Chapter 6 Counting...

Chapter 6 Counting Techniques

§ 1 The Multiplication Principle

The Multiplication Principle

ExampleA new company with just two employees, Sandoval and Ramirez,rents a floor of a building with 12 offices. How many ways are thereto assign different offices to the two employees?

What we have to think about is how many ways we can assign officesto each and then how to combine these numbers.

In how many ways can we select Sandoval’s office? 12In how many ways can we select Ramirez’ office? 11

How do we combine these?

12 · 11 = 132.

In how many ways can we select Sandoval’s office?

12In how many ways can we select Ramirez’ office? 11

12 · 11 = 132.

In how many ways can we select Sandoval’s office? 12

In how many ways can we select Ramirez’ office? 11

12 · 11 = 132.

In how many ways can we select Sandoval’s office? 12In how many ways can we select Ramirez’ office?

12 · 11 = 132.

We just applied ...

The Fundamental Counting PrincipleIf there are m possibilities for the first choice, n possibilities for thesecond choice, p possibilities for the third choice, and so on, then thetotal number of possible outcomes by selecting one from each groupis given by m× n× p and so on.

Note: this generalizes for a procedure with k tasks.

To model these, we could use tree diagrams without the probabilitieson the branches.

We just applied ...

Examples

ExampleThe chairs of an auditorium are to be labeled with a letter and apositive number not exceeding 100. What is the largest number ofchairs that can be differently labeled?

Solution: 26 · 100 = 2600

ExampleThere are 32 cluster machines in a computer center, each comprisedof 24 machines. How many machines are there in total?

Solution: 32 · 24 = 768

Examples

Solution: 26 · 100 = 2600

Solution: 32 · 24 = 768

Examples

Solution: 26 · 100 = 2600

Solution: 32 · 24 = 768

Examples

Solution: 26 · 100 = 2600

Solution: 32 · 24 = 768

Examples

ExampleChalk comes in three different lengths, eight different colors and fourdifferent diameters. How many different kind of chalk are there?

Solution: 3 · 8 · 4 = 96

ExampleHow many ways are there to make a sandwich if we have 3 choicesfor meat, 4 choices of bread and 6 choices of condiments if there areno undesirable combinations and we can only use one from eachcategory?

Solution: 3 · 4 · 6 = 72

Examples

Solution: 3 · 8 · 4 = 96

Solution: 3 · 4 · 6 = 72

Examples

Solution: 3 · 8 · 4 = 96

Solution: 3 · 4 · 6 = 72

Examples

Solution: 3 · 8 · 4 = 96

Solution: 3 · 4 · 6 = 72

Examples

ExampleHow many two-digit numbers have unique and non-zero digits?

What restrictions do we need to consider?

Solution: 9× 8 = 72

Examples

More Complicated Examples

ExampleIn a version of BASIC, the name of a variable is a string of one or twoalphanumeric characters, where the variables are not case sensitive.Moreover, the variable name must begin with a letter and must bedifferent from the 5 strings of two characters that are reserved forprogramming use. How many variable names are possible for thisversion of BASIC?

Solution: Let V be the number of different variable names. If we letV1 be the set of one character variable names and V2 be the number oftwo character variable names, then how can we represent the set of allvalid variable names? V = V1

⋃V2.

Solution: Let V be the number of different variable names.

If we letV1 be the set of one character variable names and V2 be the number oftwo character variable names, then how can we represent the set of allvalid variable names? V = V1

⋃V2.

Solution: Let V be the number of different variable names. If we letV1 be the set of one character variable names and V2 be the number oftwo character variable names, then how can we represent the set of allvalid variable names?

V = V1⋃

Solution: Let V be the number of different variable names. If we letV1 be the set of one character variable names and V2 be the number oftwo character variable names, then how can we represent the set of allvalid variable names? V = V1

⋃V2.

So we need to find the cardinality of the two subsets.

|V1| =? 26 because the first character must be a letter.

|V2| =? 26 · 36 = 936 because the second character can be a letter ora digit.

Is this all we need?

There are 5 two character strings that are not allowed, so thereforeV2 = 936− 5 = 931.

Putting this together, we have |V| = 26 + 931 = 957 differentvariables.

|V1| =?

26 because the first character must be a letter.

|V2| =?

