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.

r-Places

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)!) = ⁿP_r.

(b) Number of arrangements of n different objects taken all at a time = ⁿp_n = 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.

r-Places:

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!.

Illustration:

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 formed is 7!/4! = 7 × 6 × 5 = 210.

Illustration:

(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.

(d) In how many of these anagrams none of the vowels come together.