Combinatorial Analysis

1
Chapter 1

• Combinatorial Analysis

• Experiment
• The Basic Principle of Counting
• Multiple Principle
• Permutations
• Combinations
• Multinomial Coefficients

2
Objectives

• To understand the basic principles of counting
• To familiarize with permutations, combinations
  and multinomial coefficients

3
Experiment

• Experiment is any procedure that
• can be continuously repeated by an infinite
  number of times theoretically,
• and has a well-defined set of possible outcomes.
• Two kinds of experiments
• Deterministic experiment
• Conditions of this experiment completely
  determined the outcomes.
• Random experiment
• Conditions of this experiment only partially
  determined the outcomes.
• E.g. Repeatable experiments, experiments
  with variable outcomes, statistical regularity

4
The Basic Principle of Counting

• Suppose two experiments are performed. If the
  first experiment can result in one of m possible
  outcomes and for each outcome of experiment one
  there are n possible outcomes of the second
  experiment, the total possible outcome of the two
  overall experiments is m x n.
• Proof by enumerating all possible outcomes of
  the two experiments.
• E.g. (1, 1), (1, 2), ,(1, n)
• (2, 1), (2, 2), ,(2, n)
• .
• .
• (m, 1), (m, 2), ,(m, n)

5
The Generalized Basic Principle of Counting

• Suppose r experiments are performed, so that the
  first experiment can result in one of n1 possible
  outcomes.
• If for each outcomes of experiment one there are
  n2 possible outcomes of the second experiment ,
  and for each outcomes of first two experiments
  there are n3 possible outcomes of the third
  experiment. until the r th experiment, hence
  there are all together n1 x n2 x n3 x nr
  possible outcomes of the overall r experiments.

6

• Partition
• Partitions are collections of subsets which are
  jointly exhaustive and mutually exhaustive.
• E.g. If three dies are thrown, there are three
  defined subsets which include three as a kind,
  one pair and all different.

• Addition Principle
• The addition principle can be applied when we
  have to count the number of total outcomes in
  several partitions.
• E.g. If three as a kind includes 6 outcomes,
  one pair includes 30 outcomes and all
  different includes 20 outcomes, the total number
  of set outcomes of throwing three dies are
  63020 56.

e.g. If a coin and a die are thrown, there are
all together 2 x 6 12 outcomes
7
Multiple Principle

• The multiple principle can be applied when we
  have to count the number of total outcomes in
  several independent events.

• Ex 1.1
• A college planning committee consists of 4
  freshmen, 6 sophomores, 9 seniors and 15
  postgraduates. Now, a subcommittee of 4,
  consisting of 1 person from each class, is to be
  chosen. How many different subcommittees are
  possible?
• Ans
• We can consider each process of choosing a
  single representative from each class as
  different single experiments.
• By applying the generalized basic principle of
  counting, there are 4 x 6 x 9 x 15 3240
  possible subcommittees.

8
Permutations

• Total number of different arrangement of order
  for distinct objects
• For instance, if we have to draw r objects from
  n distinct objects without replacement, the
  sequence outcomes is nPr , or if we put r
  distinct objects into n different boxes without
  repetition, the order outcome is also nPr .
• Proof
• By the basic principle of counting, there are
  all together n(n-1)(n- 2)(n- r1) outcomes

9
Ex 1.2 Suppose we have n objects, the total
number of permutation is n! n(n-1)(n-2)1
for n3, (A, B, C) (A, C, B) (B, A, C) (B,
C, A) (C, A, B) (C, B, A) Pn n! 6. Ex 1.3
How many different orders are possible for
eight students queueing in a line? Ans 8!
10

• Ex 1.4
• Fourteen books are on a bookshelf which include
  six mathematics books, four logic books, three
  art books and one music book. If it is known that
  all the books of the same subject will be put
  next to each other, how many different
  arrangements are possible?
• Ans
• no. of arrangements of putting the books in the
  same type n!
• no. of arrangements of putting four types of
  books in the bookshelf 4!
• The total number of arrangements of all the
  books 4! (6!4!3!1!)
• Ex 1.5
• Suppose there are five colleges in CUHK. A
  principal from another university want to visit
  exactly three colleges. How many arrangements of
  tour routes can be designed to the guests?
• Ans
• 60 arrangements

