MATH2421: Probability | Dr. YU, Chi Wai
Chapter 1: Preliminary
1 BASIC PRINCIPLE OF COUNTING
In many cases with a finite sample space, we will be able to solve a probability
problem by counting the number of elements in an event without actually listing
them. The way to solve such problems is to break them down into a series of
simple tasks that are easy to count, and employ known rules of combining tasks.
We will see how to do so in next lecture note.
In the following, some fundamental principles of counting will be presented.
Two basic counting methods:
Permutation:
An ordered arrangement of all or a part of a set of distinct objects.
Combination:
An arrangement of all or a part of a set of distinct objects, regardless of the order.
1.1 PERMUTATION
Frequently, we want to know how many different ordered arrangements of the
letters a, b, and c are possible.
By direct enumeration we see that there are 6: namely, abc, acb, bac, bca, cab
and cba. Each arrangement is known as a permutation. In other words, there are
6 possible permutations of a set of 3 distinct objects.
In general, n distinct objects can be arranged in n(n - 1)(n - 2) … (3)(2)(1)
different permutations. We denote n(n - 1)(n - 2) … (3)(2)(1) by n!. Based on
the fact that the product of no numbers at all is 1, we define 0! = 1.
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MATH2421: Probability | Dr. YU, Chi Wai
EXAMPLE
In how many ways can 5 starting positions on a basketball team be filled with 8 men who can
play any of the positions?
Ans: 𝑃58 (= 6720).
QUESTION
In how many ways can 9 people be lined up to get on a bus?
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MATH2421: Probability | Dr. YU, Chi Wai
If the boy insists in sticking with his mom, then how many ways are possible?
PERMUTATION WITH SOME INDISTINGUISHABLE OBJECTS
Up to now, we have only considered permutations of distinct objects, i.e. all the
objects were distinguishable. Now we shall determine the number of
permutations of a set of n objects when certain of the objects are
indistinguishable from each other.
To set this straight in our minds, let's take a look at the following example.
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MATH2421: Probability | Dr. YU, Chi Wai
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MATH2421: Probability | Dr. YU, Chi Wai
QUESTION
A Go tournament has 5 competitors of which 2 are Japanese, 2 are from Korea, and 1 from
China. If the tournament result lists just the nationalities of the players in the order in which
they placed, how many outcomes are possible?
1.2 COMBINATION
The permutations considered before concerned the ordered choices from a
certain set. Often, the order in which the elements are selected is not relevant,
and we are interested only in the total number of possible choices, regardless of
the particular order in which they are obtained. Such choices are referred to as
combinations.
For instance, we are interested in determining the number of groups of 3 selected
from the 5 items A, B, C, D and E. More generally, we may want to determine the
number of different groups of r objects that could be formed from a total of n
objects, i.e. the number of ways of selecting r objects from n when the order of
selection is NOT considered relevant.
To answer this, we can imagine that we partition the 5 items into two cells.
One is a "selected" cell with 3 objects and another is a "not selected" cell with 2
objects. So using Theorem 5, there are 5!/(3!2!) = 10 different groups of 3
selected from the 5 items.
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MATH2421: Probability | Dr. YU, Chi Wai
QUESTION
From a group of 5 women and 7 men, how many different committees consisting of 2 women
and 3 men can be formed?
What if 2 of the men refuse to serve on the committee together?
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MATH2421: Probability | Dr. YU, Chi Wai
2 BINOMIAL AND MULTINOMIAL THEOREM
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MATH2421: Probability | Dr. YU, Chi Wai
If we replace x + y in the binomial theorem by the sum of r variables, say
𝑥1 , … , 𝑥𝑟 , then we would have the multivariable version of the binomial theorem
--- Multinomial theorem.
EXAMPLE
Expand (𝑥1 + 𝑥2 + 𝑥3 )2 .
(𝑥1 + 𝑥2 + 𝑥3 )2
=
2!
2!
2!
2!
𝑥12 𝑥20 𝑥30 +
𝑥10 𝑥22 𝑥30 +
𝑥10 𝑥20 𝑥32 +
𝑥11 𝑥21 𝑥30
2! 0! 0!
0! 2! 0!
0! 0! 2!