26 · 36 = 936 because the second character can be a letter ora digit.

|V2| =? 26 · 36 = 936 because

the second character can be a letter ora digit.

Putting this together, we have |V| =

26 + 931 = 957 differentvariables.

ExampleEach user on the network has a password, which is 6 to 8 characterslong, where each character is an uppercase letter or a digit. Eachpassword must contain at least one digit. How many passwords arepossible?

Solution: If P is the number of passwords, let Pk be the number ofpasswords of length k. How many lengths are possible for passwords?So, we have

P = P6 + P7 + P8

How do we find P6? It is difficult to do directly ...

How many total strings are possible with no restrictions? 366

How many strings of length 6 are invalid? 266

Solution: If P is the number of passwords, let Pk be the number ofpasswords of length k. How many lengths are possible for passwords?

So, we haveP = P6 + P7 + P8

P = P6 + P7 + P8

How do we find P6?

It is difficult to do directly ...

P = P6 + P7 + P8

How many total strings are possible with no restrictions?

P = P6 + P7 + P8

How many strings of length 6 are invalid?

P = P6 + P7 + P8

So, what is P6?

366 − 266 = 1, 867, 866, 560.

And P7? 367 − 267 = 70, 332, 353, 920.

Finally, P8 = 368 − 268 = 2, 612, 282, 842, 880

Putting this all together, P = P6 + P7 + P8 = 2, 684, 483, 063, 360

So, what is P6? 366 − 266 = 1, 867, 866, 560.

And P7? 367 − 267 = 70, 332, 353, 920.

Finally, P8 = 368 − 268 = 2, 612, 282, 842, 880

So, what is P6? 366 − 266 = 1, 867, 866, 560.

And P7?

367 − 267 = 70, 332, 353, 920.

Finally, P8 = 368 − 268 = 2, 612, 282, 842, 880

So, what is P6? 366 − 266 = 1, 867, 866, 560.

And P7? 367 − 267 = 70, 332, 353, 920.

Finally, P8 = 368 − 268 = 2, 612, 282, 842, 880

So, what is P6? 366 − 266 = 1, 867, 866, 560.

And P7? 367 − 267 = 70, 332, 353, 920.

Finally, P8 =

368 − 268 = 2, 612, 282, 842, 880

So, what is P6? 366 − 266 = 1, 867, 866, 560.

And P7? 367 − 267 = 70, 332, 353, 920.

Finally, P8 = 368 − 268 = 2, 612, 282, 842, 880

So, what is P6? 366 − 266 = 1, 867, 866, 560.

And P7? 367 − 267 = 70, 332, 353, 920.

Finally, P8 = 368 − 268 = 2, 612, 282, 842, 880

§ 2 Permutations

An Example

ExampleIn how many ways can we select three students from a group of fivestudents to stand in line for a picture?

How many ways to select the first student? The second? The third?So how many way are there? 5 · 4 · 3 = 60.

ExampleHow many ways can we arrange all five of the students for picture?

5 · 4 · 3 · 2 · 1 = 5! = 120

An Example

How many ways to select the first student?

The second? The third?So how many way are there? 5 · 4 · 3 = 60.

5 · 4 · 3 · 2 · 1 = 5! = 120

An Example

How many ways to select the first student? The second?

The third?So how many way are there? 5 · 4 · 3 = 60.

5 · 4 · 3 · 2 · 1 = 5! = 120

An Example

How many ways to select the first student? The second? The third?

So how many way are there? 5 · 4 · 3 = 60.

5 · 4 · 3 · 2 · 1 = 5! = 120

An Example

How many ways to select the first student? The second? The third?So how many way are there?

5 · 4 · 3 = 60.

5 · 4 · 3 · 2 · 1 = 5! = 120

An Example

5 · 4 · 3 · 2 · 1 = 5! = 120

An Example

5 · 4 · 3 · 2 · 1 = 5! = 120

An Example

5 · 4 · 3 · 2 · 1 = 5! = 120

Permutations

TheoremIf n is a positive integer and r is an integer with 1 ≤ r ≤ n, then thereare

P(n, r) =n Pr = n(n− 1)(n− 2) . . . (n− r + 1)

r-permutations of a set with n distinct elements.

