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Permutations with Some repetitions
Example
r-Combinations with Some repetitions
Example
Example
Example
Permutations and Combinations with Repetitions
Balls in Boxes
Balls in Boxes: Boxes maybe empty
Balls in Boxes: Boxes non-empty
Permutations and
Combinations with Repetitions
Sanjay Jain, Lecturer,
School of Computing
Permutations with Some Repetitions
Anagrams of MALAYSIA, INDONESIA,…
For example, the different permutations of ABA
are:
ABA
BAA
AAB
So our old method doesn’t work.
END OF SEGMENT
Permutations with Some Repetitions
Theorem: Suppose I have a collection of n objects:
n1 objects of type 1.
n2 objects of type 2.
….
nk objects of type k.
Where all the types are distinct, and
n1+n2+…+nk=n
Then, the number of distinct permutations of the n
objects is
n n n1 n n1 n2 n n1 .... nk 1 n!
* * ......*
n1 n n !n !n !.... !
2 n3 nk 1 2 3 nk
END OF SEGMENT
Proof
1 2 3 …. …. n
T1: Place the objects of type 1.
T2: Place the objects of type 2.
T3: Place the objects of type 3.
……..
Tk: Place the objects of type k.
Proof
1 2 3 …. …. n
T1: Place the objects of type 1.
T1 can be done in ways.
n
n1
Proof
1 2 3 …. …. n
T2: Place the objects of type 2.
T2 can be done in n n1 ways.
n
2
Proof
1 2 3 …. …. n
T3: Place the objects of type 3.
T3 can be done in n n1 n2
ways.
n3
Proof
1 2 3 …. …. n
Tk: Place the objects of type k.
Tk can be done in n n1 .... nk 1
ways.
nk
Permutations with Some Repetitions
Theorem: Suppose I have a collection of n objects:
n1 objects of type 1.
n2 objects of type 2.
….
….
nk objects of type k.
Thus using the multiplication rule, all the n objects can be
placed in
n n n1 n n1 n2 n n1 .... nk 1
*
n1 n *
......*
2 n3 nk
ways.
END OF SEGMENT
Permutations with Some Repetitions
n n n1 n n1 n2 n n1 .... nk 1
*
n1 n *
......*
2 n3 nk
n! (n n1 )! (n n1 n2 )! (n n1 n2 ...nk 1 )!
* * * ...... *
n1!(n n1 )! n2 !(n n1 n2 )! n2 !(n n1 n2 n3 )! nk !(n n1 n2 ...nk )!
n!
n1!n2 !n3!.... nk !
END OF SEGMENT
Example
How many anagrams of Malaysia are there?
Answer:
M ---1
A ---3
L---1
Y---1
S ---1
I --- 1
8!
1!3!1!1!1!1!
END OF SEGMENT
r-Combinations with Repetitions allowed
A r-combination, with repetition allowed, chosen
from a set X of n elements is an unordered
selection of elements taken from X, where
repetitions are allowed.
That is we can take several copies of any element
of X in the selection.
END OF SEGMENT
Theorem:
The number of r-combinations with repetition
allowed, that can be selected from a set of size
n is
r n 1
r
Proof:
1st 2nd 3rd …. nth
XX X XXX
r X’s and n-1 boundaries.
These can be arranged in any order.
The number of ways this can be done is r+n-1Cr .
Thus the number of r-combinations with
repetitions allowed is
r n 1
r
END OF SEGMENT
Example
I want to buy 5 cans of soft drink.
The possible drinks that are available are coke, sprite and
pepsi, where there are unlimited number of each.
In how many ways can I choose the 5 cans?
Answer:
This is same as the number of 5-combinations, with
repetition allowed, from a set of size 3.
(since I need to select 5 cans from a set of size 3 with
repetition allowed)
Using the theorem, this can be done in
5 3 1
5
ways.
END OF SEGMENT
Example
Library has budget to buy 20 copies of books on discrete
math.
Five different text books on discrete math are available
(unlimited number of copies) in the market.
How many ways can the library buy the books?
Answer:
This is same as the number of 20-combinations, with
repetition allowed, from a set of size 5.
(since I need to select 20 books from a set of size 5 with
repetition allowed)
Using the theorem, this can be done in
20 5 1
20
ways.
END OF SEGMENT
Example
How many integral solutions of equation
x1+x2+x3=20,
where x1,x2,x3 0
are there?
Answer:
Note that we can consider each xi as a different type.
Choosing xi=r, would mean that we are taking r copies of xi.
This is same as the number of 20-combinations, with
repetition allowed, from a set of size 3.
Using the theorem, this can be done in
20 3 1
20
ways.
