n places
" Product rule: #n × n ×$!· · · × n"
r−times
= nr distributions
Identical Objects " Distinct Containers
We have r identical objects and n distinct cells
" We choose r cells from the n in C(&
n, r ) ways
r + n − 1
If repetition is allowed, " there are
r
'
distributions

Example: Diplomats
1 How many ways are there to assign 100 different diplomats to 5
different continents?
2 How many ways if 20 diplomats must be assigned to each continent?
Solution:
1 The model for distributions tells us that:
1 the r distinct objects correspond to the 100 diplomats
2 the n distinct boxes correspond to the 5 continents
3 " assignment can be done in nr = 5100 ways
2 When distributing r different objects into n different boxes:
1 if ri objects must go in box i, for 1 # i # n then
there are P(r ; r1, . . . ,
rn) =
r !
r1! × . . . × rn!
" P(r ; r1, . . . ,
rn) = P(100; 20, 20, 20, 20, 20) =
100!
(20!)5

Example: Candy Shop
How many ways are there to distribute 20 (identical) sticks of red
licorice and 15 (identical) sticks of black licorice among 5 children?
1 Distribute the 20 red sticks in C(20 + 5 − 1, 20) = 10, 626 ways
2 Distribute the 15 black sticks in C(15 + 5 − 1, 15) = 3, 876 ways.
3 The product gives the number of ways to distribute all the candy.

Constrained Collections: A General Note
Theorem (Constrained Collections:a)
a([Tucker, 2007, Example 4, Page 195])
The number of ways to distribute r identical objects into n distinct
boxes with at least one object in each box is C(r − 1, n − 1).
Proof.
1 Put 1 object in each of the n boxes.
2 Distribute the r − n (identical) objects in n (distinct) boxes in &
(r − n) + n − 1
(r − n)
'
=
&
r − 1
(r − 1) − (r − n)
'
=
&
r − 1
n − 1
'
ways.

Application: Solving (Integer) Equations
1 How many integer solutions are there to the equation
x1 + x2 + x3 + x4 = 12, with xi $ 0.
2 How many solutions with xi > 0?
3 How many solutions with x1 $ 2, x2 $ 2, x3 $ 4, x4 $ 0?
Integer solution means (ordered) integer values for the x’s.
This kind of problems is modeled as “selection with repetition”:
each xi represents the number of (identical) 1’s in box i
(or represents the number of chosen objects of type i)
1 Here, we have r = 12 ones, and n = 4 boxes.
" The number of solutions is C(12 + 4 − 1, 12) =
&
15
12
'
= 455.
2 Here, we first put an object in each box =% r= 12 −4 = 8, n = 4.
" The number of solutions is C(8 + 4 − 1, 8) = C(11, 3) = 165.
3 Here, we first put objects in box i (as constrained):
i.e. Put 2 in x1, 2 in x2, 4 in x3
" The number of solutions is C((12 − 2 − 2 − 4) + 4 − 1, 6) = 35.

Take Home Messages:
1 Always keep in mind “WHY” the number of selections with
repetition (of r objects chosen from n (distinct) types of
objects is
r+n−1
r
)
.
2 Distributions (Part I: distinct cells):
Case/Distribution of: distinct objects identical objects
“no repetition” P(n, r ) (C(n, r )
“unlimited repetition” nr
r+n−1
r
)
“restricted repetition” P(n; r1, . . . , rt ) —
3 Remember 3 equivalent forms for selection with repetition:
1 “selecting” r objects with repetition from n different types
2 “distributing” r similar objects into n different boxes
3 “solving”
#n
i=1
xi = r (nonnegative integers)

Homework #3:
Letters & Envelopes
We have 3 identical letters, and we want to placea them into 4 different
colored envelopes: yellow, blue, red, and green.
How many ways can the 3 letters be placed into the 4 envelopes?
a(N.B. It is only possible to introduce one letter into each different envelope.)
Think:
Do we need
(4
3
)
,
(4+3−1
4
)
, or
(3+4−1
3
)
ways?

Homework #3:
Shirts & Children
In how many different ways can 3 identical green shirts and 3 identical
red shirts be distributed among 6 children such that each child receives a
shirt?
Think:
Are we “selecting” 3 children in C(6, 3) = 6!
3! 3! = 20 ways?
Are we “arranging” 6 shirts in P(6; 3, 3) = 6!
3! 3! = 20 ways?
Why/Why not?

References:
Tucker, A. (2007).
Applied combinatorics.
Wiley, 5th edition.

CS301 Algorithmic Combinatorics I Lecture 3

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