Generalized Permutations Combinations Selected Exercises Exercise 10 a

Generalized Permutations & Combinations: Selected Exercises

Exercise 10 (a) A croissant shop has 6 kinds of croissants: plain, cherry, chocolate, almond, apple, & broccoli. How many ways are there to choose 12 croissants? Abstract version of the problem There is an infinite supply of 6 kinds of objects. How many ways are there to choose 12 of them? I. e, how many multisets of them are there of size 12? Copyright © Peter Cappello 2

Exercise 10 (a) Solution Item 1 Item 2 Item 3 Item 4 Item 5 Item 6 1. How many distinct orders 12 items are there from a menu of 6 kinds? 2. How many binary strings of length 12 + 6 – 1 are there with exactly 12 0 s? (= # of 17 -bit binary strings with exactly 5 1 s. ) C( 12 + 6 – 1, 6 – 1 ) = C( 12 + 6 – 1, 12 ). 1 -to-1 correspondence between these binary strings & orders. Copyright © Peter Cappello 3

Exercise 10 (b) A croissant shop has 6 kinds of croissants: plain, cherry, chocolate, almond, apple, & broccoli. How many ways are there to choose 36 croissants? Abstract version of the problem There is an infinite supply of 6 kinds of objects. How many ways are there to choose 36 of them? Copyright © Peter Cappello 4

Exercise 10 (b) Solution How many binary strings of length 36 + 6 – 1 are there with exactly 36 0 s? (= # of length-41 binary strings with 5 1 s. ) C( 36 + 6 – 1, 6 – 1 ) = C( 36 + 6 – 1, 36 ). How many ways are there to order 36 items from a menu of 6 kinds of items? 1 -to-1 correspondence between binary strings & orders. Copyright © Peter Cappello 5

Exercise 10 (c) A croissant shop has 6 kinds of croissants: plain, cherry, chocolate, almond, apple, & broccoli. How many ways are there to choose 24 croissants with ≥ 2 of each kind? Abstract version of the problem There is an infinite supply of 6 kinds of objects. How many ways are there to choose 24 of them with ≥ 2 of each kind? Copyright © Peter Cappello 6

Exercise 10 (c) Solution How many ways are there to order 24 – 2. 6 = 12 items from a menu of 6 kinds of items? There is a 1 -to-1 correspondence between this set of orders (multisets) & the original set of orders: For each order of 12 items: For each kind of item, increment the order of that kind by 2. The resulting order has 24 items, ≥ 2 of each kind of item. Answer: C( 24 – 2*6 + 6 - 1, 6 – 1 ). Copyright © Peter Cappello 7

Exercise 10 (d) A croissant shop has 6 kinds of croissants: plain, cherry, chocolate, almond, apple, & broccoli. How many ways are there to choose 24 croissants with 2 broccoli? Abstract version There is an infinite supply of 6 kinds of objects. How many ways are there to choose 24 of them with 2 of kind 1? Copyright © Peter Cappello 8

Exercise 10 (d) Solution Use the sum rule: Partition the set of orders based on the # of broccoli croissants: 1. Count the solutions with exactly 0 broccoli croissants C( 24 + 5 - 1, 5 – 1 ). 2. Count the solutions with exactly 1 broccoli croissant Pick 23 croissants from the remaining 5 kinds of croissants: C( 24 - 1 + 5 - 1, 5 – 1 ). 3. Count the solutions with exactly 2 broccoli croissants Pick 22 croissants from the remaining 5 kinds of croissants: C( 24 - 2 + 5 - 1, 5 – 1 ). Copyright © Peter Cappello 9

Exercise 10 (d) Better Solution 1. Count all orders of 24 croissants: C( 24 + 6 – 1, 6 – 1 ) 2. Subtract the “bad” orders: Order 24 - 3 croissants from all 6 varieties: C( 24 - 3 + 6 – 1, 6 – 1 ) Answer: C( 24 + 6 – 1, 6 – 1 ) – C( 24 - 3 + 6 – 1, 6 – 1 ) Copyright © Peter Cappello 10

