Permutations (Arrangement of Objects):
The number of
permutations of n objects, taken r at a time, is the total number of
arrangements of r objects, selected from n objects where the order of the
arrangement is important.
Without repetition:
(a) Arranging n
objects, taken r at a time is equivalent to filling r places from n things.
The number of ways
of arranging = The number of ways of filling r places
= n(n – 1) (n – 2)
….. (n – r + 1) .
= 
=
= nPr
(b) The number of arrangements of n
different objects taken all at a time = npn = n!
With repetition:
(a) The number of
permutations (arrangements) of n different objects, taken r at a time, when
each object may occur once, twice, thrice …. upto r times in any arrangement =
The number of ways of filling r places where each place can be filled by any
one of n objects .
The number of permutations = The number
of ways of filling r places = (n)r
(b) The number of arrangements that can be
formed using n objects out of which p are identical (and of one kind), q are
identical (and of another kind), r are identical (and of another kind) and the
rest are distinct is 
.
Example.1
How many 7 - letter words can be formed using the
letters of the words
(a) BELFAST (b)
ALABAMA
Solution
(a) BELFAST has all different letters.
Hence the number of
words = 7P7 = 7! = 5040.
(b) ALABAMA has 4 A's but the rest are all
different. Hence the number of words that can be formed is
= 7 x 6
x 5 = 210.
Example.2
(a) How many anagrams can be made by using
the letters of the word HINDUSTAN .
(b) How many of these anagrams begin and end
with a vowel.
(c) In how many of these anagrams, all the
vowels come together.
(d) In how many of these anagrams, none of
the vowels come together.
(e) In how many of these anagrams, do the
vowels and the consonants occupy the same relative positions as in HINDUSTAN.
Solution
(a) The total number of anagrams
= Arrangements of nine letters taken
all at a time
= 
=
181440.
(b) We have 3 vowels and 6 consonants, in
which 2 consonants are alike. The first place can be filled in 3 ways and the
last in 2 ways. The rest of the places can be filled in 
ways.
Hence the total number of anagrams = 3 ´
2 ´ = 15120
(c) Assume the vowels (I, U, A) as a single
letter. The letters (IUA) H, D, S, T, N, N can be arranged in ways.
Also IUA can be arranged among themselves in 3! = 6 ways.
Hence the total number of anagrams = ´ 6 = 15120
(d) Let us divide the task into two parts.
In the first, we arrange the 6 consonants as shown below in 
ways.
´
C ´ C ´ C ´ C ´
C ´ C ´(C stands for consonants and ´ stands for blank spaces in between them)
Now 3 vowels
can be placed in 7 places (in between the consonants) in 7p3 = = 210
ways.
Hence the
total number of anagrams = ´ 210 = 75600.
(e) In this case,
the vowels can be arranged among themselves in 3! = 6ways.
Also, the consonants
can be arranged among themselves in ways.
Hence the total number of anagrams = ´ 6 = 2160.
Example.3
How many 3 digit
numbers can be formed using the digits 0, 1,2,3,4,5 so that
(a) digits may not be
repeated
(b) digits may be repeated
Solution
(a) Let
the 3-digit number be XYZ
Position (X) can be
filled by 1,2,3,4,5 but not 0. So it can be filled in 5 ways. .
Position (Y) can be filled in 5 ways again.
(Since 0 can be placed in this position). .
Position (Z) can be filled in 4 ways.
Hence by the fundamental principle of
counting, total number of ways is 5 x 5 x 4 = 100 ways.
(b) Let the 3 digit number be XYZ
Position (X) can be filled in 5 ways
Position (Y) can be filled in 6ways.
Position (Z) can be filled in 6 ways.
Hence by the fundamental principle of
counting, total number of ways is 5 x 6 x 6 = 180.
