Posted on - 18-02-2017

IIT JEE

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.

(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) .

= =
= ^{n}P_{r}

(b) The number of arrangements of n
different objects taken all at a time = ^{n}p_{n} = n!

(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 .

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 that can be formed is

= 7 x 6
x 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

= = 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 7p_{3} = = 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.

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

(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.