Think Like a Mathematician C( 28, 4 ) + C( 27, 4 ) + C( 26, 4 ) = C( 29, 5 ) – C( 26, 5 ) What is a possible generalization of this equation? 1. Generalize with respect to total number of objects (croissants) 2. Generalize with respect to number of kinds of objects (croissants) 3. Generalize with respect to the number limit of kind 1 (broccoli). Can you give a combinatorial argument for it? Copyright © Peter Cappello 11

Exercise 10 (e) A croissant shop has 6 kinds of croissants: plain, cherry, chocolate, almond, apple, & broccoli. How many ways are there to choose 24 with ≥ 5 chocolate & ≥ 3 almond? Abstract version There is an infinite supply of 6 kinds of objects. How many ways are there to choose 24 of them with ≥ 5 of kind 1 & ≥ 3 of kind 2? Copyright © Peter Cappello 12

Exercise 10 (e) There is a 1 -to-1 correspondence between 1. Orders of 24 – 5 – 3 objects of 6 kinds 2. Orders of 24 objects of 6 kinds with ≥ 5 of kind 1 & ≥ 3 of kind 2. Correspondence: Add 5 objects of kind 1 & 3 objects of kind 2 to each order of type 1 to get an order of type 2. We count the # of orders of type 1: Answer: C( 24 – 5 – 3 + 6 – 1, 6 – 1 ) Copyright © Peter Cappello 13

Exercise 10 (f) A croissant shop has 6 kinds of croissants: plain, cherry, chocolate, almond, apple, & broccoli. How many ways are there to choose 24 with: • ≥ 1 plain & ≥ 2 cherry & ≥ 3 chocolate & ≥ 1 almond & ≥ 2 apple • 3 broccoli? Abstract version There is an infinite supply of 6 kinds of objects. How many ways are there to choose 24 of them with: – ≥ 1 of kind 1 & ≥ 2 of kind 2 & ≥ 3 of kind 3 & ≥ 1 of kind 4 & ≥ 2 of kind 5 – 3 of kind 6 Copyright © Peter Cappello 14

Exercise 10 (f) 1. Put 1 plain & 2 cherry & 3 chocolate & 1 almond & 2 apple to the side (9 objects to the side) 2. There are 24 – 9 = 15 left to distribute w/o restriction on broccoli. C( 15 + 6 – 1, 6 – 1 ) 3. Subtract the “bad” orders (≥ 4 broccoli): C( 15 - 4 + 6 – 1, 6 – 1) Answer: C( 15 + 6 – 1, 6 – 1 ) – C( 15 - 4 + 6 – 1, 6 – 1) Copyright © Peter Cappello 15

Exercise 20 How many integer solutions are there to the inequality x 1 + x 2 + x 3 11, for x 1 , x 2 , x 3 ≥ 0? Use the sum rule, partitioning the set of solutions. Alternatively: Introduce an auxiliary variable x 4 such that x 1 + x 2 + x 3 + x 4 = 11, for x 1 , x 2 , x 3 , x 4 ≥ 0 Copyright © Peter Cappello 16

Exercise 20 Solution There is a 1 -to-1 correspondence between: The set of solutions to the equality The set of solutions to the inequality x 1 + x 2 + x 3 from the equality satisfy the inequality. Equivalent to counting solutions to the equality: How many ways are there to order 11 items from a menu of 4 kinds of items? Answer: C( 11 + 4 – 1, 4 – 1 ). Copyright © Peter Cappello 17

Think Like a Mathematician C(2, 2) + C(3, 2) + C(4, 2) +. . . + C(13, 2) = C(14, 3) What is a possible generalization of this equation? 1. Generalize with respect to number of objects (11) 2. Generalize with respect to number of kinds (3) Can you give a combinatorial argument for it? Copyright © Peter Cappello 18

Exercise 30 How many different strings can be made from the letters in MISSISSIPPI,

