Lecture #3: Algorithmic Combinatorics I "#FOSCS301"

Computer Science Division: CS301

Course CS301, Part:“ALGORITHMIC COMBINATORICS I”

( 3rd Year Students )

Prepared by:

http://www.researchgate.net/profile/Ahmed_Abdel-Fattah

Lecture #3N.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 di!erent 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 fromn (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 di!erent domino faces are possible?

We use the theorem to find the number of ways to construct an unorderedpair of 2 numbers (with doubles allowed) between 0 and 6. This means:

1 We need to select r = 2 numbers,

2 With possible repetition, and3 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. 12pieces) chosen from 5 di!erent varieties with the requirement that atleast 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) =

!

117

"

= 330



Example: Constrained Collections

How many ways are there to pick a collection of exactly 10 balls from apile 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! 15

"

= 21 ways

2 Count the complementary set:!

10 + 3! 110

"

= 66 ways to pick 10 balls without restriction

!

4 + 3! 14

"

= 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

If repetition is allowed, " there are

&r + n ! 1

r

'

distributions



Example: Diplomats

1 How many ways are there to assign 100 di!erent diplomats to 5di!erent 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 diplomats2 the n distinct boxes correspond to the 5 continents3 " assignment can be done in n

r = 5100 ways

2 When distributing r di!erent objects into n di!erent 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 redlicorice 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 NoteTheorem (Constrained Collections:a)

a([Tucker, 2007, Example 4, Page 195])

The number of ways to distribute r identical objects into n distinctboxes 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 equationx1 + 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 ofobjects 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 di!erent types2 “distributing” r similar objects into n di!erent 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 di!erentcolored 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 di!erent envelope.)

Think:

Do we need(43

)

,(4+3"1

4

)

, or(3+4"1

3

)

ways?



Homework #3:Shirts & Children

In how many di!erent ways can 3 identical green shirts and 3 identicalred shirts be distributed among 6 children such that each child receives ashirt?

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.


Education

Lecture #3: Algorithmic Combinatorics I "#FOSCS301"