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ECON*2740 Class Notes, Winter 2011
Tuesday, 1 February
Course Instructor: David Prescott
Permutations and Combinations
These concepts are helpful in calculating many types of probabilities. For example, 5
tokens are marked 1 to 5. They are selected randomly one after the other.
Q1: Find probability that they are selected in the specific order 1, 2, 3, 4, 5 (event A)
Q2: Find the probability the first two to be drawn are 5 and 4 in that order (event B)
First, define the sample space – the set of basic outcomes that are all equally probable.
For Q1, determine how many ways the 5 tokens can be placed in order.
The first token can be any one of the 5. Once this has been selected, the second can be
chosen in any one of 4 ways, the third in any one of 3 ways and so one. The total
number of ordered arrangements of 5 different tokens is
5x4x3x2x1 = 120.
This product is usually written as 5!
The event that the tokens are drawn in the specific order 1 to 5 represents just 1 of the
120 possible orderings, the probability of event A is 1/120
In general,
n
different objects can be ordered in n(n
‐
1)(n
‐
2).
.(2)(1) = n! different ways.
The term
n
! is referred to as
n
‐
factorial
Example, consider how many ways 3 tokens numbered 1, 2, 3 can be arranged. The
answer is 3! = 3x2x1 = 6:
(1)
1 2 3
(2) 1 3 2
(3) 2 1 3
(4) 2 3 1
(5) 3 1 2
(6) 3 2 1
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Q2: Find the probability the first two tokens are 5 and 4 in that order (event B)
Event B allows the 3
rd
, 4
th
and 5
th
tokens to be in any order provided the 1
st
and 2
nd
are 5
and 4 respectively. Token 3 can be selected in any of 3 ways, token 4 in any of 2 ways
and token 5 will be the one left over. That’s 3x2x1 possible orderings.
Hence the sequence 5, 4, *, *, *
(where * means “any token”) can occur in 3! ways out of the 120 possible orderings.
Hence P(B) = 6/120 = 1/20 = 0.05
Permutations:
The number of ways
m
different objects can be ordered when selected
randomly from
n
>
m
objects is called
the permutations of m objects drawn from n
.
The first of the
m
ordered objects can be any one of the original
n
objects and the
second can be any one of (
n
‐
1
) and so on. The
m
th
can be selected in any of (
n
‐
m+1
)
ways. Hence the number of permutations of
m
objects drawn from
n
is :
݊ሺ݊െ1ሻሺ݊െ2ሻ … ሺ݊െ݉1ሻ
This can also be written as:
ܲ
ൌ
݊ሺ݊െ1ሻ.
. ሺ݊െ݉1ሻሺ݊െ݉ሻሺ݊െ݉െ1ሻ…1
ሺ݊െ݉ሻሺ݊െ݉െ1ሻ….1
ൌ
݊!
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 Fall '11
 Prescott

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