Quantitative Aptitude |

In this page read permutation and combination question bank, you will also find permutation and combination questions for cat, basic mathematics permutation and combination questions, permutation and combination questions for campus placement, MBA, CAT |

Permutation And Combination-2 : Next- Pipes And Cistern |

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 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 : 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) 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 ) 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 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 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 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 Next- Pipes And Cistern |