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Permutations (Arrangements of Objects):
The number of permutations of n objects, taken r at a time, is the total number of arrangements of n objects, in groups of r where the order of the arrangement is important.
(i) Without repetition
(a) Arranging n objects, taking r at a time in every arrangement, is equivalent to filling r places from n things.
Number of Choices: n n-1 n-2 n-3 n - (r - 1)
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)((n-r)!))/((n-r)!) = n!/((n-r)!) = nPr.
(b) Number of arrangements of n different objects taken all at a time = npn = n!
(ii) With repetition
(a) 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.
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
7P7 = 7! = 5040.
(b) ALABAMA has 4 A's but the rest are all different. Hence the number of words formed is 7!/4! = 7 × 6 × 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?