Section 5 4 Distributions By Colleen Raimondi Tucker

Section 5. 4 Distributions By Colleen Raimondi Tucker, Section 5. 4 1

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 2

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…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; r 1, r 2, …, rn) distributions. r distinct objects n different boxes Tucker, Section 5. 4 Red Blue Green 3

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 Red Blue Green Tucker, Section 5. 4 4

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 x 1+x 2+…+xn=r. r identical objects = 7 Red Blue Green 7= 2 + 2 Tucker, Section 5. 4 + 3 5

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; r 1, r 2, …, 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; r 1, r 2, …, 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) = 100!/(20!)5 ways. Tucker, Section 5. 4 8

Example 2 In bridge, the 52 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; r 1, r 2, …, rm) Tucker, Section 5. 4 C(n+r-1, r) ----- 9

Example 2 (continued) There are P(52; 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) 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 10

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) 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 ( 4 ) =0. 105 11

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 r 1 balls in the first box, at least r 2 balls in the second box, …, and at least rn balls in the nth box, the number is C(r - r 1 - r 2 -…- rn + n – 1, n-1). Distribution of distinct objects No repetition Unlimited Repetition P(n, r) nr Restricted Repetition P(n; r 1, r 2, …, 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 r-n 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 – r 1 – r 2 - …- rn balls in any way into the n boxes. There are C((r-r 1 -r 2 -…-rn)+n-1, (r-r 1 -r 2 -…-rn)) = C((r-r 1 -r 2 -…-rn) +n-1, n-1) ways. Tucker, Section 5. 4 13

Example 4 (Pg. 204) How many integer solutions are there to the equation x 1 + x 2 + x 3 + x 4 = 12, with xi > 0? How many solutions with xi > 1? How many solutions with x 1 > 2, x 2 > 2, x 3 > 4, x 4 > 0? Distribution of distinct objects No repetition Unlimited Repetition P(n, r) nr Restricted Repetition P(n; r 1, r 2, …, 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 x 1 + x 2 + x 3 + x 4 = 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 x 1 = 2, x 2 = 3, x 3 = 3, x 4 = 4. We can model this problem as a distribution-of-identical-objects problem or as a selectionwith-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 15

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 x 1>2, x 2>2, x 3>4, x 4>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 16

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; r 1, r 2, …, 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 18

Class Problem 2 How many binary sequences of length 10 are there consisting of a (positive) number of 1 s, followed by a number of 0 s, followed by a number of 1 s, followed by a number of 0 s? An example would be 111000. Hint: No repetition Unlimited Repetition Distribution of distinct objects P(n, r) nr Restricted Repetition P(n; r 1, r 2, …, rm) Tucker, Section 5. 4 Distribution of identical objects C(n, r) C(n+r-1, r) ----- 19

Class Problem 2 First create four distinct boxes, the first box for the initial set of 1 s, the second box for the following set of 0 s, 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 0 s or of 1 s 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 C(6+4 -1, 4 -1) Tucker, Section 5. 4 20