Lecture5 Arrangement With Repetition Indistinguishable

8/20/2019 Lecture5 Arrangement With Repetition Indistinguishable

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Arrangement with Repetition

(Indistinguishable Objects)

DR. HAJAR SULAIMAN Discrete Mathematics 1

MSS 318





Topic’s Learning Outcome

At the end of this lesson, students should be able to

• Find the number of linear arrangements of objects

involving indistinguishable ones• Use C(n,r) to solve arrangement problems with

indistinguishable objects

• Compute arrangements with indistinguishableobjects where certain restrictions are set

• Apply the knowledge in solving path problems

DR. HAJAR SULAIMAN SEM 1 (2014/2015) 3



Preliminaries


Problem:

Let us see how many ways there is to arrange

the letters A, A, B, B, B.

A A B B B

Let us list all possibilities:

A B B A B

A B B B A

B A A B B

A B A B B

B A B A B

B B A A B

B B A B A

B B B A A

B A B B A




A A B B B

A B B A B

A B B B A

B A A B B

A B A B B

B B A A B

B B A B A

B B B A A

B A B B A

B A B A B

There are 10 different

arrangements.

Recall that if the letters are distinct, then

there are 5!=120 ways to arrange.

120 5!Observation: 10

12 2!3!

and there are in fact 2 A’s and 3 B’s!!!

IS THIS THE FORMULA?



Observe: Consider one of the arrangements:

A1 A2 B3 B1 B2 A1 A2 B1 B2 B3

A1 A2 B1 B3 B2

A1 A2 B2 B1 B3

A1 A2 B2 B3 B1

A1 A2 B3 B2 B1

A2 A1 B1 B2 B3

A2 A1 B1 B3 B2

A2 A1 B2 B1 B3

A2 A1 B2 B3 B1

A2 A1 B3 B1 B2

A2 A1 B3 B2 B1

A A B B B

Suppose the repeated letters are distinct; then the

following are different arrangements.




A1 A2 B1 B2 B3

A1 A2 B1 B3 B2

A1 A2 B2 B1 B3

A1 A2 B2 B3 B1

A1 A2 B3 B1 B2

A1 A2 B3 B2 B1

Similarly, fixing the B’s, below are different

arrangements of the A’s and there are 2! of them.

A1 A2 B1 B2 B3 A2 A1 B1 B2 B3

Fixing the A’s, below are different arrangements

of the B’s and there are 3! of them.

However, when the A’s and the B’s are the same, the

permutations are considered as 1 arrangement.

Again, when the A’s and the B’s are the same, thisis considered as 1 arrangement.




Thus, the computation resulting to 120 (i.e. 5!)

considers the A’s and the B’s to be different.

Hence, those possibilities of 2! and 3! must be

excluded to give us the exact number of ways to

arrange the given letters. This is done by division.

Therefore, the number of ways to arrange the

letters A, A, B, B, B is

5! 5.410.

2!3! 2.1





Can we usecombination?



Using Combination


Idea: Instead of thinking about the number of ways

to arrange the 5 letters, we consider selecting theboxes to put the letters.

There is an alternative solution to the previousproblem.

Explanation: This is because we want to arrange allthe letters, so there is no choice on the letters.

So, it is a matter of choosing the boxes to put A’s

or B’s in.



Using Combination


Can you see that it is the same answer?

Suppose we want to put the A’s first.

After that, how many ways to put the B’s?

5! 5.4(5,2) 103!2! 2.1

C Then there are ways to do this.

What happens if we choose to put the B’s first?

1 way.

There will be C (5,3) ways (and we know C (5,3) = C (5,2)).



Formulation


Suppose there are N total objects, n1 of type 1,

n2 of type 2, n3 of type 3, and so on up to nk of

type k , where N = n1 + n2 + n3 +…..+ nk .

Then there are

1 2 3

!

! ! ! !k

N

n n n n

ways to place the objects.





Example 1Suppose a couple who recently married wishes tohave three sons and four daughters. How manybirth-order arrangements are possible? (Assume

that there are no occurrence of multiple births suchas twins, triplets etc.)

Solution

Let us use the symbols SSS and DDDD torepresent the three sons and four daughters.





Suppose we choose to place the D’s.

Using the same idea as the previous alternativesolution:

We want to place SSS and DDDD in the 7 boxes.

7! 7.6(7,4)3!4!

C Then .53.2

35.1

.

So, there are 35 birth order arrangements forthe couple.



Example 2

In how many ways can all of the letters in the word

SASKATOON be arranged?

If all 9 letters were different, we could arrange them in 9!ways, but because there are 2 identical S’s, 2 identical A’s,

and 2 identical O’s,we can arrange the letters in:

1 2 3

! 9!

45,360! ! ! ... 2! 2! 2!

