# Handout 2 - Winona State University ```As we’ve seen above, listing out every possible outcome/element for a particular scenario/sample space can be
impossible and/or unnecessary at times. The following methods will help us to determine the number of possible
outcomes/elements for a particular scenario/sample space without having to list out every possibility in the process.
There are four main methods for counting the number of possible outcomes/elements.
Without Replacement
With Replacement
Ordered
Unordered
Example: Let’s suppose we have a bowl containing four balls numbered 1, 2, 3, and 4 as shown below.
Suppose we want to draw two balls from the bowl. For example,
the second ball drawn was a “3.”
means the first ball drawn was a “2” and
Questions:
16. What situation from the table above do you think would result in the possible outcomes given below?
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17. How does the situation depicted below differ from the one in Question 16?
18. What situation from the table above do you think would result in possible outcomes given in Question 17?
19. How does the situation depicted below differ from the one in Question 16?
20. What situation from the table above do you think would result in possible outcomes given in Question 19?
21. How does the situation depicted below differ from the one in Question 16?
22. What situation from the table above do you think would result in possible outcomes given in Question 21?
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Example: Suppose that in a given state’s lottery, there are 44 numbers and each player chooses 6 numbers. How
many tickets can be made? This question can be answered in a number of ways, depending on the rules of the
lottery system.

