COMP239: Mathematics for Computer Science...

COMP239: Mathematics for Computer Science II

Prof. Chadi [email protected]

EV7.635

Generalized Permutations and Combinations

Permutations with RepetitionExample:

For {1, 2, 3, 4}, (2,2,1) and (2,2,3) are permutations with repetition allowed

Example: How many strings of length r can be formed from the English alphabet?

Theorem: The number of r-permutations of a set of n objects with repetition allowed is nr.

Combinations with Repetition

Example: How many ways are there to select 5 bills from a box containing 1$, 2$, 5$, 10$, 20$, 50$, and 100$ bills? Assuming that the order in which the bills are chosen does not matter and there are at least 5 bills of each type?

Combinations with Repetition

$100 $50 $20 $10 $5 $2 $1

Cash box with seven types of bills

• Each compartment in the cash box is holing one type of bill

• Compartments are separated by dividers• The choice of 5 bills corresponds to placing 5 markers in

the compartments

Combinations with Repetition$100 $50 $20 $10 $5 $2 $1

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Selecting 5 bills corresponds to the number of ways to arrange six bars and 5 stars, which is indeed the number of ways to select the positions of (unordered) 5 stars from 11 possible positions = C(11,5)

Combinations with RepetitionTheorem:

There are C(n + r − 1, r) r-combinations from a set with n elements when repetition is allowed.

Proof Similar to before, use n-1 bars (to represent ncells) and r stars. Selecting the positions of rstars out of n – 1 + r positions equals to C(n + r − 1, r)

Combinations with RepetitionExample:

How many solutions does the equation: x1 + x2 + x3 + x4 = 11 have, where x1, x2, x3, x4 are nonnegative integers?

Solution: A solution corresponds to selecting 11 items from a set with 4 elements (x1, x2, x3, x4) so that x1 items of type 1, x2 items of type 2, x3 items of type 3, and x4 items of type 4 chosen; total number of solutions is number of 11-combinations, repetition allowed, from a set with 4 elements = C(4 + 11 − 1, 11)

Combinations with RepetitionExample:

How many solutions does the equation: x1 + x2 + x3 + x4 = 11 have, where x1 ≥ 1, x2 ≥ 1, x3 ≥ 2, x4 ≥ 4?

Solution: A solution corresponds to selecting 11 items so that x1 items of type 1, x2 items of type 2, x3 items of type 3, and x4 items of type 4 chosen;In addition, there is at least: 1 item of type 1, 1 item of type 2, 2 items of type 3, and 4 items of type 4.Then select 3 (unordered) additional items from the set of 4 elements; C(4 + 3 − 1, 3)

Summary

C(n+r-1,r)Yesr-combinations

nrYesr-permutations

C(n,r)Nor-combinations

P(n,r)Nor-permutations

FormulaRepetition Allowed?

Type

Permutations with Indistinguishable Objects

Example: How many different strings can be made by reordering the letters of the word SUCCESS?

Theorem: The number of different permutations of nobjects, where there are n1 indistinguishable objects of type 1, n2 indistinguishable objects of type 2, ..... , and nk indistinguishable objects of type k, is:

!!....!!

21 knnnn

Distributing Objects into Boxes

Distinguishable objects and distinguishable boxes

Example: How many ways are there to distribute 5 cards to each of four players from a deck of 52 cards?

Solution: in class


Theorem:The number of ways to distribute ndistinguishable objects into k distinguishable boxes so that ni objects are placed into box i, i = 1, 2, ..., k equals:

Proof (in class)

!!....!!

21 knnnn


Indistinguishable objects and distinguishable boxes

Example: How many ways are there to place 10 indistinguishable balls into eight distinguishable bins?

Solution: in class

Advanced Counting Techniques

Recurrence Relations

Some problems cannot be solved using counting techniques we explained earlier.

Example: How many bit strings of length n do not contain two consecutive 0?

Such problems can be solved by finding relationships, called recurrence relations.


Example: The number of bacteria in a colony doubles every hour. If a colony begins with 5 bacteria, how many will be present in n hours?