N

n n n

Solution



Therefore, there are 45,360 different ways the letters can

be arranged.



Example 3In how many different orders is it possible to write

the letters of the word MISSISSIPPI such that no

two I’s are adjacent?

In MISSISSIPPI, there are 4 I’s, 4 S’s, 2 P’s & 1 M.

In this case, since we do not want the 4 I’s to be together,

then we arrange the other letters first.

Solution



The number of arrangements of 4 S’s, 2 P’s & 1 M is:

7!105.

4! 2! 1!





So, suppose we have arranged the 7 letters shown as thefollowing in boxes. What do we do about the 4 I’s?

Since we don’t want the I’s to be together, then we can slotthem in between the already arranged letters.

I I I I

(8,4) 8! 8.7.6.5There are or (8,4) 70 ways to arrange the I's.

4! 4!4! 1.2.3.4

P C





105 70 7350.

So, by the product rule, the number of

ways to arrange the letters is



Example 4 A coin is flipped and it is recorded whether it lands

heads or tails. This is done 10 times. What are the

possible arrangements of getting four heads and

six tails for the 10 flips?

Let H = head, T = tail.

In this case, we want an arrangement of 4 H’s and 6 T’s

(with regards to the order of the flips).

Solution



The number of arrangements of 4 H’s & 6 T’s is:10!

210.

4! 6!



Arrangements where CertainLetters are Fixed







A A B B B

A B B A B

A B B B A

B A A B BA B A B B B B A A B

B B A B A

B B B A A

B A B B A

B A B A B

Out of the 10 arrangements,

four starts with A.

So, if a question asks, how many ways to arrange the

letters A, A, B, B, B which starts with A, then the answer

would be 4.

How do we get this answer by computation? Consider:

A

A, B, B, B




If we consider the remaining letters, there are 3 B’s and 1

A. So, we can confidently say that there are

4!4

3!1!

since there are no repetition of A after one of the A’s is

fixed. So, “4” is in fact the correct answer.

However, what if we don’t know which letter is fixed at

the beginning?

A

A, B, B, B



Example 5How many arrangements of the letters in the word

CANN that begins with a consonant?

We would be asking “which consonant”? It can either be C

or N. If N, then we will be left with 1 N for arrangements of

the remaining 3 letters; but if it is a C, then will be left with 2

N for arrangements of the remaining 3 letters.

Solution



If the letters are all different, then how would we do it?

So, let’s distinguish the N’s with subscripts.

So, how?



Say, C A N1 N2?Then, to choose a consonant for the first place, there will be

C(3,1) ways. And then to arrange the rest, there will be 3!

ways.



So, we know that there will be 18 ways.But this includes words like N2 C A N1 and N1 C A N2 with

the same starting consonant N and same sequence of

letters that follow.

Hence, we still need to divide by 2! even though one of theN’s has already been chosen to be the consonant at the

beginning.

Which means number of ways will be 9.



Example 6

Find the number of arrangements of the letters from the

word CALCULATE such that each word starts and ends

with a consonant.

There are 2 C’s, 2 A’s, 2 L’s, 1 U, 1 T and 1 E.

Solution



Then, P(7,7) ways to arrange the remaining letters; after

which we divide by the repetition. Which gives:

5.4.7!12600 arrangements.

2! 2!2!1!1!1!

So, there are 5 consonants.

There are C(5,1)C(4,1) to choose the beginning and end

consonants.



Application in a Path Problem




Example 7How many different paths exist from one vertex ofthe grid (shown below) to the vertex diagonallyopposite the first?

North

East





North

East

One possible path would be represented by thearrangement NNEENENNENNNENE.

We can represent each unit up asN and each unit right as E.

Basically, we have to travel 6 E’sand 9 N’s

The question then becomes one to determine the

number of arrangements of 15 letters where 9 areN’s and 6 are E’s.

Therefore, there are15!

(15,6) .6!9!

C or different paths




EXERCISES

1. How many arrangements can be made out of the letters ofthe word ENGINEERING?

2. In how many different ways can the letters of the word

'CORPORATION' be arranged so that the vowels always cometogether?

3. How many arrangements of the word RESTRICTIONS can be

made if, the word must start with a consonant and end witha vowel.




EXERCISES

4. A man works in a building located seven blocks east and

eight blocks north of his home. Thus in walking to work

each day he goes fifteen blocks. Denote the north-south

streets by A, B, C, …, H and the east-west streets by 1st,

2nd, …, 9th. Presume that the man lives at the corner of 1st

and A, and works at the corner of 9th and H. Given the

information that all streets are available for walking with

one exception, namely that E street from 5th to 6th is not

cut through, in how many different paths can the man walk

from home to work, walking only fifteen blocks?




THE END


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Lecture5 Arrangement With Repetition Indistinguishable