Corollary

If n and r are integers with 0 ≤ r ≤ n, then nPr =n!

(n−r)!

Note: By convention, 0! = 1

Permutations

P(n, r) =n Pr = n(n− 1)(n− 2) . . . (n− r + 1)

Corollary

(n−r)!

Permutations

P(n, r) =n Pr = n(n− 1)(n− 2) . . . (n− r + 1)

Corollary

(n−r)!

Back to Examples

ExampleHow many ways are there to choose the first prize winner, secondprize winner and third prize winner from a contest with 100 differentpeople?

P(100, 3) =100!

(100− 3)!=

100!97!

= 100 · 99 · 98 = 970, 200

ExampleSuppose a saleswoman has to visit 8 different cities. She must beginher trip in a specified city, but then she can visit the other seven citiesin whatever order she wishes. In how many different orders can shevisit these cities?

7! = 5040

Back to Examples

P(100, 3) =100!

(100− 3)!=

100!97!

= 100 · 99 · 98 = 970, 200

7! = 5040

Back to Examples

P(100, 3) =100!

(100− 3)!=

100!97!

= 100 · 99 · 98 = 970, 200

7! = 5040

Back to Examples

P(100, 3) =100!

(100− 3)!=

100!97!

= 100 · 99 · 98 = 970, 200

7! = 5040

A Harder Example

ExampleTen job applicants have been invited for interviews, five having beentold to come in the morning and five being told to come in theafternoon. In how many ways can the interviews be scheduled? Is thisthe same number as if all ten interviews were in the morning?

If all were in the morning, then there would be 10! = 3628800 waysto schedule the interviews.

What must we consider if we are looking at the other scenario?People were already told which block to appear at.

We have 5! · 5! = 14400 ways.

What if we didn’t preselect who was interviewing when?

A Harder Example

If all were in the morning, then there would be

10! = 3628800 waysto schedule the interviews.

We have 5! · 5! = 14400 ways.

A Harder Example

We have 5! · 5! = 14400 ways.

A Harder Example

What must we consider if we are looking at the other scenario?

People were already told which block to appear at.

We have 5! · 5! = 14400 ways.

A Harder Example

We have 5! · 5! = 14400 ways.

A Harder Example

We have

5! · 5! = 14400 ways.

A Harder Example

We have 5! · 5! = 14400 ways.

A Harder Example

We have 5! · 5! = 14400 ways.

§ 3 Combinations

A Different Kind of Problem

ExampleHow many different three student committees can be formed from agroup of four students?

What is different? The order in which we select the students does notmatter.

We can select a three student committee by excluding one of thestudents, and there are four ways to do so. Therefore, there are 4committees possible.

What is different?

The order in which we select the students does notmatter.

Combinations

Theorem

The number of r-combinations of a set with n elements, where ∈ Z+

and 0 ≤ r ≤ n, equals

C(n, r) =n Cr =n!

r!(n− r)!

How does this formula relate to that for r-permutations?

Since the difference between permutations and combinations is order,we destroy the order of a permutation by dividing by the number ofways to permute the r selected elements ...

Combinations

Theorem

C(n, r) =n Cr =n!

r!(n− r)!

Combinations

Theorem

C(n, r) =n Cr =n!

r!(n− r)!

Back to Examples

ExampleHow many 5 card poker hands are there if we are using a standarddeck?

52C5 = 52!5!·47! =

52·51·50·49·485·4·3·2·1 = 2, 598, 960

ExampleSuppose we wanted to select 47 cards from that standard deck. In howmany ways can we do this?

52C47 = 52!47!·5!

Why are these answers the same?

Back to Examples

52C5 = 52!5!·47! =

52·51·50·49·485·4·3·2·1 = 2, 598, 960

52C47 = 52!47!·5!

Back to Examples

52C5 = 52!5!·47! =

52·51·50·49·485·4·3·2·1 = 2, 598, 960

52C47 = 52!47!·5!

Back to Examples

52C5 = 52!5!·47! =

52·51·50·49·485·4·3·2·1 = 2, 598, 960

52C47 = 52!47!·5!

Back to Examples

52C5 = 52!5!·47! =

52·51·50·49·485·4·3·2·1 = 2, 598, 960

52C47 = 52!47!·5!

