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# PERMUTATION & COMBINATION

## PERMUTATION & PROBABILITY MCQ SET-10

Q.1 A farmer, has 9 buffaloes whose ages are 1,2,3,4…9  years respectively. Each buffalo gives milk daily, the quantity of which is proportional to the buffalo’s age. In how many ways can Farmer  allocate the buffaloes among his three sons, such that each son gets the same quantity of milk each day ?   A) 6   B) 7   C) 8   D) 9   Ans. A   ?(1+2+…+9)=45 so,three sets of 15=(1,6,8),(2,4,9),(3,5,7) so,no of ways 3!=6    Q.2 In how many different ways can books A, B, C, D be piled up such that A comes exactly in between C and D ?   A) 4 B) 2 C) 1 D) 3   Ans. A Assuming C, A, D as one group, they can be grouped in two ways, CAD or DAC. This group and B again can be arranged in 2 ways.        So = 2 × 2 = 4 ways.   Q.3 Out of 12 batsmen and 8 bowlers, a cricket team of 11 players has to be selected such that there should be at least 5 batsmen and 5 bowlers. In how many ways this can be done?   A) 73,920 B) 72,640 C) 40,420 D) 46,400   Ans. A   5 bowlers and 5 batsmen mean the 11th can either be a batsman or a bowler. So, 6 batsmen and 5 bowlers or 5 batsmen and 6 bowlers. = 12C6 × 8C5 + 12C5 × 8C6 = 73,920   Q.4 An university has to select an examiner from a list of 60 persons. Of 25 are women and 35 men. 15 know French, 45 do not, 20 of them are teachers and 40 are not. What is the probability that the University selects a French knowing woman teacher?   A)  B)  C)  D) Ans. B   Probability of selecting a teacher = P (T) = 20/60 = 1/3 Probability of selecting a woman = P (W) = 25/60 = 5/12 Probability of selecting a French knowing candidate = P (F) = 15/60 = 1/4 Required probability = P (T) P (W) P (F) = (1/3) (5/12) (1/4) = 5/144.

## PERMUTATION & PROBABILITY MCQ SET-9

Q.1  50 students were admitted to a Nursery Class. Some students can speak only Hindi. Ten students can speak both Hindi & English. If the no. of students who can speak English is 21, then how many students can speak Hindi & how many can speak only English?   A) 39, 29 & 11 respectively   B) 37, 27 & 13 respectively   C) 28, 18 & 22 respectively   D) 21, 11 & 29 respectively   Ans. A   Circle X represent English. Circle Y represent Hindi   A = Students speaking English only B = Students speaking both English & Hindi C= Students speaking Hindi only Total no. of students =   A+B+C = 50……..(i) B = 10………………(ii) A+B=21……………….(iii) Now, A=(A+B)-B =21-10 =11 Thus the no. of students who can speak English only=11 No. of students who can speak Hindi= (B+C)=(A+B+C)-A =50-11=39……………..iv No. of students who can speak Only Hindi = C From iv (B+C)=39,Thus C=39-10 =29 Therefore, Number of students speaking Hindi=39 No. of students speaking Hindi only = 29 No. of students speaking English only = 11   Q.2 Out of 20 tickets sold in a lottery, 5 will have different prizes. In how many different ways, a person who purchased 4 tickets can win the prize(s)?   A) 24   B) 5   C) 31   D) 20   Ans. C   He can win zero, one, two, three or four prizes. So, the answer is 5C0 + 5C1 + 5C2 + 5C3 + 5C4 = 31.   Q.3 In how many ways 4 persons out of 8 (A, B, C, D, E, F, G, H) can be selected such that if B is selected, D should not be selected?   A) 30   B) 64   C) 36   D) 55   Ans. D   If B is selected, D should not be. So, 3 more have to be selected from (8 – 2) = 6. Or if B is not selected, 4 more have to be selected from (8 – 1) = 7 So, = 6C3 + 7C4 = 55   Q.4 In how many ways 8 persons can stand in a row, if one refuses to stand at the extreme end? A) 6 × 7! B) 6 × 8! C) 7 × 8! D) 7 × 7! Ans. A One person refuses to stand at the extreme. So, the remaining 7 persons can stand in 7! ways. The refused person can stand in 6 ways in between these 7 persons. So, the answer is 6 × 7!   Q.5 Five persons have to be selected out of ten (A to J). In how many ways can it be done such that if ‘I’ is selected, D and E cannot be selected and ‘H’ has to be selected? A) 20 B) 120 C) 146 D) 126 Ans. C   If I is selected, three more (except I and H) have to be selected from the remaining (except I, D, E and H). The different ways are 6C3. If I is not selected, then the five have to be selected from the remaining 9. This can be done in 9C5 ways. So, the total number of ways are  6C3 + 9C5 = 146.