Let an be the number of bacteria at the end of n hours an = 2an-1 (n>0). Knowing the initial condition (a0 = 5), we can solve for an .


Definition: A recurrence relation for the sequence {an} is an equation that expresses an in terms of one or more of the previous terms of the sequence.

A sequence is called a solution of the recurrence relation if its terms satisfy the recurrence relation


Example: (Compound Interest at 8%) Let Pn denote the amount in a bank account after n years. Recurrence relation:Pn = Pn-1 + 0.08Pn-1 = (1.08) Pn-1

If the initial condition is P0, thenPn = (1.08)n P0

If P0 =$10,000, then after 30 years, the account contains $100,626.56


Example: Fibonacci Numbers A young pair of rabbits (one of each sex) is placed on an island. A pair of rabbits does not breed until they are two months old. Thereafter, each pair produces another pair each month. Find a recurrence relation for the number of pairs of rabbits on the island after n months (assume no rabbits ever die).

Recurrence RelationsExample: Towers of Hanoi

Move disks to peg 2; never place a larger disk on top of a smaller one. Set up a recurrence relation for the number of moves needed for a tower with n disks.

Peg 1 Peg 2 Peg 3

Recurrence RelationsLet Hn be the number of moves needed.

Transfer n - 1 disks to peg3 (using Hn-1 moves); move the largest disk to peg 2; then transfer the n - 1 disks from Peg3 to Peg 2 (using Hn-1 moves).

Peg 1 Peg 2 Peg 3

Recurrence RelationsHn = Hn-1 + 1 + Hn-1 = 2 Hn-1 +1

= 2 (2 Hn-2 +1) + 1 = 22 Hn-2 + 2 + 1= 22 (2 Hn-3 +1) + 2 + 1 = 23 Hn-3 + 22 + 2 + 1...= 2n-1 H1 + 2n-2 + 2n-3 + ….+ 2 + 1 (Note, H1 = 1)= 2n-1 + 2n-2 + 2n-3 + ….+ 21 + 20

= 2n - 1


Example:How many bit strings of length n do not contain 2 consecutive 0’s?

Solution: (assume n > 2)Denote an the number of such strings.an = the number of such bits strings ending

with 0 + the number of such bit strings ending with 1.

Recurrence RelationsThe number of bits strings of length n, ending with 1 and do not have two consecutive 0s is the number of bit strings of length n - 1 with no two consecutive 0s and a 1 added at the end

hence, there is an-1 such bit strings

The number of bits strings of length n, ending with 0 and do not have two consecutive 0s must have 1 as their (n - 1)th bit these bit strings are equivalent to the bit strings of length n - 2 with no two consecutive 0s and 10 added at the end

hence, there is an-2 such bit strings


Hence, we have the following recurrence relation: an = an-1 + an-2 (n > 2)with initial conditions: a1 = 2 and a2 = 3

Find the number of bit strings of length 5 that do not contain two consecutive 0sa3 = a2 + a1 = 3 + 2 = 5a4 = a3 + a2 = 5 + 3 = 8a5 = a4 + a3 = 8 + 5 = 13


Example:How many strings of n decimal digits (0-9) contain an even number of 0’s?

Solution:Let an be the number of such valid strings:a1 = 9

There are two ways to construct valid strings of length n from strings of length n-1.

Recurrence RelationsThere are two ways to construct valid strings of length n from strings of length n-1.1)- Start from a valid string of length n-1 and append a

digit other than 0 (to obtain a valid string of length n). Hence a valid string can be formed in this manner in 9an-1 ways.

2)- Start from a non-valid string (a string with odd number of 0s) of length n-1 and append a 0 (hence you obtain a string of length n with even number of 0s). In how many ways can this be done?


In how many ways can this (case 2) be done?the same as the number of invalid strings

of length n –1.The number of strings of length n –1 is:

10n-1

and the number of valid strings of length n –1 is an-1

Hence, the number of invalid strings of length n –1 is 10n-1 – an-1

an = 9an-1 + 10n-1 – an-1 = 8an-1 + 10n-1

COMP239: Mathematics for Computer Science...

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