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Permutation of r objects taken
from n different object
A permutation of r objects taken from n
different objects without repetition is an
arrangement of the objects in a specific order
For example, there 12 permutations for the
letters A, B, C and D taken 2 at a time.
These are : AB BA CA DA
AC BC CB DB
AD BD CD DC
Using the multiplication principle
The number of permutations of 4 objects
taken two at a time = 4 x 3 = 12. Similarly, the
number of permutations of 10 objects taken 3
at a time = 10 X 9 X 8 = 720.
In general, the number of permutations of n
objects taken r at a time
= n(n  1)(n  2)...(n  r  1)
=
=
n(n  1)( n  2)( n  3)...( n  r  1)(n  r )(n  r  1)...3x2 x1
(n  r )(n  r  1)...3x2 x1
n!
( n  r )!
The number of permutations of r objects
chosen from a set of n different objects is
denoted by
n
Pr
=
n!
(n - r)!
Example
Suppose you have 4 different flags. How
many different signals could you make using
(i) 2 flags
(ii) 2 or 3 flags
Solution
(i) n = 4
r=2
There are 12 different signals using 2
flags from 4 flags.
n
Pr =
4
P
2
4!
4 x 3 x 2! = (4)(3) = 12
=
=
2!
2!
(ii) 2 or 3 flags
n=4 r=2
4
P
+
2
4
P
3
or
4!
= ____
2!
+
4!
1!
n=4 r=3
4x3x1!
= 4x3x2! +
1!
2!
= 12 + 24
=
36
There are 36 different signals using 2 or
3 flags from 4 flags.
Example
How many arrangements of the letters of the
word B E G I N are there if
(i) 3 letters are used
(ii) all of the letters are used
Solution
(i) n = 5
r=3
5
P = 5! = (5)(4)(3) = 60
3
2!
The arrangements of the letters of the
word BEGIN is 60 if
3 letters are used.
ii) n = 5
Number of arrangements if all of the letters are used.
5
P
5
=
5! = (5)(4)(3)(2)(1)
=
120
Example
A relay team has 5 members. How many ways can a coach
arrange 4 of them to run a 4x100 m race.
The order of the four runners is important.
Number of arrangements the coach can make
5
=
P4
=
120
Example
How many three-digit numbers can be made from the
integers 2, 3, 4, 5, 6 if
(i) each integer is used only once?
(ii) there is no restriction on the number of times each
integer can be used?
Solution
(i) n= 5
r= 3
n
P
5
= P =
r
3
5!
2!
=
(5)(4)(3)(2)(1)
= (5)(4)(3) =60
2!
There are 60 different arrangements.
ii) there is no restriction on the number of times each
integer can be used?
Solution:
Number of ways of making the three-digit numbers
= 5 x
=
125
5 x 5
(Repetition is allowed)
Example
Find the number of arrangements of 4 digits taken from the
set { 1, 2, 3, 4}. In how many ways can these numbers be
arranged so that
(a) The numbers begin with digit ‘1’
(b) The numbers do not begin with digit ‘1’
Solution
Number of arrangements of 4 digits = 4! = 24
(a) If the arrangements begin with digit ‘1’, then the number of
ways the 3 remaining digits can be arranged
=
3!
=
6
(b) The number of arrangements that do not begin
with digit ‘1’
= 24 - 6
=
8
Example
Four sisters and two brothers are arranged in different ways in
a straight line for several photographs to be taken. How many
different arrangements are possible if
(a) there are no restrictions
(b) the two brothers must be separated
Solution
(a) 6! = 720
(b) First, find the numbers of arrangements with the two
brothers standing next to each other. In these arrangements,
the two brothers move together as one unit and this is
equivalent to the arrangement of 5 objects except that they
are able to switch positions with each other.
Number of arrangements with two brothers next to each
other
= 5! x 2!
= 120 x 2
= 240
Number of arrangements with the two brothers separated
= 720 – 240 = 480
Example
Arrange 6 boys and 3 girls in a straight line so that the girls
are separated. In how many ways can this be done?
Solution
Consider this arrangement : o B o B o B o B o B o B o
Let the 6 B’s represent the 6 boys and the
spaces for the girls.
‘o’
represent the
Number of arrangements for the boys = 6!
Number of arrangements for the girls
=
7
P
3
= 210
(7 spaces available for the 3 girls)
Total number of arrangements of 6 boys and 3 girls where
the girls are separated
= 6! x 210
= 151200
EXAMPLE
There are 10 students out of whom six are females. How many
possible arrangements are there if
a) they are arranged in a row?
b) males always sit on one side and female on the other side?
Solution
a) The NOP =
10!
= 3628800
b) The NOP = 2! x 6! x 4! = 34560
Example
A witness to a hit-and-run accident told the police that the plat number
contained the letters PDW followed by 3 digits, the first of which is 5. If the
witness cannot recall the last 2 digits, but is certain that all 3 digits are different,
find the minimum number of automobile registrations that the police may have
to check.
Solution
PDW
5
The NOP =
1 x 9x 8
= 72 ways
Example
In how many ways can 4 girls and 5 boys sit in a row if
boys and girls must sit alternate to each other?
Solution
B
The NOP
B
= 5! x 4!
B
B
= 2880 ways
B
the
Example
Four digit numbers are to be formed from the digits 0, 1, 2, 3, 4, 5, 6 without repetition .
How many numbers can be formed if each number
a) is less than 5000
b) begins with digit 4 or 6
c) is between 2000 and 6000
d) is an odd number
Solution
a)
b)
The NOP =
4
or
4 x 6 x 5 x 4 = 480 ways
6
The NOP = 2 ( 1 x 6 x 5 x 4 )
= 240 ways
c)
The NOP = 5 x 5 x 4 x 3 = 300 ways
EExample
HHow many four-digit even numbers can be formed from the
digits, 1, 2, 3, 4, 5, 6, 7 to make up between 2000 and 6000
A
aa) without repetition
Solution:
consider the last position by two parts : “0” and “not 0”
ends with 0
not ends with 0
0
The NOP
or
= (4 x 6 x 5 x 1 ) + ( 3 x 6 x 5 x 3)
= 390 ways
b) With repetition
The NOP = 4 x 8 x 8 x 4 = 1024 ways.
for
Example
Three married couples have bought 6 seats in the same row
seated
a concert In how many different ways can they be
a) with no restrictions
b) If each couple is to sit together
c) If all the men sit together to the right of all the women
Solution
a)
The NOP = 6! = 720
b)
H1 W 1
H2 W 2
H 3 W3
The NOP = 3! X 2! X 2! X 2! = 48 ways
c) If all the men sit together to the right of all the women
W1 W2 W3
H1H2H3
The NOP = 1 x 3! X 3! = 36 ways.