Transcript Section 5.4 Distributions

Section 5.4 Distributions
By
Colleen Raimondi
Tucker, Section 5.4
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Distributions
A distribution problem is an arrangement or
selection problem with repetition.
Specialized distribution problems must be
broken up into subcases that can be counted
in terms of simple permutations and
combinations (with and without repetition).
General guidelines for modeling distributions:
Distributions of distinct objects are
equivalent to arrangements and Distributions
of identical objects are equivalent to
selections.
Tucker, Section 5.4
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Basic Models for Distributions
Distinct Objects:
The process of distributing r distinct objects into n different
boxes is equivalent to putting the distinct objects in a row and
then stamping one of the n different box names on each object.
The resulting sequence of box names is an arrangement of
length r formed from n items with repetition.
Thus there are n x n x…x n (r ns) = n r distributions of the
r distinct objects.
If ri objects must go in box i, 1< i < n, then there are P(r;
r1, r2, …, rn) distributions.
r distinct objects
n different boxes
Tucker, Section 5.4
Red Red Blue Green
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Basic Models for Distributions
Identical Objects:
The process of distributing r identical objects into
n different boxes is equivalent to choosing an
(unordered) subset of r box names with repetition
from among the n choices of boxes.
Thus there are C(r+n-1, r) = (r+n-1)!/r!(n-1)!
distributions of the r identical objects.
r identical
objects
Red Red Blue Blue Green Green Green
Tucker, Section 5.4
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Equivalent Forms for Selection with Repetition
1. The number of ways to select r objects with
repetition from n different types of objects.
2. The number of ways to distribute r identical
objects into n distinct boxes.
3. The number of nonnegative integer
solutions to x1+x2+…+xn=r.
r identical
objects = 7
Red Red Blue Blue Green Green Green
7=
2
+
2
Tucker, Section 5.4
+
3
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Ways to Arrange, Select, or Distribute r Objects
from n Items or into n Boxes
Arrangement
(order outcome)
No repetition
Unlimited Repetition
Combination
(unordered outcome)
or
or
Distribution of
distinct objects
Distribution of
identical objects
P(n,r)
nr
Restricted Repetition P(n; r1, r2, …, rm)
Tucker, Section 5.4
C(n,r)
C(n+r-1, r)
----6
Example 1
(pg. 203)
How many ways are there to assign 100 different
diplomats to 5 different continents? How many
ways if 20 diplomats must be assigned to each
continent?
For part one we want to use the distinct objects with unlimited
repetition model from below.
For the second part we want to use the distinct objects with
restricted repetition model from below.
Distribution of
distinct objects
No repetition
Unlimited Repetition
P(n,r)
nr
Restricted Repetition P(n; r1, r2, …, rm)
Tucker, Section 5.4
Distribution of
identical objects
C(n,r)
C(n+r-1, r)
----7
Example 1
(Continued)
These are distinct objects according to the model
for distributions. Following that model that means it
equals the number of sequences of length 100 involving
5 continents. 5100 sequences.
If you add the constraint of assigning 20
diplomats to each continent, that means that each
continent name should appear 20 times in a sequence.
You can do this P(100; 20, 20, 20, 20, 20) =
100!/(20!)5 ways.
Tucker, Section 5.4
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Example 2
In bridge, the52 cards of a standard card deck are
randomly dealt 13 apiece to players North, East, South,
and West. What is the probability that West has all 13
spades? That each player has one Ace?
No repetition
Unlimited Repetition
Distribution of
distinct objects
Distribution of
identical objects
P(n,r)
C(n,r)
nr
Restricted Repetition P(n; r1, r2, …, rm)
Tucker, Section 5.4
C(n+r-1, r)
-----
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Example 2
(continued)
There are P(52; 13, 13, 13, 13) distributions of
the 52 cards into 13-card hands.
Distributions in which West gets all the spades may be
counted as the ways to distribute West all the spades, 1
way, times the ways to distribute the 39 non-spade cards
among the 3 other hands, P(39; 13, 13, 13) ways.
52! =1
52
39!
52! =1
( 13)
(13!)3 (13!)4
13!39!
A simpler answer to the question, what is the
probability that West has all 13 spades, can be directly
obtained by looking at West’s possible hands alone.
So the unique hand of 13 spades has the probability
of 1/C(52,13).
Tucker, Section 5.4
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Example 2
(Continued)
What is the probability that each player has one Ace?
To count the ways that each player gets one
Ace, we divide the distribution into and Ace part, 4!
ways to arrange the 4 Aces among the 4 players, and
a non-Ace part, P(48; 12, 12, 12, 12) ways to
distribute the remaining 48 non-Ace cards. So the
probability that each player gets an Ace is:
4!48!
52!
(12!)4
(13!)4
=
13!4
12!4
x
4!48!
52!