## PERMUTATION & PROBABILITY MCQ SET-7

Q.1 When the persons shake hands with one another , in how many ways is it possible ?   A) 25   B) 40   C) 20   D) 45   Ans. D   10C2 = 10!/8! 2! = 10*9*8!/8! * 2 =10*9/2 =90/2 = 45 Or 10*9/2*1 = 45   Q.2 A person has four notes of rupee 1,2,5,10 denomination. The number of different sums of money he can form from them is:   A) 15   B) 12   C) 16   D) 8   Ans. B   4! / 2 = 4*3*2*1 / 2 =12   Q.3 In how many ways can four children be made to stand in a line such that two of them A & B are always together ?   A) 12   B) 6   C) 18   D) 24   Ans. A   AB + 2 = 3 persons Therefore, 3! = 3 * 2 ways = 6 ways  A & B can interchange in 2 ways Total no. of arrangement = 6 * 2 = 12 ways   Q.4 How many three digit nos. can be generated from 1,2,3,4,5,6,7,8,9 such that digits are in ascending order ?   A) 84   B) 81   C) 80   D) 83   Ans. A   Nos. are = 1,2,3,4,5,6,7,8,9 from which three digit nos are to be formed in ascending order Taking 1 as fixed, no of 3 digits in ascending order = 8c2 Now,Taking 2 as fixed no of 3 digits in ascending order = 7c2 In this way, if 3,4,5,6,7 used as fixed nos then the nos of 3 digits in ascending order are = 6c2, 5c2, 4c2, 3c2, & 2c2 Therefore required nos in ascending order are:- 8c2 + 7c2+6c2+5c2+4c2+3c2+ 2c2 = 28+21+15+10+6+3+1 = 84   Q.5 A person has 4 coins of different denominations. What is the number of different sums of money person can form using one or more coins at a time?   A) 15   B) 12   C) 11   D) 16   Ans. A   4c1+4c2+4c3+4c4 = 4+6+4+1 = 15

## PERMUTATION & PROBABILITY MCQ SET-6

Q.1 In how many ways can you send 5 children to 7 classroom  :   I) When there is no restriction ? A) 16807 ways B) 2401 ways C) 117649 ways D) 343 ways Ans. A Ist child can go in any 7 classroom 2nd child can go in any 7 classroom 3rd child also in any 7 classrooms Therefore, 7*7*7*7*7 = 16807 ways   II) Such that two children are in the same class ? A) 2520 ways B) 5040 ways C) 840 ways D) 740 Ans. A 1st children in 7 classes 2nd children in 6 classes 3rd in 5 classes 4th in 4 classes 5th children in 3 classes Therefore, 7*6*5*4*3 = 2520 ways   III) Such that atleast two children are in the same class ? A) 14287 B) 2639 C) 116809 D) 397 Ans. A This you can easily calculate just by subtracting above two i.e. all the ways – ways in which restriction is there Therefore, 16807 – 2520 = 14287 ways   Q.2  How many number of 5 digits can be formed with the digits 0,1,4,5,6,7 & 8 ? A) 2160 ways B) 1800 ways C) 2220 ways D) 1440 ways Ans. A At first place of 5 digit nos., we can’t put 0 So, 6P1 = 6 And, for remaining 4 digits we are left with 6 digits, then 6P4 = 6*5*4*3 = 360 Therefore number of ways = 360*6 = 2160   Q.3  How many words can be formed out of the letters of the word ARTICLE so that the vowels occupy even places ? A) 144 ways B) 120 ways C) 168 ways D) None of the above Ans. A There are 3 vowels & can be placed at 3 even places = 3p3 = 3*2*1 = 6 ways There are 4 consonants which can be put in 4 places. Then, 4P4 = 24 ways Or 24*6=144 ways   Q.4 In how many ways can 8 children sit on 8 seats number A,B,C,D,E,F,G,H ?   I) With no restriction A) 40320 ways B) 6720 ways C) 20160 ways D) None of the above Ans. A 8*7*6*5*4*3*2*1 = 40320 ways (Write in this way keeping in mind 1st child can sit at 8 seats)   II) such that 8 children can sit on 8 seats such that Ramesh sits on seat C ? A) 7! B) 8! C) 6! D) 9! Ans. A Since Seat C is fixed then it can be arranged in 7! ways   III) Such that Ramesh is just left of Suresh ? A) 7! B) 6! C) 8! D) 5! Ans. A (1 + 6!)   Q.5 Ranjit likes to take a different route to work each day. He always walks alongside part of the park & only ever travels south or east. In the diagram lines represent streets in this area. How many different routes are there by which Ranjit can walk from home  to the Work ?                           H                                       PARK    w           A) 4   B) 3   C) 6   D) 8   Ans. C   When he will go downwards then ways can be d d d (3 times downward)   After then when he will go right then ways will be R R R (3 times right)   Therefore, 6 Will be answer   If it has asked to find the no. of ways then,   You should apply in this way 6! / 3! 3!