END OF SEGMENT
Permutations and Combinations with
Repetitions
Suppose we are choosing k elements from a set of size n.
Many problems can be categorized into the following four
forms:
Order matters and repetitions are allowed: nk
Order doesn’t matter and repetitions are k n 1
k
allowed:
Order matters and repetitions are not allowed: P(n,k)
n
Order doesn’t matter and repetitions are not allowed:
k
END OF SEGMENT
Balls in Boxes
Many problems can be formulated as a balls in boxes
problem.
Suppose we have n balls which are to be placed in m
boxes.
Balls are distinguishable or not?
Boxes are distinguishable or not?
Boxes have limited capacity or infinite capacity?
Boxes are required to be non-empty?
Examples:
Number of functions from X to Y.
X has k elements, and Y has r elements.
We can consider elements of X as balls, and elements of Y
as boxes.
So this problem is same as how many ways we can
distribute k balls in r boxes, where both balls and boxes
are distinguishable.
Note that in the above problem we cannot take elements of
Y to be balls and elements of X to be boxes:
“ball” goes into only one “box”.
A “box” may contain several “balls”.
Elements of X may be mapped to only one element of Y.
Elements of Y may be an image of several elements of X.
Examples:
All functions: capacity unlimited.
1--1 functions: capacity of boxes is 1.
Onto functions: Boxes should be non-empty.
Bijective functions: Boxes should be non-empty and
capacity is one.
END OF SEGMENT
Balls in Boxes
We will only consider the cases where Boxes are
distinguishable. The question when boxes are
indistinguishable is hard.
Case I: n distinguishable balls are to be placed in m
distinguishable boxes with infinite capacity.
We can divide the job into n tasks.
Ti: place the i-th ball. (i=1 to n)
Ti can be done in m ways (one can select any of the
boxes).
Thus using multiplication rule, number of ways to do the
job is
m n.
END OF SEGMENT
Balls in Boxes
Case II: n indistinguishable balls are to be placed in m
distinguishable boxes with infinite capacity.
1st 2nd 3rd …. mth
XX X XXX
Choosing n-combination from a set of size m with
repetition allowed.
n m 1
This can be done in n
ways by the theorem done
earlier.
END OF SEGMENT
Balls in Boxes
Case III: n distinguishable balls are to be placed in m
distinguishable boxes with capacity =1.
m<n 0 ways
m n P(m,n) ways
END OF SEGMENT
Balls in Boxes
Case IV: n indistinguishable balls are to be placed in m
distinguishable boxes with capacity =1.
m < n 0 ways
m
mn
n
END OF SEGMENT
Balls in Boxes: non-empty boxes.
We will only be considering the case of distinguishable
boxes and indistinguishable balls.
Case V: n indistinguishable balls are to be placed in m
distinguishable boxes, with unlimited capacity such
that each box is non-empty.
n<m: 0 ways.
n m
Give one ball to each box.
Now distribute the remaining n-m balls in m boxes,
without the non-empty constraint.
n m m 1 n 1
n m
nm
END OF SEGMENT
Balls in Boxes: non-empty boxes.
Case VI: n indistinguishable balls are to be placed in m
distinguishable boxes, with capacity=1, such that each
box is non-empty.
If m = n: 1 way.
If m n : 0 ways.
END OF SEGMENT
Balls in Boxes:
We can generalize the technique to several other cases.
For example, suppose we want to place 10
indistinguishable balls into 5 distinguishable boxes such
that box B1 gets exactly two balls, and each of the
remaining boxes get at least one ball.
How many ways can we do it?
First place 2 balls in B1, and one ball in each of the other
boxes.
Thus we are left with 4 balls, and 4 boxes.
(balls: indistinguishable, boxes distinguishable, no other
constraint)
4 4 1
4
END OF SEGMENT
Example:
Suppose we want distribute 20 eclairs to 3 children, Tom,
Mary and John. Suppose Tom should get atleast 4
eclairs.
In how many ways can we do it?
Note: Children are distinguishable. Eclairs are not.
We can first give away 4 eclairs to Tom.
Then we can distribute 16 remaining eclairs with no
constraints.
16 3 1
16
END OF SEGMENT
Example:
Suppose library wants to buy 20 copies of discrete math
book. Suppose there are 5 discrete math text books
available in the market.
Suppose there is a constraint that library should buy at
least one of each of the text books.
How many ways can the library buy the books?
Answer: Different copies of same text books are not
distinguishable. However different text books are
distinguishable. We can consider text books as boxes,
and the number of copies bought as balls.
Thus we need to distribute 20 indistinguishable balls into 5
distinguishable boxes, with each box being non-empty.
20 1
20 5
END OF SEGMENT