Tucker, Section 5.4
=
134
52 =0.105
(4)
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Example 3
(Pg. 204)
Show that the number of ways to distribute r
identical balls into n distinct boxes with at least
one ball in each box is C(r-1, n-1). With at least r1
balls in the first box, at least r2 balls in the second
box, …, and at least rn balls in the nth box, the
number is C(r - r1 - r2 -…- rn + n – 1, n-1).
Distribution of
distinct objects
No repetition
Unlimited Repetition
P(n,r)
nr
Restricted Repetition P(n; r1, r2, …, rm)
Tucker, Section 5.4
Distribution of
identical objects
C(n,r)
C(n+r-1, r)
----12
Example 3
(continued)
The requirement of at least one ball in each box can
be though of as a section-with-repetition model. First we
want to put one ball in each box. Now it remains to count
the ways to distribute with our restriction the remaining rn balls into the n boxes. You can do this in
C((r-n)+n-1, (r-n)) =
[(r-n)+n-1]!
(r-n)!(n-1)!
= C(r-1, n-1) ways.
In the case where at least ri balls must be in the i th box, we
first put ri balls into the i th box, and then distribute the
remaining r – r1 – r2 - …- rn balls in any way into the n boxes.
There are C((r-r1-r2-…-rn)+n-1, (r-r1-r2-…-rn))
= C((r-r1-r2-…-rn) +n-1, n-1) ways.
Tucker, Section 5.4
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Example 4
(Pg. 204)
How many integer solutions are there to
the equation x1 + x2 + x3 + x4 = 12, with
xi > 0? How many solutions with xi >
1? How many solutions with x1 > 2, x2
> 2, x3 > 4, x4 > 0?
Distribution of
distinct objects
No repetition
Unlimited Repetition
P(n,r)
nr
Restricted Repetition P(n; r1, r2, …, rm)
Tucker, Section 5.4
Distribution of
identical objects
C(n,r)
C(n+r-1, r)
----14
Example 4
(Continued)
How many integer solutions are there to the
equation x1 + x2 + x3 + x4 = 12, with xi > 0?
By an integer solution to this problem we mean an
ordered set of integer values for the xis summing to 12,
such as x1 = 2, x2 = 3, x3 = 3, x4 = 4.
We can model this problem as a distribution as a
distribution-of-identical-objects problem or as a
selection-with-repetition problem.
Let xi represent the number of (identical) objects in
box i or the number of objects of type i chosen. Using
either of these models, we see that the number of
integer solutions is C(12+4-1, 12) = 455.
Tucker, Section 5.4
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Example 4
(Continued)
Solutions with xi > 1 correspond in these models to
putting at least one object in each box or choosing at
least one object of each type.
Solution with x1>2, x2>2, x3>4, x4>0 correspond
to putting at least 2 objects in the first box, at least 2 in
the second, and at least 4 in the third, and any number in
the fourth.
You can use the selection-with-repetition model. The
answer for xi > 1 is C(12-1, 4-1) = 165 and the other
answer is C((12-2-2-4)+4-1, 4-1) = C(7,3) = 35.
Tucker, Section 5.4
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Class Problem 1
How many ways are there to distribute 20
(identical) sticks of red licorice and 15 (identical)
sticks of black licorice among five children?
Hint:
Distribution of
distinct objects
No repetition
Unlimited Repetition
P(n,r)
nr
Restricted Repetition P(n; r1, r2, …, rm)
Tucker, Section 5.4
Distribution of
identical objects
C(n,r)
C(n+r-1, r)
----17
Class Problem 1
Use the identical objects model for distribution. The
ways to distribute 20 identical sticks of red among 5
children equals the ways to select a collection of 20 names
(or destinations) from a set of 5 different names with
repetition.
There are C(20+5-1, 20) = 10,626 ways.
Use the same modeling argument to distribute the 15
identical sticks of black. There are C(15+5-1, 15) = 3,876
ways. The product of these is the number of ways to distribute
the red and the black licorice. 10,626 x 3,876 = 41,186,376
Tucker, Section 5.4
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Class Problem 2
How many binary sequences of length 10 are
there consisting of a (positive) number of 1s,
followed by a number of 0s, followed by a number
of 1s, followed by a number of 0s? An example
would be 1110111000.
Hint:
No repetition
Unlimited Repetition
Distribution of
distinct objects
P(n,r)
nr
Restricted Repetition P(n; r1, r2, …, rm)
Tucker, Section 5.4
Distribution of
identical objects
C(n,r)
C(n+r-1, r)
-----
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Class Problem 2
First create four distinct boxes, the first box for the initial
set of 1s, the second box for the following set of 0s, and so
on. Then we have 10 identical markers ( call them xs) to
distribute into the four boxes. Each box must have at least
one marker, since each subsequence of 0s or of 1s must be
nonempty. The number of ways to distribute 10 xs into 4
boxes with no box empty is C(10-1, 4-1) = 84. So there
are 84 such binary sequences.
In other words put a 1 or a 0 in each box, which leaves 6
symbols that need to be put into 4-1 dividers.
1
0
1
0
C(6+4-1, 4-1)
Tucker, Section 5.4
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