And More Examples

ExampleHow many ways are there to select a 5 student contingent from theschool’s 10 person tennis team to make up the travel team for atournament?

10C5 = 10!5!·5! = 252

ExampleA group of 30 people have been trained as astronauts to go on the firstmission to Mars. How many ways are there to select six people to goon the mission, assuming they can all do all required tasks?

30C6 = 30!6!·24! = 593, 775

And More Examples

10C5 = 10!5!·5! = 252

30C6 = 30!6!·24! = 593, 775

And More Examples

10C5 = 10!5!·5! = 252

30C6 = 30!6!·24! = 593, 775

And More Examples

10C5 = 10!5!·5! = 252

30C6 = 30!6!·24! = 593, 775

§ 4 Mixed Counting Problems

When They Become More Interesting

ExampleThere are 9 faculty members in mathematics and 11 in computerscience. How many ways are there to select a committee to develop astatistics course if the committee must consist of three facultymembers from each department?

Number of ways to select the math faculty 9C3 = 9!3!·6!

Number of ways to select the CS faculty 11C3 = 11!3!·8!

9!3! · 6!

· 11!3! · 8!

= 84 · 165 = 13, 869

Number of ways to select the math faculty

9C3 = 9!3!·6!

9!3! · 6!

· 11!3! · 8!

= 84 · 165 = 13, 869

9!3! · 6!

· 11!3! · 8!

= 84 · 165 = 13, 869

Number of ways to select the CS faculty

11C3 = 11!3!·8!

9!3! · 6!

· 11!3! · 8!

= 84 · 165 = 13, 869

9!3! · 6!

· 11!3! · 8!

= 84 · 165 = 13, 869

9!3! · 6!

· 11!3! · 8!

= 84 · 165 = 13, 869

9!3! · 6!

· 11!3! · 8!

= 84 · 165 = 13, 869

More Examples

ExampleHow many ways are there to form a joint congressional committeewith 3 senators and 5 representatives?

100C3 ·435 C5

ExampleHow many different ‘words’ can we make using the letters of ‘REDSOX’ if we can use at least 3 but no more than 5 letters?

6P3 +6 P4 +6 P5 = 120 + 360 + 720 = 1200

More Examples

100C3 ·435 C5

6P3 +6 P4 +6 P5 = 120 + 360 + 720 = 1200

More Examples

100C3 ·435 C5

6P3 +6 P4 +6 P5 = 120 + 360 + 720 = 1200

More Examples

100C3 ·435 C5

6P3 +6 P4 +6 P5 = 120 + 360 + 720 = 1200

Car Fleet Example

ExampleA fleet is to be chosen from a set of 7 different make foreign cars and4 different make domestic cars. How many ways can we choose afleet (provided we can only have at most one of each make) if

The fleet has five cars, three foreign and two domestic?

7C3 ·4 C2 = 35 · 6 = 210

The fleet can be of any size but must contain the same number offoreign and domestic cars?

7C1 ·4 C1 +7 C2 ·4 C2 +7 C3 ·4 C3 +7 C4 ·4 C4

= 7 · 4 + 21 · 6 + 35 · 4 + 35 · 1= 28 + 126 + 140 + 35

Car Fleet Example

The fleet has five cars, three foreign and two domestic?7C3 ·4 C2 = 35 · 6 = 210

7C1 ·4 C1 +7 C2 ·4 C2 +7 C3 ·4 C3 +7 C4 ·4 C4

= 7 · 4 + 21 · 6 + 35 · 4 + 35 · 1= 28 + 126 + 140 + 35

Car Fleet Example

7C1 ·4 C1 +7 C2 ·4 C2 +7 C3 ·4 C3 +7 C4 ·4 C4

= 7 · 4 + 21 · 6 + 35 · 4 + 35 · 1= 28 + 126 + 140 + 35

Car Fleet Example

7C1 ·4 C1 +7 C2 ·4 C2 +7 C3 ·4 C3 +7 C4 ·4 C4

= 7 · 4 + 21 · 6 + 35 · 4 + 35 · 1= 28 + 126 + 140 + 35

Car Fleet Example

The fleet has four cars and one is a Chevy?

