Lecture 20
Andrei Antonenko
March 26, 2003
1
Permutations
Let we have
n
numbers from 1 to
n
: 1
,
2
, . . . , n
.
If we change their order we get their
per
mutation
. We will write these numbers in brackets, for example, (34152) is a permutation of
numbers from 1 to 5.
Example 1.1.
There are 6 different permutations of numbers
1
,
2
and
3
:
(123)
(132)
(213)
(231)
(312)
(321)
Let’s compute the total number of permutations of numbers from 1 to
n
. On the first place
in the permutation we can put any of
n
elements. On the second place we can put any element
which is not equal to the element on the first place — so we have (
n

1) possibilities; than to
the third place we can put any element which is not on the first and the second place, so totally
we have (
n

2) possibilities, etc. Finally, we will have only one possibility to put an element on
the last
n
th place. So, total number of permutations of
n
elements is
n
(
n

1)(
n

2)
· · ·
2
·
1 =
n
!
(By definition,
n
! is a product of all numbers from 1 to
n
. It is called
n
factorial
).
So, we see that the number of permutations increases very fast. We have just 3! = 6 different
permutations of numbers from 1 to 3, 4! = 24 different permutations of numbers from 1 to 4,
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 Spring '03
 Andant
 Group Theory, Permutations, Prime number, Parity

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