More sophisticated counting

1
More sophisticated counting

• Advanced CountingText sections 4.5 4.6

2
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).

3
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.

4
Generalized Permutations Take 1

• Lets 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 Is, the second quotient
  of 4! mods out the effect of the unordered Ss,
  and the last quotient of 2! mods out the effect
  of the unordered Ps.

5
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 Is 2. Choose 4 slots from the
  remaining 7 for the Ss 3. Choose 2 slots
  from the remaining 3 for the Ps 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!).

6
Generalized Permutation Theorem

• Theorem Suppose a collection consists of n
  objects of whichn1 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 otherand
  n1 n2 ... nk n. Then the number of
  distinct permutations of the n objects
  isC(n,n1)C(n-n1,n2)C(n-n1-n2,n3)...C(nk,nk)
  n! / (n1!n2!n3!...nk!).

7
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 dont
  want order to matter, but we will allow
  repetitions to occur.
• This will complete the matrix of counting
  formulae, indexed by order and repetition.

8
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.

9
A Motivating Example (contd.)

• Now, each event will contain fifteen 1s, 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 1s things to be chosen and 0s
  transitions between categories.

10
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).

11
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(10029,100) 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(7029,70) 99!/70!29!

12
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.

13
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
  solutions are there to the equation a b c
  d 100.
• In this case, we could have 100 as or 99 as and
  1 b, or 98 as and 2 ds, etc.
• We see that the slots 100 and we are ranging
  over 4 categories, so transitions 3.
• Therefore, there are C(1003,100) 103!/100!3!
  non-negative solutions to a b c d 100.

14
Integer Solutions with Restrictions

• How many integer solutions are there to a b
  c d 15, when a ³ 3, b ³ 0, c ³ 2 and d ³
  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(93,9) 12!/(9!3!) solutions.

15
More Integer Solutions Restrictions

• How many integer solutions are there to a b
  c d 15, when a ³ -3, b ³ 0, c ³ -2 and d ³
  -1?
• In this case, we alter the restrictions and
  equation so that the restrictions go away. To
  do this, we need each restriction ³ 0 and balance
  the number of slots accordingly.
• Hence a ³ -33, b ³ 0, c ³ -22 and d ³
  -11,yields a b c d 15321 21
• So, there are C(213,21) 24!/(21!3!) solutions.

16
Summary

• Theorem The number of integer solutions to a1
  a2 a3 ... an r, when a1 ³ b1, a2 ³ b2,
  a3 ³ b3 , ..., an ³ bn isC(rn-1-b1-b2-b3-...-bn
  , r-b1-b2-b3-...-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).