10C3 = 120

The fleet has four cars, two of each type, and it cannot have aChevy and a Honda?How many ways if there is no Chevy?7C2 ·3 C2 = 21 · 3 = 63How many ways if there is no Honda?6C2 ·4 C2 = 15 · 6 = 90How do we combine these? 63 + 90 = 153

Car Fleet Example

The fleet has four cars and one is a Chevy?10C3 = 120

Car Fleet Example

The fleet has four cars, two of each type, and it cannot have aChevy and a Honda?

How many ways if there is no Chevy?7C2 ·3 C2 = 21 · 3 = 63How many ways if there is no Honda?6C2 ·4 C2 = 15 · 6 = 90How do we combine these? 63 + 90 = 153

Car Fleet Example

The fleet has four cars, two of each type, and it cannot have aChevy and a Honda?How many ways if there is no Chevy?

7C2 ·3 C2 = 21 · 3 = 63How many ways if there is no Honda?6C2 ·4 C2 = 15 · 6 = 90How do we combine these? 63 + 90 = 153

Car Fleet Example

The fleet has four cars, two of each type, and it cannot have aChevy and a Honda?How many ways if there is no Chevy?7C2 ·3 C2 = 21 · 3 = 63

How many ways if there is no Honda?6C2 ·4 C2 = 15 · 6 = 90How do we combine these? 63 + 90 = 153

Car Fleet Example

The fleet has four cars, two of each type, and it cannot have aChevy and a Honda?How many ways if there is no Chevy?7C2 ·3 C2 = 21 · 3 = 63How many ways if there is no Honda?

6C2 ·4 C2 = 15 · 6 = 90How do we combine these? 63 + 90 = 153

Car Fleet Example

The fleet has four cars, two of each type, and it cannot have aChevy and a Honda?How many ways if there is no Chevy?7C2 ·3 C2 = 21 · 3 = 63How many ways if there is no Honda?6C2 ·4 C2 = 15 · 6 = 90

How do we combine these? 63 + 90 = 153

Car Fleet Example

The fleet has four cars, two of each type, and it cannot have aChevy and a Honda?How many ways if there is no Chevy?7C2 ·3 C2 = 21 · 3 = 63How many ways if there is no Honda?6C2 ·4 C2 = 15 · 6 = 90How do we combine these?

63 + 90 = 153

Car Fleet Example

ExampleSo there are 153 ways to have a fleet of 4 cars with no Chevy and noHonda. Right?

What are we neglecting to consider?6C2 ·3 C2 = 15 · 3 = 45So the total is 153− 45 = 108 ways.

Car Fleet Example

ExampleSo there are 153 ways to have a fleet of 4 cars with no Chevy and noHonda. Right?What are we neglecting to consider?

6C2 ·3 C2 = 15 · 3 = 45So the total is 153− 45 = 108 ways.

Car Fleet Example

ExampleSo there are 153 ways to have a fleet of 4 cars with no Chevy and noHonda. Right?What are we neglecting to consider?6C2 ·3 C2 = 15 · 3 = 45

So the total is 153− 45 = 108 ways.

Car Fleet Example

ExampleSo there are 153 ways to have a fleet of 4 cars with no Chevy and noHonda. Right?What are we neglecting to consider?6C2 ·3 C2 = 15 · 3 = 45So the total is 153− 45 = 108 ways.

Car Fleet Example

An alternate way to consider this?

How many different fleets could we have if there are no makerestrictions but we want two foreign and two domestic cars?

7C2 ·4 C2 = 21 · 6 = 126

How many have Hondas and Chevy’s?

6C1 ·3 C1 = 6 · 3 = 18

So, we have 126− 18 = 108 ways.

Car Fleet Example

7C2 ·4 C2 = 21 · 6 = 126

6C1 ·3 C1 = 6 · 3 = 18

So, we have 126− 18 = 108 ways.

Car Fleet Example

7C2 ·4 C2 = 21 · 6 = 126

6C1 ·3 C1 = 6 · 3 = 18

So, we have 126− 18 = 108 ways.

Car Fleet Example

7C2 ·4 C2 = 21 · 6 = 126

6C1 ·3 C1 = 6 · 3 = 18

So, we have 126− 18 = 108 ways.

Car Fleet Example

7C2 ·4 C2 = 21 · 6 = 126

6C1 ·3 C1 = 6 · 3 = 18

So, we have 126− 18 = 108 ways.