## PERMUTATION & PROBABILITY MCQ SET-5

Q. 1 There are 5 boys and 4 girls in a class. A committee of 5 is to be selected such that there are 3 boys and 2 girls in the committee. In how many ways the committee can the committee be selected?. Also in how many ways can the committee be selected if there are at least one boy in the committee.   A) 40 ways   B) 50 ways   C) 60 ways   D) 70 ways   Ans. D   5C3 * 4C2 =10*6 =60 ways   Q.2  Distribute 50 kids into the groups of size 5,10,15 & 20 ? A) 50! /( 5!*10!*20!15!) B) 50! * 2! /( 5!*10!*20!15!) C) 49!/2!( 5!*10!*20!15!) D) None of the above Ans. A 50C5                       *  45C10                                             *   35C15                                 *            20C20 (50 of size 5)   (Remaining 45 from 10)     (Remaining 35 from 15) Or 50! / (45!5!) * 45! (35!10!) * 35! (20! 15!) * 1 Or 50! / 5!*10! * 20! * 15!   Q.3 A person has eighth letters & eigth addressed envelopes corresponding to those letters. In how many ways can he put the letters in the envelopes such that exactly 5 of them get delivered correctly ? A) 112ways B) 113ways C) 114ways D) 115ways Ans. A Since five got delivered correct=8C5 ways Therefore, 3 get delivered incorrectly i.e. dearrangement (3) = 2 ways i.e. total ways = 2 * 8C5 = 2 * 56 = 112ways   Q.4 In how many ways can you distribute 7 identical chocolates in 5 kids :   I) When there is no restriction ? A) 11C4 B) 10C4 C) 11C3 D) 10C3 Ans. A Formula used: (N+r-1)! / C(r-1) (where, n= no. of chocolates & r=kids) Therefore, 11! / 7! 4! = 11C4   II) in such a way each kid should get atleast one chocolate ? A) 6C4 B) 6C3 C) 6C5 D) 6C2 Ans. A Formula used: (n-1) / C (r-1) i.e. 6C4   Q.5 In how many ways can you distribute 5 distinct rings :   I) in four boxes; A) 1024 ways B) 1025 ways C) 256 ways D) 4096 ways Ans. A 4                            *    4                     *            4         *     4      *     4 = 1024 ways (1st ring in 4 boxes)  (2nd ring in 4boxes) as on…   II) in 4 fingers A) 8C3 B) 8C2 C) 8C4 D) 8C5 Ans. A Formula used: (N+r-1) / C(r-1) Or (5+4-1) / C(4-1) Or 8C3

