Permutation And Combination-2

Permutation And Combination-2

Ex. In how many different ways can the letters of the word COMPUTER can be arranged in such a way that vowels may occupy only odd positions ?

Solution : Here odd and even positions are :

 C    O   M    P    U    T   E   R
(O) (E) (O) (E) (O) (E) (O) (E)

Now 3 vowels O, U , E



Also remaning 5 places can be arranged by C, M, P, T, R



So, required number of ways= 120 * 60 = 7200

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Ex. In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women ?

Solution :

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Ex. In a group of 5 boys and 3 girls, 3 childrens are to be selected. In how many different ways can they be selected such that at east 1 boy should be there ?

Solution : (1boy and 2 Girls) or ( 2 boys and 1 girl ) or ( 3 boys)

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Ex. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one lack ball is to be included in the draw ?

Solution : No of ways = (drawing 1 black AND 2 others ) OR (drawing 2 black AND 1 others ) OR (drawing 3 blacks )

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Ex. There are 5 boys and 5 girls. In how any ways they can be seated in a row so that all the girls do not sit together ?

Solution : There are 5 boys and 5 girls, so total number of ways of sitting will be 10 ! in a row.

Now, when all girls sit together, then 5 boys and (group of 5 girls as one person ) , so total numbers will of six persons, also 5 girls can be arrenged in 5! ways,

No of ways when 5 girls sit together=6!×5!

So total no. of ways when all 5 girls do not sit together= total number of ways of sitting 10 boys and girls - No of ways when 5 girls sit together

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Ex. In a party every guest shakes hand with every other guest. If there was total of 105 handshakes in the party, find the number of persons persent in the party ?

Solution : For every handshake two persons are required, let n be the number of persons persent in the party. So,


so   n=15 , -14

Total number of persons present in the party were 15

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Ex. Five digits are given as 3, 1, 0, 9, 5

(1) From these digits how many five digits numbers can be formed, without repetition of the digits ?

(2) How many of them are divisible by 5 ?

(3) How many of them are not divisible by 5 ?

Solution : (1) Total number of 5 digit numbers will be 5! but when 0 be at last place then it will become 4 digits so ,

Total numbers will be : 5 ! - 4 ! = 96

(2) For divisibility with 5, at unit place number should be 0 or 5

(a) when unit place has 0,(ex. 39510) then remaning 4 numbers can be arrenged in 4! = 24 ways

(b) when unit place has 5 (ex, 90135 ) then remaning 4 numbers can be arrenged in 4! = 24 ways , but when 0 wll be at last place (ex. 09315) then
Total number of ways reduced to 4! - 3! = 18

So , Total numbers divisible by 5 will be = 24 + 18 = 42

(3) Numbers not divisible by 5 = ( Total numbers - Numbers divisible by 5 ) = 96 - 42 = 54

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Ex. From the word MATHEMATICS

(a) How many different arrangements can be made by using all the letters in the word MATHEMATICS ?

Solution : Word MATHEMATICS has total 11 letters out of which 2Ms, 2As, 2Ts, rest all single




(b) How many of them begin with I ?

Solution : when I will be fixed at first place , then there will be 10 letters left having 2Ms, 2As, 2Ts



(c) How many of them begin with M ?

Solution : when M will be fixed at first place , then there will be 10 letters left having 2As, 2Ts