Car Fleet Example

7C2 ·4 C2 = 21 · 6 = 126

6C1 ·3 C1 = 6 · 3 = 18

So, we have 126− 18 = 108 ways.

Poker Example

ExampleHow many ways are there to get a full house with a standard deck?

Denomination for the set? 13C1 = 13 And which cards? 4C3 = 4.

Denomination for the pair? 12C1 = 12 and which cards? 4C2 = 6.

So, the number of full houses would therefore be 13 · 12 · 4 · 6 = 3744.

Poker Example

Denomination for the set?

13C1 = 13 And which cards? 4C3 = 4.

Poker Example

Denomination for the set? 13C1 = 13

And which cards? 4C3 = 4.

Poker Example

Denomination for the set? 13C1 = 13 And which cards?

4C3 = 4.

Poker Example

Denomination for the pair?

12C1 = 12 and which cards? 4C2 = 6.

Poker Example

Denomination for the pair? 12C1 = 12

and which cards? 4C2 = 6.

Poker Example

Denomination for the pair? 12C1 = 12 and which cards?

4C2 = 6.

Poker Example

Back to the Interviews

ExampleTen job applicants have been invited for interviews with 5 to bescheduled in the morning and 5 to be scheduled in the afternoon. Inhow many ways can the interviews be scheduled? Is this the samenumber as if all ten interviews were in the morning?

We know the number of ways to arrange the interviews once we knowhow many are in each time slot are is 5! · 5! = 14400.

How do we choose who goes when? 10C5 = 252

How do we combine?

10C5 ·5 P5 ·5 P5 = 3628800

We know the number of ways to arrange the interviews once we knowhow many are in each time slot are is

5! · 5! = 14400.

How do we combine?

10C5 ·5 P5 ·5 P5 = 3628800

How do we combine?

10C5 ·5 P5 ·5 P5 = 3628800

How do we choose who goes when?

10C5 = 252

How do we combine?

10C5 ·5 P5 ·5 P5 = 3628800

How do we combine?

10C5 ·5 P5 ·5 P5 = 3628800

How do we combine?

10C5 ·5 P5 ·5 P5 = 3628800

How do we combine?

10C5 ·5 P5 ·5 P5 = 3628800

Permutations Revisited

ExampleHow many words can we make with the letters in ‘SPRING’?

6! = 720

ExampleHow many words can we make with the letters in ‘SUMMER’?

It isn’t 720 ... Why not?

We have to factor in that there are two ‘Ms’ but that we can’t tell themapart.

6! = 720

Another Way For This Type

ExampleHow many words can we make with the letters of ‘SUCCESS’?

We can first consider that we have 7 letters and we are going to choosewhere to place S’s. How many ways are there for us to do that?

And now the C’s. There two of them in the remaining four letters.How many ways are there to arrange them?

And now, we have two letters left, so let’s look at how many ways toarrange the U.

And finally, we are down to one letter.

Putting this all together, we have

7C3 ·4 C2 ·2 C1 ·1 C1

3!4!· 4!

2!2!· 2!

1!1!· 1!

3! 6 4!· 6 4!

2! 6 2!· 6 2!

1! 6 1!· 6 1!

3!2!1!1!= 420

What are the numbers in the denominator?

7C3 ·4 C2 ·2 C1 ·1 C1

3!4!· 4!

2!2!· 2!

1!1!· 1!

3! 6 4!· 6 4!

2! 6 2!· 6 2!

1! 6 1!· 6 1!

3!2!1!1!= 420

7C3 ·4 C2 ·2 C1 ·1 C1

3!4!· 4!

2!2!· 2!

1!1!· 1!

3! 6 4!· 6 4!

2! 6 2!· 6 2!

1! 6 1!· 6 1!

3!2!1!1!= 420

7C3 ·4 C2 ·2 C1 ·1 C1

3!4!· 4!

2!2!· 2!

1!1!· 1!

3! 6 4!· 6 4!

2! 6 2!· 6 2!

1! 6 1!· 6 1!

3!2!1!1!= 420

ExampleHow many ‘words’ can be made from the letters in MATHEMATICS?

Is our answer 11!?

We need to take into account that the repeated letters areindistinguishable.

