1. 4/5/05Tucker, Sec. 4.31 Applied Combinatorics, 4rth Ed. Alan Tucker Section 5.4 Distributions Prepared by Jo Ellis-Monaghan

2. 4/5/05Tucker, Sec. 4.32 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 –Distributions of identical objects are equivalent to selections.

3. 4/5/05Tucker, Sec. 4.33 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. Thus there are n * n *…*n = n r distributions of the r distinct objects. r distinct objects n different boxes Red Red Blue Green

4. 4/5/05Tucker, Sec. 4.34 Specified amount in each box If r i objects must go in box i, then there are P(r; r 1, r 2, …, r n ) distributions. 6 distinct objects 3 12 How many in each box

5. 4/5/05Tucker, Sec. 4.35 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 Red Blue Blue Green Green Green r identical objects

6. 4/5/05Tucker, Sec. 4.36 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 +…+x n =r. Red Red Blue Blue Green Green Green 7 identical objects into 3 distinct boxes 7 = 2 + 2 + 3

7. 4/5/05Tucker, Sec. 4.37 Ways to Arrange, Select, or Distribute r Objects from n Items or into n Boxes Arrangement (order outcome) or Distribution of distinct objects Combination (unordered outcome) or Distribution of identical objects No repetition P(n,r) C(n,r) Unlimited Repetition n r C(n+r-1, r) Restricted Repetition P(n; r 1, r 2, …, r m ) -----

8. 4/5/05Tucker, Sec. 4.38 Example 1 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? Distribution of distinct objects Distribution of identical objects No repetition P(n,r) C(n,r) Unlimited Repetition n r C(n+r-1, r) Restricted Repetition P(n; r 1, r 2, …, r m ) ----- 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.

9. 4/5/05Tucker, Sec. 4.39 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. 5 100 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. P(100; 20, 20, 20, 20, 20) = 100!/(20!) 5 ways.

10. 4/5/05Tucker, Sec. 4.310 Distribution of distinct objects Distribution of identical objects No repetition P(n,r) C(n,r) Unlimited Repetition n r C(n+r-1, r) Restricted Repetition P(n; r 1, r 2, …, r m ) ----- 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?

11. 4/5/05Tucker, Sec. 4.311 Example 2 (part 1) Alternatively, ask the probability that West has all 13 spades (one way) out of all ways West can be dealt 13 cards: 1/C(52,13). Count the ways West can get all spades-- 1 way. Count the ways to distribute the 39 non-spade cards among the 3 other hands-- P(39; 13, 13, 13) ways. To get the probability, divide by the distributions of the 52 cards into 13-card hands-- P(52; 13, 13, 13, 13) ways. 39! (13!) 3 52! (13!) 4 = 1 52! 13!39! = 1 ( ) 52 13

12. 4/5/05Tucker, Sec. 4.312 Example 2 (part 2) What is the probability that each player has one Ace? First distribute the Aces– 4! Ways. Then distribute the 48 non-Aces— P(48; 12, 12, 12, 12) = 48! / (12!) 4 ways. So the probability that each player gets an Ace is:

13. 4/5/05Tucker, Sec. 4.313 Example 3 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 r n balls in the n th box, the number is C(r - r 1 - r 2 - …- r n + n – 1, n-1). Distribution of distinct objects Distribution of identical objects No repetition P(n,r) C(n,r) Unlimited Repetition n r C(n+r-1, r) Restricted Repetition P(n; r 1, r 2, …, r m ) -----

14. 4/5/05Tucker, Sec. 4.314 Example 3 (at least one ball in each box) First put one ball in each of the r boxes. Then count the ways to distribute without 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. Recalling that the number of ways to distribute R things without restriction into N boxes is C(N+R-1, R). back

15. 4/5/05Tucker, Sec. 4.315 C((r-r 1 -r 2 -…-r n )+n-1, (r-r 1 -r 2 -…-r n )) = C((r-r 1 -r 2 -…-r n ) +n-1, n-1) ways. Example 3 (r i balls in the i th box) Recalling that the number of ways to distribute R things without restriction into N boxes is C(N+R-1, R). First, for each I, put r i balls in the i th box. Note that there are now r-r 1 -r 2 -…-r n balls left. Finally, count the ways to distribute without restriction the remaining r-r 1 -r 2 -…-r n balls into the n boxes. This can be done in back

16. 4/5/05Tucker, Sec. 4.316 Example 4 How many integer solutions are there to the equation x 1 + x 2 + x 3 + x 4 = 12, with x i > 0? How many solutions with x i > 1? How many solutions with x 1 > 2, x 2 > 2, x 3 > 4, x 4 > 0? Distribution of distinct objects Distribution of identical objects No repetition P(n,r) C(n,r) Unlimited Repetition n r C(n+r-1, r) Restricted Repetition P(n; r 1, r 2, …, r m ) -----

17. 4/5/05Tucker, Sec. 4.317 Example 4 (Continued) An example of one solution is: x 1 = 2, x 2 = 3, x 3 = 3, x 4 = 4. Two ways to think about this: 1.Let x i be the number of (identical) objects in box i, or 2.Let x i be the number of objects of type i chosen. Either way, the number of integer solutions is C(12+4-1, 12) = 455. How many integer solutions are there to the equation x 1 + x 2 + x 3 + x 4 = 12, with x i > 0? Box 1 Box 2 Box 3 Box 4 12 identical objects

18. 4/5/05Tucker, Sec. 4.318 Example 4 (Continued) Solutions with x i > 1: First put one object in each box, and then solve x 1 + x 2 + x 3 + x 4 = 12-4 = 8. C(12-4 +(4-1), 4-1) = C(8+4-1, 4-1) = 165 ways. Solution with x 1 >2, x 2 >2, x 3 >4, x 4 >0: First put 2 objects in the first and second boxes, 4 objects in the third box, and then solve x 1 + x 2 + x 3 + x 4 =12 – (2 + 2+4) = 4. C (12 –(2+2+4) + (4-1), 4-1) = C(4+4-1, 4-1) = 35 ways. FORMULA

19. 4/5/05Tucker, Sec. 4.319 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? Distribution of distinct objects Distribution of identical objects No repetition P(n,r) C(n,r) Unlimited Repetition n r C(n+r-1, r) Restricted Repetition P(n; r 1, r 2, …, r m ) ----- Hint:

20. 4/5/05Tucker, Sec. 4.320 Solution First distribute the red sticks among the 5 children: There are C(20+5-1, 20) = 10,626 ways. Then distribute the 15 identical sticks of black. The product gives the number of ways to distribute all the candy: 10,626 * 3,876 = 41,186,376 There are C(15+5-1, 15) = 3,876 ways.

21. 4/5/05Tucker, Sec. 4.321 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? E. g. 1110111000. Distribution of distinct objects Distribution of identical objects No repetition P(n,r) C(n,r) Unlimited Repetition n r C(n+r-1, r) Restricted Repetition P(n; r 1, r 2, …, r m ) ----- Hint:

22. 4/5/05Tucker, Sec. 4.322 Solution C(6+4-1, 4-1) = C(10-1, 4-1) = 84. Thus, there are 84 such binary sequences. 1---- 0---- 1----- 0----- 1’s 0’s 4 Boxes 0’s 1’s Put one digit in each box. Then distribute the remaining 10 – 4 digits into the 4 boxes: __ __ __ (Corresponds to 1100111000.)