## PERMUTATION & PROBABILITY MCQ SET-4

Q.1 A two member committee comprising of one male & one female member is to be constituted out of five males & three females. Amongst the females A refused to be a member of the committee in which Mr. B is taken as the member. In how many different ways can the committee be constituted ? A) 11 B) 12 C) 13 D) 14 Ans. D Females(3)      Males (5) A                        B C                        D E                        F                           G                           H Since A can’t go with B, it will make team with four males in four ways AD, AF, AG, AH. Since there is no compulsion with female C & E, may combine with 5 males in 5 different ways each. Therefore, total number of ways = 4+5+5=14   Q.2 In a question of a test paper there are five items each under List-A & List-B. The examinees are required to match each item under List-A with its corresponding correct item under List-B . Further, it is given that i) No examinee has given the correct answer ii) answers of no two examinees are identical What is the maximum number of examinees who took this test ? A) 119 B) 26 C) 129 D) 24 Ans. A Total number of combination = 5*4*3*2*1=120 Since it has given two fact, i.e. No examinee has given the correct answer We know in there must be correct answer in out of 120 combination given in the question So, number of examinees = 120-1 = 119   Q.3 Two friends decided to meet at a place between 2 to 3 pm, both can arrive any time b/w 2-3,they decided that person arriving first will not wait for more than 10 min for the other. What is the prob. they will meet? A) 11/36 B) 12/36 C) 11/35 D) 12/35 Ans. A This is a question of area probability. IF we plot each 10 minute intervals on x-axis and y axis we will get a square with lines x=y=0 and x=y=6 now. we need to find the area |x-y|<1 thus the probability is (36-25)/36=11/36   Q.4 In a tournament 14 teams play league matches. If each team plays against every other team once only then how many matches are played ?   A) 85   B) 91   C) 105   D) 78   Ans. B   No. of matches played= 14C2 = 14 * 13 / 2 = 91   Q.5 If there are 12 persons in a party, and each of them shakes hands with each other, how many handshakes happen in the party ?   A) 55   B) 66   C) 88 D) 44   Ans. B   12*11/2=66

## PERMUTATION & PROBABILITY MCQ SET-3

Q.1 Amit has five friends 3 girls & 2 boys . Amit’s wife also has 5 friends: 3 boys & 2 girls. In how many maximum no. of different ways can they invite 2 boys & 2 girls such that two of them are Amit’s friends & two are his wife’s ? A) 38 B) 24 C) 46 D) 58 Ans. C      AMIT’S FRIENDS      AMIT’S WIFE FRIENDS    BOYS (2) GIRLS (3) BOYS (3) GIRLS (2) WAYS OF SELECTION CASE I    2  0  0  2  2C2*3C0*3C0*2C2 =1*1*1*1=1 CASE II   0  2  2  0  2C0*3C2*3C2*2C0 =1*3*3*1=9 CASE III  1  1  1  1  2C1*3C1*3C1*2C1 =2*3*3*2=36   Total selection = 1+9+36 = 46   Q.2 Groups each containing 3 boys are to be formed out of 5 boys A,B,C,D & E such that no group can contain both C & D together. What is the maximum number of different groups ? A) 5 B) 6 C) 7 D) 8 Ans. C Maximum no. of different groups = ABC, ABD, ABE, BCE,BDE,CEA,DEA =7   Q.3 Each of the 3 persons is to be given some identical items such that no product of the number of items received by the each of the three persons is equal to 30.In how many maximum different ways can this distribution be done ? A) 27 B) 21 C) 24 D) 33 Ans. A Items can be distributed in such a manner that product of the number of items is equal to 30 (3)3 = 27 ways   Q.4 Each of the two women & three men is to occupy one chair out of eight chairs , each of which numbered from 1 to 8. First women are to occupy any two chairs from those numbered 1 to 4; & then the three men would occupy any three chairs  out of the remaining six chairs. What is the maximum number of different ways in which this can be done ? A) 1440 B) 132 C) 40 D) 3660 Ans. A One Women can  sit in four different ways Second women can  sit on anyone chair in three different ways So, the different ways of sitting becomes 4*3 = 12 The ways of sitting of three men out of the remaining six chairs is 6P3 i.e 6*5*4 = 120 Therefore, total number of ways = 12*120 = 1440   Q.5 A mixed double tennis game is to be played between the two teams (Consisting of one male & one female) There are four married couples. No team is to consist of husband & his wife. What is the maximum number of games that can be played ? A) 48 B) 36 C) 12 D) 42 Ans. A Married couples= MF  MF  MF  MF                              AB   CD  EF  GH Possible Teams= AD CB EB  GB                             AF  CF  ED  GD Team AD can play Only with CB,CF,CH,EB,EH,GB,GF Team AD cannot play with : AF,AH, ED & GD This same can apply with all the teams , Therefore, total number of matches =12*7 = 84 But every match includes two teams , thus actual number of matches = 84/2 = 48