MATHE MATICS

So, we have 11!2!·2!·2! ways to make a ‘word’

Is our answer 11!?

MATHE MATICS

Is our answer 11!?

MATHE MATICS

Is our answer 11!?

MATHE MATICS

Is our answer 11!?

MATHE MATICS

So, we have

11!2!·2!·2! ways to make a ‘word’

Is our answer 11!?

MATHE MATICS

Permutations with Non-Unique Elements

Repetition PermutationsIf we want to arrange n objects (where order matters) in which thereare ki indistinguishable elements of type i, then the number ofpermutations of this set would be

n!k1! · k2! · . . . · kk!

Notice that this works when we have all indistinguishable objectsbecause 1! = 1.

Permutations with Non-Unique Elements

Repetition PermutationsIf we want to arrange n objects (where order matters) in which thereare ki indistinguishable elements of type i, then the number ofpermutations of this set would be

n!k1! · k2! · . . . · kk!

Notice that this works when we have all indistinguishable objectsbecause 1! = 1.

Generalized Permutations

ExampleHow many ways are there to deal 4 poker hands for 5 card draw froma standard deck of cards?

Thoughts?

52!5!5!5!5!32!

Thoughts?

52!5!5!5!5!32!

Thoughts?

52!5!5!5!5!32!

A Different Kind of Example

ExampleHow many ways can I assign 3 tasks to people from a class of 10students if people can be assigned more than one task?

How many think the answer is 10P3?

How many think the answer is 10C3?

Suppose the people are different slots and let ♣ indicate who a taskwas assigned to.

1 2 3 4 5 6 7 8 9 10

We can assign all three tasks to the same person.

1 2 3 4 5 6 7 8 9 10

♣♣♣

We can assign the three tasks to three different people.

1 2 3 4 5 6 7 8 9 10

♣♣♣

We can assign two tasks to one person and one to another.

1 2 3 4 5 6 7 8 9 10

♣ ♣ ♣

Let’s remove the two end lines.

♣ ♣ ♣

What we are trying to figure out is how many ways we can arrangethe 9 (n− 1) dividers and 3 (r) tasks.

Does order matter? i.e. Can you tell the difference between differentdividers?

♣ ♣ ♣

Since your answer is no, we are looking at some kind of combination.

So we are looking to arrange n− 1 + r objects where order doesn’tmatter.

So the idea is that we are randomly placing 12 objects without regardfor which type is which, but we do need to take into account howmany are of each type. So for our example, we would have

12C3 =12 C9

ways to assign the job.

12C3 =12 C9

Generalized Combinations

TheoremThere are n+r−1Cr =n+r−1 Cn−1 r-combinations from a set with nelements when repetition of elements is allowed.

ExampleSuppose a cookie shop has 4 different kinds of cookies. In how manyways can 6 cookies be selected for purchase? Assume that only thetype of cookie, and not the individual cookies or the order in whichthey are chosen, matters.

There are 9C6 = 84 ways to select this half dozen.

Solving Equations

ExampleHow many solutions does the equation

x1 + x2 + x3 = 11

have, where x1, x2 and x3 are nonnegative integers?

How is this related to what we just did?

The solution is found by noting that we are looking for a way ofselecting 11 items from a set with three elements so that x1 items oftype 1, x2 item of type 2 and x3 items of type 3 are chosen. So, thesolution is equal to the number of 11-combinations with repetitionallowed from a set of three elements.

C(11 + 3− 1− 1, 2) = C(13, 2) = 78

Solving Equations

x1 + x2 + x3 = 11

C(11 + 3− 1− 1, 2) = C(13, 2) = 78

Solving Equations

x1 + x2 + x3 = 11

The solution is found by noting that we are looking for a way ofselecting 11 items from a set with three elements so that x1 items oftype 1, x2 item of type 2 and x3 items of type 3 are chosen.

So, thesolution is equal to the number of 11-combinations with repetitionallowed from a set of three elements.

C(11 + 3− 1− 1, 2) = C(13, 2) = 78

Solving Equations

x1 + x2 + x3 = 11

C(11 + 3− 1− 1, 2) = C(13, 2) = 78

Solving Equations

x1 + x2 + x3 = 11

C(11 + 3− 1− 1, 2) = C(13, 2) = 78

Chapter 6 Counting...

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