11

• Ex 1.6 How many different letter arrangements
  can be formed by BCCCABD?
• Ans
• There are 7! permutations if all seven letters
  are different. Considering a single permutation,
  for example ABBCCCD, there are total of 3!2!1!1!
  repeated permutations.
• AB1B2C1C2C3D AB2B1C1C2C3D
• AB1B2C2C1C3D AB2B1C2C1C3D
• AB1B2C3C2C1D AB2B1C3C2C1D
• AB1B2C1C3C2D AB2B1C1C3C2D
• AB1B2C3C1C2D AB2B1C3C1C2D
• AB1B2C2C3C1 D AB2B1C2C3C1D
• Hence there are 7!/3! 2! 84 different letter
  arrangements
• Generally, there are n!/(n1!n2!...nr! )
  permutations of n objects, of which n1 are alike,
  n2 are alike, ,nr are alike

12
Extension

• For instance, if we have to draw r objects
  from n distinct objects with replacement, the
  sequence outcome is nr, or if we put r distinct
  objects into n different boxes with repetition,
  the order outcome is also nr
• Proof
• By the basic principle of counting, there are
  all together n(n)(n)(n) nr outcomes
• Ex 1.7
• If seven dies are thrown, how many sequence
  outcomes are there?
• Ans 67

13
Combination

• Total number of different ways of drawing
  distinct objects
• For instance, if we have to draw r objects from
  n distinct objects without replacement, the set
  outcome is nCr , or if we put r distinct objects
  into n different boxes without repetition, the
  order outcome is also nCr .
• Proof
• By the basic principle of counting, there are
  all together n(n-1)(n-2)(n-r1) different ways
  that a group of r objects selected from n objects
  when the order of selection is relevant.
• For each way of selecting r objects, there are
  r! Arrangements of order for the r objects
• Hence the total outcomes

14
Ex 1.8 How many different ways are possible
for selecting three representatives from eight
students? Ans Ex 1.9 How many
different ways are there if we draw three balls
from a box containing eleven distinct balls?
Ans
15
Ex 1.10 There are n students. We want to select
r of them to form a team. How many different
teams we can form. A useful
combinatorial identity select r-1 from
n-1, (student n is in the team) select r
from n-1, (student n not in the team)
16
Binomial Theorem. Ex 1.11
17
Multinomial Coefficients

• Considering there is a set of n distinct items,
  they are divided into r distinct groups of sizes
  n1, n2, nr ,where . How many
  different divisions are possible?
• No. of choices for the first group with sizes n1
• For each choice of the first group
• No. of choices for the second group with sizes
  n2
• For each choice of the first two group
• No. of choices for the third group with sizes n3
• ...

18

• Hence, by the generalization of the basic
  counting principle, the number of possible
  divisions are
• The case of expanding ( x1 x2 xr )n is
  similar to the previous example.

Multinomial coefficients
19
On the distribution of balls in urns.
distinguishable rn possible
outcomes. Each ball can be distributed into any
of r possible urns.
1
2
3
n
1
2
r
20
What if n balls are indistinguishable? xi
number of balls in the i th urn. Then the
problem reduces to finding the number of distinct
positive integer-valued solution (xi gt 0 for all
i ) x1x2xr n For n 8, r 3 0
0 0 0 0 0 0 0 x13, x23, x32 Select r-1
divisions out of n-1 spaces between adjacent
objects The number of selections is
21
For xi 0, Consider the balls are divided as a
sequence of 0, 1 string. 0011011101 There
are nr-1 numbers, and n out of these numbers
are 0, the total number of combination is Ex
1.12 How many non-negative integer solution
of which is the number of positive solutions
of (let ).
Thus There are
selections