1! 1! 0!
2!
2!
+
𝑥11 𝑥20 𝑥31 +
𝑥10 𝑥21 𝑥31
1! 0! 1!
0! 1! 1!
= 𝑥12 + 𝑥22 + 𝑥32 + 2𝑥1 𝑥2 + 2𝑥1 𝑥3 + 2𝑥2 𝑥3
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MATH2421: Probability | Dr. YU, Chi Wai
3 PROBLEM ON THE DISTRIBUTION OF BALLS IN BOXES
Note that if 𝑛 "distinguishable" balls are distributed into 𝑟 different boxes, then
there are 𝑟 𝑛 possible outcomes, because each ball can be distributed into any of 𝑟
possible boxes. Now suppose that all 𝑛 balls are "indistinguishable" from each
other, then how many different outcomes are possible in this case?
Since the balls are indistinguishable, it follows that the outcomes of the
experiment of distributing the 𝑛 balls into 𝑟 boxes can be described by a vector
(𝑥1 , … , 𝑥𝑟 ), where 𝑥𝑖 denotes the number of balls that are distributed into 𝑖 𝑡ℎ box.
Thus, the problem is equivalent to finding the number of distinct nonnegative
integer-valued solutions of
𝑥1 + ⋯ + 𝑥𝑟 = 𝑛.
To get the answer easily, we first consider the question with the number of
positive integer-valued solutions. Suppose that there are 𝑛 indistinguishable
objects lined up and that we want to divide them into 𝑟 nonempty groups. To do
so, we can select 𝑟 − 1 of the 𝑛 − 1 spaces between adjacent objects as our
dividing points.
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MATH2421: Probability | Dr. YU, Chi Wai
For instance, if we have 𝑛 = 8 and 𝑟 = 3, and then choose the 2 divisors as
shown
ooo|ooo|oo
then the solution (𝑥1 , 𝑥2 , 𝑥3 ) is (3, 3, 2). Similarly, if we choose the 2 divisors as
shown
o|ooo|oooo
then the solution (𝑥1 , 𝑥2 , 𝑥3 ) is (1, 3, 4), and so on.
If we list all possible outcomes, then we would see that there are 21 different
solutions. This number can be obtained by the following result:
𝑛≥𝑟
EXAMPLE
4
5−1
If 𝑛 = 5 and 𝑟 = 3, then there are (
) = ( ) = 6 possible positive integer-valued
2
3−1
solutions of 𝑥1 + 𝑥2 + 𝑥3 = 5.
Indeed, we have 5 = 1+1+3 = 1+3+1 = 3+1+1= 1+2+2= 2+2+1= 2+1+2.
Now we can go back to our original question: to obtain the number of
nonnegative solutions (𝑥1 , … , 𝑥𝑟 ) of 𝑥1 + ⋯ + 𝑥𝑟 = 𝑛. Note that the number of
nonnegative solutions of 𝑥1 + ⋯ + 𝑥𝑟 = 𝑛 is equivalent to the number of
positive solutions of 𝑦1 + ⋯ + 𝑦𝑟 = 𝑛 + 𝑟, where 𝑦𝑖 = 𝑥𝑖 + 1 for 𝑖 = 1, … , 𝑟.
Hence, from the above result, we have
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MATH2421: Probability | Dr. YU, Chi Wai
EXAMPLE (CONT’D)
7
5+3−1
If 𝑛 = 5 and 𝑟 = 3, then there are (
) = ( ) = 21 possible nonnegative integer2
3−1
valued solutions of 𝑥1 + 𝑥2 + 𝑥3 = 5.
Indeed, we have 5 = 1+1+3 = 1+3+1 = 3+1+1= 1+2+2= 2+2+1= 2+1+2
= 0+0+5 = 0+5+0 = 5+0+0
= 1+0+4 = 1+4+0 = 0+1+4 = 0+4+1 = 1+0+4 = 4+0+1
= 2+0+3 = 2+3+0 = 0+2+3 = 0+3+2 = 2+0+3 = 3+0+2
QUESTION
20 identical red balls are distributed into 4 different boxes. How many different divisions are
possible if (a) No box is empty, and (b) Boxes can be empty.
Challenging question for students:
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