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Generalized Permutations & Combinations: Selected Exercises

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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?

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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

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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?

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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.

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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?

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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 ).

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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?

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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 ).

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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 )

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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?

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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?

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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 )

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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

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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)

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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

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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 ).

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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?

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Copyright © Peter Cappello19 Exercise 30 How many different strings can be made from the letters in MISSISSIPPI, using all 11 letters?

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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 “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!

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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.)

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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?

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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!

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Copyright © Peter Cappello24 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 ).

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Copyright © Peter Cappello25 End

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Copyright © Peter Cappello 201126 Exercise 50 How many ways are there to distribute 5 distinguishable objects in 3 indistinguishable boxes?

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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!

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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?

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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! / (4!) 4 (3!) 11 (2!) 16

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