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Generalized Permutations & Combinations: Selected Exercises Slide 2 Copyright Peter Cappello2 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? Slide 3 Copyright Peter Cappello3 Exercise 10 (a) Solution How many binary strings of length 12 + 6 1 are there with exactly 12 0s? (= # of 17-bit binary strings with exactly 5 1s.) C( 12 + 6 1, 6 1 ) = C( 12 + 6 1, 12 ). How many ways are there to order 12 items from a menu of 6 kinds of items? 1-to1 correspondence between these binary strings & orders. Item 1Item 2Item 3Item 4Item 5Item 6 Slide 4 Copyright Peter Cappello4 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? Slide 5 Copyright Peter Cappello5 Exercise 10 (b) Solution How many binary strings of length 36 + 6 1 are there with exactly 36 0s? (= # of length-41 binary strings with 5 1s.) 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-to1 correspondence between binary strings & orders. Slide 6 Copyright Peter Cappello6 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? Slide 7 Copyright Peter Cappello7 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 ). Slide 8 Copyright Peter Cappello8 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? Slide 9 Copyright Peter Cappello9 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 ). Slide 10 Copyright Peter Cappello10 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 ) Slide 11 Copyright Peter Cappello11 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? Can you give a combinatorial argument for it? Slide 12 Copyright Peter Cappello12 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? Slide 13 Copyright Peter Cappello13 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 & 2 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 ) Slide 14 Copyright Peter Cappello14 Exercise10 (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 Slide 15 Copyright Peter Cappello15 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) Slide 16 Copyright Peter Cappello16 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 Slide 17 Copyright Peter Cappello17 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 ). Slide 18 Copyright Peter Cappello18 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? Can you give a combinatorial argument for it? Slide 19 Copyright Peter Cappello19 Exercise 30 How many different strings can be made from the letters in MISSISSIPPI, using all 11 letters? Slide 20 Copyright Peter Cappello20 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 Is go: C( 10, 4 ) 3.Pick the position in the 6 remaining positions where the 4 Ss go: C( 6, 4 ) 4.Pick the position in the 2 remaining positions where the 2 Ps go: C( 2, 2 ) = 1 Answer: 11! / 1!4!4!2! Slide 21 Copyright Peter Cappello21 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... n k objects of them are of type t k 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 n k-1, n k ) = n! / (n 1 ! n 2 ! n k !) (This equality is simple to verify algebraically.) Slide 22 Copyright Peter Cappello22 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? Slide 23 Copyright Peter Cappello23 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! Slide 24 Copyright Peter Cappello24 Exercise 40 This is a generalization of Pascals 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 ). Slide 25 Copyright Peter Cappello25 End Slide 26 Copyright Peter Cappello 201126 Exercise 50 How many ways are there to distribute 5 distinguishable objects in 3 indistinguishable boxes? Slide 27 Copyright Peter Cappello 201127 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! Slide 28 Copyright Peter Cappello 201128 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? Slide 29 Copyright Peter Cappello 201129 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! /