Permutations (Arrangements of Objects)
 Without repetition
 Arranging n objects, taking r at a time in every arrangement, is equivalent to filling r places from n things.
n n1 n2 n3 n  (r1)
 Arranging n objects, taking r at a time in every arrangement, is equivalent to filling r places from n things.
Number of ways of arranging = Number of ways of filling r places
= n(n  1)(n  2) ... (n  r + 1)
= (n(n  1)(n  2) ... (n  r + 1)((nr)!))/((nr)!) = n!/((nr)!) =^{n}P_{r}.
 Number of arrangements of n different objects taken all at a time = ^{n}p_{n} = n!
(ii) With repetition
 Number of permutations (arrangements) of n different objects, taken r at a time, when each object may occur once, twice, thrice ...... up to r times in any arrangements.
= Number of ways of filling r places, each out of n objects.
rPlaces:
1 
2 
3 
4 

r 
Number of Choices : n n n n n
Number of ways to arrange = Number of ways to fill r places = (n)^{r}.
(iii) 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 one kind) and rest are different = n!/p!q!r!.
How many 7  letter words can be formed using the letters of the words:
(a) BELFAST, (b) ALABAMA
(a)BELFAST has all different letters.
Hence, the number of words
^{7}P_{7} = 7! = 5040.
(b)ALABAMA has 4 A's but the rest are all different. Hence the number of words formed is 7!/4! =7 Ã— 6 Ã— 5 = 210.
(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?
(a) The total number of anagrams
= Arrangements of nine letters taken all at a time = 9!/2! = 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 7!/2! ways. Hence the total number anagrams = 3 Ã— 2 Ã— 7!/2! = 15210.
(c) Assume the vowels (IUA) as a single letter. The letters (IUA) H, D, S, T, N, N can arranged in 7!/2! ways. Also IUA can be arranged, among themselves, in 3! = 6 ways.
Hence the total number of anagram = 7!/2! Ã— 6 = 15120.
(d) Let us divide the task in two parts. In the first, we arrange the 6 consonants as shown in 6!/2! ways.
Ã— C Ã— C Ã— C Ã— C Ã— C Ã— C Ã— (C stands for consonants and Ã— stand for blank spaces between them)
3 vowels can be arranged in 7 places (between the consonants) in 7p_{3} = 7!/2! =210.
(e) In this case the vowels are arranged among themselves in 3! = 6 ways. Also the consonants are arranged among themselves in 6!/2! ways.
Hence the total number of anagrams = 6!/2! Ã— 6 = 2160.
How many 3 digit numbers can be formed using the digits 0, 1, 2, 3, 4, 5 where
(a) digits may not be repeated,
(b) digits may be repeated
(a)Let the 3 digit number by 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 principal of counting, total number of ways is 5 Ã— 5 Ã— 4 = 100 ways.
â€‹(b)Let the 3 digit number be XYZ.
Position (X) can be filled in 5 ways.
Position (Y) can be filled in 6 ways
Position (Z) can be filled in 6 ways.
Hence, by the fundamental principle of counting, total number of ways is 5 Ã— 6 Ã— 6 = 180.
Find the number of ways in which 6 letters can be posted in 10 letterboxes.
For every letter, we have 10 choices (i.e. 10 letterboxes).
Hence the total number of ways = 10^{6} = 1,000,000.