lecture5 arrangement with repetition indistinguishable
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Arrangement with Repetition
(Indistinguishable Objects)
DR. HAJAR SULAIMAN Discrete Mathematics 1
MSS 318
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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
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Preliminaries
DR. HAJAR SULAIMAN Discrete Mathematics 4
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
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DR. HAJAR SULAIMAN Discrete Mathematics 5
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?
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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.
DR. HAJAR SULAIMAN Discrete Mathematics 6
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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.
DR. HAJAR SULAIMAN Discrete Mathematics 7
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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
DR. HAJAR SULAIMAN Discrete Mathematics 8
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DR. HAJAR SULAIMAN Discrete Mathematics 9
Can we usecombination?
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Using Combination
DR. HAJAR SULAIMAN Discrete Mathematics 10
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.
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Using Combination
DR. HAJAR SULAIMAN Discrete Mathematics 11
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)).
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Formulation
DR. HAJAR SULAIMAN Discrete Mathematics 12
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.
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Arrangement with Repetition
DR. HAJAR SULAIMAN Discrete Mathematics 13
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.
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Arrangement with Repetition
DR. HAJAR SULAIMAN Discrete Mathematics 14
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.
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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
Arrangement with Repetition
DR. HAJAR SULAIMAN Discrete Mathematics 15
Therefore, there are 45,360 different ways the letters can
be arranged.
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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
Arrangement with Repetition
DR. HAJAR SULAIMAN Discrete Mathematics 16
The number of arrangements of 4 S’s, 2 P’s & 1 M is:
7!105.
4! 2! 1!
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Arrangement with Repetition
DR. HAJAR SULAIMAN Discrete Mathematics 17
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
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Arrangement with Repetition
DR. HAJAR SULAIMAN Discrete Mathematics 18
105 70 7350.
So, by the product rule, the number of
ways to arrange the letters is
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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
Arrangement with Repetition
DR. HAJAR SULAIMAN Discrete Mathematics 19
The number of arrangements of 4 H’s & 6 T’s is:10!
210.
4! 6!
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Arrangements where CertainLetters are Fixed
DR. HAJAR SULAIMAN Discrete Mathematics 20
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DR. HAJAR SULAIMAN Discrete Mathematics 22
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
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DR. HAJAR SULAIMAN Discrete Mathematics 23
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
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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
Arrangement with Repetition
DR. HAJAR SULAIMAN Discrete Mathematics 24
If the letters are all different, then how would we do it?
So, let’s distinguish the N’s with subscripts.
So, how?
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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.
Arrangement with Repetition
DR. HAJAR SULAIMAN Discrete Mathematics 25
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.
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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
Arrangement with Repetition
DR. HAJAR SULAIMAN Discrete Mathematics 26
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.
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Application in a Path Problem
DR. HAJAR SULAIMAN Discrete Mathematics 27
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Example 7How many different paths exist from one vertex ofthe grid (shown below) to the vertex diagonallyopposite the first?
North
East
Arrangement with Repetition
DR. HAJAR SULAIMAN Discrete Mathematics 28
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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
DR. HAJAR SULAIMAN Discrete Mathematics 29
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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.
DR. HAJAR SULAIMAN Discrete Mathematics 30
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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?
DR. HAJAR SULAIMAN Discrete Mathematics 31
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THE END
DR. HAJAR SULAIMAN Discrete Mathematics 32