Ordered, without replacement
Example:
Find the number of possible lottery tickets:
Proof: Let’s assume we want to select r objects from an urn containing n distinguishable objects.
Step 1: Select the first object.
Step 2: Select the second object.
Step 3: Select the third object.
Step r: Select the rth object.
Formula:
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
Ordered, with replacement –
Example:
Find the number of possible lottery tickets:
Proof: Let’s assume we want to select r objects from an urn containing n distinguishable objects.
Step 1: Select the first object. Return the object to the urn.
Step 2: Select the second object. Return the object to the urn.
Step 3: Select the third object. Return the object to the urn.
Step r: Select the rth object. Return the object to the urn.
_____ x _____ x _____ x
x _____ x _____ =
Formula:
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
Unordered, without replacement –
Example:
Find the number of possible lottery tickets:
Proof: Let’s assume we want to select r objects from an urn containing n distinguishable objects.
Let’s start with an example. Suppose we have an urn which consists of n = 8 balls, numbered from 1 to 8. From this
urn we would like to select r = 3 balls without replacement where order is important.
Question:
23. How many possible permutations are possible?
Therefore, an exhaust list of all possible permutations would give us a table that looks something like this:
(1,2,3)
(1,3,2)
(2,1,3)
(2,3,1)
(3,1,2)
(3,2,1)
(1,2,4)
(1,4,2)
(2,1,4)
(2,4,1)
(4,1,2)
(4,2,1)
⋮
⋮
⋮
⋮
⋮
⋮
(2,5,8)
(2,8,5)
(5,2,8)
(5,8,2)
(8,2,5)
(8,5,2)
⋮
⋮
⋮
⋮
⋮
⋮
(6,7,8)
(6,8,7)
(7,6,8)
(7,8,6)
(8,6,7)
(8,7,6)
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Looking at the table above, we can see that
# permutations =
Formula:
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
Unordered, with replacement –
Example:
Find the number of possible lottery tickets:
Proof: Let’s assume we want to select r objects from an urn containing n distinguishable objects.
In order to show how this works, let’s assume n = 4 and r = 3.
We are going to start out by considering the question of counting all arrangements in a line that consists of (n – 1) 0’s
and (n – 2 + r) X’s, such that no two 0’s are adjacent.
We are going to take two approaches towards solving this problem. The plan is to find the answer twice and equate
the answers. The two answers may look different, but logically they have to be equal. (It is logically impossible to
have two unequal but correct answers.) The idea of equating two seemingly different answers is a standard method
for proving combinatorial identities. In this problem we will use this approach to derive a formula, which is a bit
unusual, but mathematically valid.
Approach 1
Step 1: Put down the (n – 2 + r) X’s.
Step 2: Put an empty urn before, between and after each X. Therefore, a total of _________
urns are needed.
Step 3: Distribute the identical 0’s into the urns subject to the restriction that at most one 0 can
be in each urn. This restriction is necessary because we require that no two 0’s can be
Note: The urns in Step 3 are distinguishable from each other because we can
distinguish between the patterns 0XX and X0X, for example. This reduces
Step 3 to determining how many ways there are to assign n - 1 identical
objects to n + r – 1 urns, which is a matter of determining the number of
combinations.
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Approach 2
Step 1: Put down the (n – 1 ) 0’s.
Step 2: Put an empty urn before, between and after each 0. A total of _______ will be needed.
Step 3: Distribute the identical X’s into the urns subject to the restriction that at least one X in
every urn except the urns on either end. The urns on either end could be (but are not
necessarily) empty. These restrictions are necessary because we require that no two 0’s
Note: To accomplish Step 3, it helps to break it down further.
Step 3a: Put a single X in every urn EXCEPT the urns on either end.
Step 3b: Distribute the remaining (n– 2 + r) – (n – 2) = ____ X’s into the
n urns without any restrictions.
Formula:
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Without Replacement
With Replacement
Ordered
This is known as a permutation.
Unordered
This is known as a combination.
Questions:
24. A brand of automobile comes in five different styles, with four types of engines, with two types of
transmissions, and in eight colors.
a. How many autos would a dealer have to stock if he included one for each style-engine-transmission
combination?
b. How many would a distribution center have to carry if all colors of cars were stocked for each
combination in part a?
25. A personnel director for a corporation has hired ten new engineers. If three (distinctly different) positions
are open at a Cleveland plant, in how many ways can she fill the positions?
26. In how many ways can 3 novels, 2 mathematics books and 1 chemistry book be arranged on a shelf if…
c. the books can be arranged in any order?
d. the mathematics books must be together and the novels must be together?
e. the novels must be together, but the other books can be arranged in any order?
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27. Ten teams are playing in a basketball tournament. In the first round, the teams are randomly assigned to
games 1, 2, 3, 4, and 5. In how many ways can the teams be assigned to games?
28. An experimenter wishes to investigate the effect of three variables – pressure, temperature, and the type of
catalyst – on yield in a refining process. If the experimenter intends to use three settings each for
temperature and pressure and two types of catalysts, how many experimental runs will have to be
conducted if he wishes to run all possible combinations of pressure, temperature and types of catalysts?
29. A fleet of nine taxis is to be dispatched to three airports in such a way that three go to airport A, five go to
airport B and one goes to airport C. In how many distinct ways can this be accomplished?
30. A student is to answer 7 out of 10 questions in an examination.
f.
How many choices does the student have?
g. How many choices does the student have if they must answer at least 3 of the first 5 questions?
31. How many ways can you arrange the letters in FLUKE?
32. How many ways can you arrange the letters in MISSISSIPPI?
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Binomial Theorem
n
n
(x + y)n =   xk yn-k
k=0  r 
Questions:
33. Expand (x + y)3.
34. If we wish to expand (x + y)8, what is the coefficient of x5y3?
The number of possible divisions of n distinct objects into r distinct groups of respective sizes n1, n2, …, nr equals
n!
.
n1!n2! nr!
Therefore, the number of distinct “words” that can be formed using n total letters of r distinct types if you have n1
letters of the first type, n2 letters of the second type, …, nr letters of the r type so that
n = n1 + n2 + … + nr equals
n


n!
.

=
n
,
n
,
n
n
!n
nr!
r
 1 2
1 2!
Questions:
35. A police department in a small city consists of 10 officers. If the department policy is to have 5 of the officers
patrolling the streets, 2 of the officers working full time at the station and 3 of the officers on reserve at the
station, how many different divisions of the 10 officers into the 3 groups are possible?
36. How many different signals, each consisting of 9 flags hung in a line, can be made from a set of 4 white flags,
3 red flags and 2 blue flags if all the flags of the same color are also identical?
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Multinomial Theorem
 x1 + x 2 +
+ xr  =
n

(n1 ,n2 ,
n1 + n2 +
,nr ):
+ nr = n
n!
x1n1 xn22
n1!n2! nr!
xrnr
That is, the sum is over all nonnegative integer-valued vectors (n1, n2, …, nr) such that n1 + n2 + … + nr = n.
Question:
37. Expand (x1 + x2 + x3)2.
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