Exercise 30 Solution Use the product rule: 1. Pick the position in the 11 -letter string where the letter “M” goes: C( 11, 1 ) 2. Pick the 4 positions in the 10 remaining positions where the 4 “I”s go: C( 10, 4 ) 3. Pick the position in the 6 remaining positions where the 4 “S”s go: C( 6, 4 ) 4. Pick the position in the 2 remaining positions where the 2 “P”s go: C( 2, 2 ) = 1 Answer: 11! / 1!4!4!2! Copyright © Peter Cappello 20

Generalizing Exercise 30 If you have n objects such that: n 1 objects of them are of type t 1 n 2 objects of them are of type t 2. . . nk objects of them are of type tk The # of arrangements of these objects is C(n, n 1 ) C( n - n 1 , n 2 ) C( n – n 1 – n 2 , n 3 ) … C( n – n 1 – … – nk-1, nk) = n! / (n 1! n 2! … nk!) (This equality is simple to verify algebraically. ) Copyright © Peter Cappello 21

Exercise 40 How many ways are there to travel in xyzw space from the origin ( 0, 0, 0, 0 ) to ( 4, 3, 5, 4 ) by taking steps: 1 unit in the positive x direction 1 unit in the positive y direction, 1 unit in the positive z direction 1 unit in the positive w direction? Copyright © Peter Cappello 22

Exercise 40 Solution Any path from (0, 0, 0, 0) to (4, 3, 5, 4) is a sequence with: 4 x steps 3 y steps 5 z steps 4 w steps. Equivalent problem: How many 4 + 3 + 5 + 4 = 16 letter sequences of x, y, z, & w are there with exactly 4 x, 3 y, 5 z, 4 w? Answer: 16! / 4! 3! 5! 4! Copyright © Peter Cappello 23

Exercise 40 • This is a generalization of Pascal’s triangle, viewed as block walking. • Travel from ( 0, 0 ) to ( j, k ) by taking steps: 1 unit in the positive x direction 1 unit in the positive y direction Sequences of j xs & k ys: ( j + k )!/j!k! = C( j + k, j ) = C( j + k, k ). Copyright © Peter Cappello 24

Exercise 50 How many ways are there to distribute 5 distinguishable objects in 3

Exercise 50 Solution Use the sum rule to partition the set of solutions: 1 box used: # solutions with 5 objects in 1 box: 1. 2 boxes used: # solutions with 4 objects in 1 box, 1 object in 2 nd: C(5, 1). # solutions with 3 objects in 1 box, 2 objects in 2 nd: C(5, 2). 3 boxes used: # solutions with 3 objects in 1 box, 1 object in 2 nd box, 1 object in 3 rd box: C(5, 3). # solutions with 2 objects in 1 box, 2 objects in 2 nd box, 1 object in 3 rd box: C(5, 1) C(4, 2)/2! Answer: 1 + C(5, 1) + C(5, 2) + C(5, 3) + C(5, 1) C(4, 2)/2! Copyright © Peter Cappello 2011 27

60 Suppose a basketball league has 32 teams, split into 2 conferences of 16 teams each. Each conference is split into 3 divisions. Suppose that the North Central Division (NCD) has 5 teams. Each team in this division plays: 4 games against each of the other 4 teams in its division 3 games against each of the 11 remaining teams in its conference 2 games against each of the 16 teams in the other conference. In how many different orders can the games of a team in the NCD be scheduled? Copyright © Peter Cappello 2011 28

60 Solution Let the 4 other teams in the NCD be x 1, x 2, x 3, x 4. Let the 11 other teams in the conference be y 1, y 2, …, y 11. Let the 16 teams in the other conference be z 1, z 2, …, z 16. The total # of games that a team in the NCD plays is 4. 4 + 11. 3 + 16. 2 = 81 The number of 81 -“letter’’ sequences with: 4 each of x 1, x 2, x 3, x 4 3 each of y 1, y 2, …, y 11 2 each of z 1, z 2, …, z 16 is 81! / (4!)4(3!)11(2!)16 Copyright © Peter Cappello 2011 29