The Division Rule• Theorem: Suppose a set A has n elements and is

partitioned by the collection {A1, A2, ..., Ap}, where each partition set has m elements. Then:

p = n/m.

• In other words, if a set is partitioned into equal-sized partition sets, then the number of partition sets is the quotient of the size of the set with the size of any partition set.

• For example, if a set has 100 elements and is partitioned in 20-element subsets, then there must be 5 subsets (equivalence classes).

Generalized Permutations• Permutations teach us how to count the number of

orderings of the letters of COMPUTER (8!). What about the number of orderings of the letters of MISSISSIPPI?

• In this case, we note that not all the letters are distinct. In particular,

MISSISSIPPI IIIISSSSPPM, so although we are still searching for an ordering structures, there are sub-unorderings present, induced by the repeated letters, for which we have to account.

Generalized Permutations Take 1• Let’s apply the Division Rule to negate the effect

of the unordering portions of the overall order problem.

• This leaves us with a total count of 11!/4!4!2!.

• Here, the first quotient of 4! “mods” out the effect of the unordered I’s, the second quotient of 4! “mods” out the effect of the unordered S’s, and the last quotient of 2! “mods” out the effect of the unordered P’s.

Generalized Permutations Take 2• If we model this problem, purely as a combination,

and not a permutation at all, we can reason the task as:

1. Choose 4 slots from 11 for the I’s;2. Choose 4 slots from the remaining 7 for the S’s;3. Choose 2 slots from the remaining 3 for the P’s;4. Place the M (only 1 way remaining).

• This yields: C(11,4)C(7,4)C(3,2)C(1,1) =(11!7!3!)/(7!4!4!3!2!1!) = 11!/(4!4!2!).

Generalized Permutation Theorem• Theorem: Suppose a collection consists of n

objects of which:n1 are of type 1, indistinguishable from each other; n2 are of type 2, indistinguishable from each other; ...nk are of type k, indistinguishable from each other;and n1 + n2 + ... + nk = n. Then the number of distinct permutations of the n objects is:C(n,n1)C(nn1,n2)C(nn1n2,n3)...C(nk,nk) =

n! / (n1!n2!n3!...nk!).

Combinations with Repetition

• In the last section, we saw how to count combinations, where order does not matter, based on permutation counts, and we saw how to count permutations where repetitions occur.

• Now, we shall consider the case where we don’t want order to matter, but we will allow repetitions to occur.

• This will complete the matrix of counting formulae, indexed by order and repetition.

A Motivating Example• How many ways can I select 15 cans of soda

from a cooler containing large quantities of Coke, Pepsi, Diet Coke, Root Beer and Sprite?

• We have to model this problem using the chart: Coke Pepsi Diet Coke Root Beer Sprite

A: 111 111 111 111 111 =15

B: 11 111111 111111 1 =15

C: 1111 1111111 1111 =15

• Here, we set an order of the categories and just count how many from each category are chosen.

A Motivating Example (cont’d.)• Now, each event will contain fifteen 1’s, but we

need to indicate where we transition from one category to the next. If we use 0 to mark our transitions, then the events become:

A: 1110111011101110111

B: 1100111111011111101

C: 0011110111111101111

• Thus, associated with each event is a binary string with #1’s = #things to be chosen and #0’s = #transitions between categories.

Counting Generalized Combinations

• From this example we see that the number of ways to select 15 sodas from a collection of 5 types of soda is C(15 + 4,15) = C(19,15) = C(19,4).

• Note that #zeros = #transitions = #categories 1.

• Theorem: The number of ways to fill r slots from n catgories with repetition allowed is:

C(r + n 1, r) = C(r + n 1, n 1).

• In words, the counts are:C(#slots + #transitions, #slots)

or C(#slots + #transitions, #transitions).

Another Example

• How many ways can I fill a box holding 100 pieces of candy from 30 different types of candy?

Solution: Here #slots = 100, #transitions = 30 1,so there are C(100+29,100) = 129!/(100!29!) different ways to fill the box.

• How many ways if I must have at least 1 piece of each type?

Solution: Now, we are reducing the #slots to choose over to (100 30) slots, so there are C(70+29,70) = 99!/70!29!

When to Use Generalized Combinations

• Besides categorizing a problem based on its order and repetition requirements as a generalized combination, there are a couple of other characteristics which help us sort:

– In generalized combinations, having all the slots filled in by only selections from one category is allowed;

– It is possible to have more slots than categories.

Integer Solutions to Equations• One other type of problem to be solved by the

generalized combination formula is of the form:How many non-negative integer solutionsare there to the equation a + b + c + d =

100.• In this case, we could have 100 a’s or 99 a’s and 1

b, or 98 a’s and 2 d’s, etc.• We see that the #slots = 100 and we are ranging

over 4 categories, so #transitions = 3.• Therefore, there are C(100+3,100) = 103!/100!3!

non-negative solutions to a + b + c + d = 100.

Integer Solutions with Restrictions

• How many integer solutions are there to:a + b + c + d = 15,

when a 3, b 0, c 2 and d 1?

• Now, solution “strings” are 111a0b011c01d, where the a,b,c,d are the remaining numbers of each category to fill in the remaining slots.

• However, the number of slots has effectively been reduced to 9 after accounting for a total of 6 restrictions.

• Thus there are C(9+3,9) = 12!/(9!3!) solutions.

More Integer Solutions & Restrictions

• How many integer solutions are there to:a + b + c + d = 15,

when a 3, b 0, c 2 and d 1?

• In this case, we alter the restrictions and equation so that the restrictions “go away.” To do this, we need each restriction 0 and balance the number of slots accordingly.

• Hence a 3+3, b 0, c 2+2 and d 1+1,yields a + b + c + d = 15+3+2+1 = 21

• So, there are C(21+3,21) = 24!/(21!3!) solutions.

Summary

• Theorem: The number of integer solutions to:a1 + a2 + a3 +...+ an = r,

when a1 b1, a2 b2, a3 b3 , ..., an bn isC(r+n1b1b2b3...bn , rb1b2b3...bn).

• Theorem: The number of ways to select r things from n categories with b total restrictions on the r things is C(r + n 1 b , r b).

• Corollary: The number of ways to select r things from n categories with at least 1 thing from each category is C(r 1 , r n) (set b = n).