Generalized Permutations & Combinations: Selected Exercises

Generalized Permutations & Combinations:

Selected Exercises

Copyright © Peter Cappello 2

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?


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 1 Item 2 Item 3 Item 4 Item 5 Item 6


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?



How many ways are there to choose 36 of them?


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.


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?



How many ways are there to choose 24 of them with ≥ 2 of each kind?


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 ).


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


How many ways are there to choose 24 of them with 2 of kind 1?


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 ).


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 )


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?


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


How many ways are there to choose 24 of them with ≥ 5 of kind 1 & ≥ 3 of kind 2?


Exercise 10 (e)

There is a 1-to-1 correspondence between

• Orders of 24 – 5 – 3 objects of 6 kinds

• 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 )


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


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


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)


Exercise 20

How many integer solutions are there to the inequality

x1 + x2 + x3 11, for x1 , x2 , x3 ≥ 0?

Use the sum rule, partitioning the set of solutions.

Alternatively: Introduce an auxiliary variable x4 such that

x1 + x2 + x3 + x4 = 11, for x1 , x2 , x3 , x4 ≥ 0


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

x1 + x2 + x3 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 ).


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?


Exercise 30

How many different strings can be made from the

letters in MISSISSIPPI, using all 11 letters?



Use the product rule:

– Pick the position in the 11-letter string where the letter “M” goes:

C( 11, 1 )

– Pick the 4 positions in the 10 remaining positions where the 4 “I”s go:

C( 10, 4 )

– Pick the position in the 6 remaining positions where the 4 “S”s go:

C( 6, 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!


Generalizing Exercise 30

If you have n objects such that:

n1 objects of them are of type t1

n2 objects of them are of type t2

. . .

nk objects of them are of type tk

The # of arrangements of these objects is

C(n, n1 ) C( n - n1 , n2 ) C( n – n1 – n2 , n3 ) … C( n – n1 – … – nk-1, nk)

= n! / (n1! n2! … nk!)

(This equality is simple to verify algebraically.)


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?



Any path from (0, 0, 0, 0) to (4, 3, 5, 4) is a sequence with:4 x steps3 y steps5 z steps4 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!


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 ).


End

Copyright © Peter Cappello 2011 26

Exercise 50

How many ways are there to distribute 5

distinguishable objects in 3 indistinguishable

boxes?



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 2nd: C(5, 1).

# solutions with 3 objects in 1 box, 2 objects in 2nd: C(5, 2).

3 boxes used:

# solutions with 3 objects in 1 box, 1 object in 2nd box, 1 object in 3rd box:

C(5, 3).

# solutions with 2 objects in 1 box, 2 objects in 2nd box, 1 object in 3rd 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!


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?


60 Solution

Let the 4 other teams in the NCD be x1, x2, x3, x4.

Let the 11 other teams in the conference be y1, y2, …, y11.

Let the 16 teams in the other conference be z1, z2, …, z16.

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 x1, x2, x3, x4

3 each of y1, y2, …, y11

2 each of z1, z2, …, z16 is

81! / (4!)4(3!)11(2!)16

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Generalized Permutations & Combinations: Selected Exercises