## PERMUTATION & PROBABILITY MCQ SET-2

Q.1 Out of 120 applications for a post, 70 are male and 80 have a driver’s license. What is the ratio between the minimum to maximum number of males having driver’s license? A) 1 to 2 B) 2 to 3 C) 3 to 7 D) 5 to 7 Ans. C n(A U B) = n(A) + n (B) – n (A n B) 120 = 70 + 80 – n (A n B) n (A n B)= 150-120 =30 Therefore required ratio = 30:70 =3:7 Ans.   Q.2 In an examination there are three subjects A, B & C.A student has to pass in each subject. 20% students failed in A, 22% students failed in B & 16% failed in C. The total number of students passing the whole examination lies between : A) 58% & 84% B) 42% & 78% C) 42% & 84% D) 58% & 78% Ans. C Total % of students failed individually in each subject : 20+22+16 = 58 Total % of students failed commonly in subjects: 16 Therefore total no. of students passing the whole examination lies between 100-58 & 100-16 i.e. 42% & 84%   Q.3 In an examination 70% of the students passed in the paper I & 60% of the students passed in the paper II. 15% of the students failed in both the papers while 270 students passed in both the papers. What is the total no. of students ? A) 560 B) 580 C) 540 D) 600 Ans. D Total % of failed students = 15+15+25 = 55% Total % of students passed in both the papers = 45% If total no. of students = x X*45/100 = 270 Or 45x=270*100 X= 270*100 / 45 = 600   Q.4 In how many ways can four books A,B,C & D be arranged one above other in a vertical order such that the books A & B are never in continuous position ? A) 14 B) 18 C) 12 D) 9 Ans. C Total no. of arrangements – (no. of arrangements in which A & B are together) 4! – 3! * 2! = 24-12 = 12   Q.5 Five balls of different colours are to be placed in three different boxes such that any box contains atleast one ball. What is the maximum no. of different ways in which this can be done ? A) 120 B) 180 C) 150 D) 90 Ans. C n= 5 r=3 Therefore, no. of ways = rn – rC1 (r-1)n+rC2 (r-2)2 + … 35 – 3C1(3-1)5 + 3C2 (3-2)2 243-3*32 + 3*1 =243-96+3 = 150

## PERMUTATION & PROBABILITY MCQ SET-1

Q.1 A bag contains four black and five red balls, if three balls are picked at random one after another WITH replacement, what is the chance that they’re all black?   A) 64/729   B) 64/730   C) 63/729   D) 63/730   Ans. A   With replacement means you pick up a ball note down its color and then put it back in the bag again. So Total Number of balls remain same for each event. And Hence probability of picking a black ball (4/9) remains the same in every case. 1st Pick up:4/9; 2nd Pick up: 4/9 ;because we put the ball back in the bag, So probability is same as “1st Pick”. 3rd Pick up: 4/9 ;because we put the ball back in the bag. So final probability =1st x 2nd x 3rd =4/9 x 4/9 x 4/9 =Cube of 4/9 =64/729   Q.2 If you have a spinning wheel with 3 green sectors, 1 blue sector and 1 red sector, what is the probability of getting a green sector? What is the probability of getting a non blue sector?   A) Green = 3/5 ; Non-blue = 4/5   B) Green = 2/5 ; Non-blue = 3/5   C) Green = 4/5 ; Non-blue = 3/5   D) Green = 3/5 ; Non-blue = 2/5   Ans. A           Total number of events = 5   Q.3 A bag has 4 red balls and 2 yellow balls. (The balls are identical in all respects other than colour). A ball is drawn from the bag without looking into the bag. What is probability of getting a red ball?   A) 2/3   B) 2/4   C) 2/5   D) 2/6   Ans. A   There are in all (4 + 2 =) 6 outcomes of the event. Getting a red ball consists of 4 outcomes.   Therefore, the probability of getting a red ball is 4/6 = 2/3   Q.4 There are 100 students in a particular class. 60% students play cricket, 30% student play football and 10% students play both the games. What is the number of students who play neither cricket nor football?   A) 25   B) 20   C) 18   D) 15   Ans. B   100- (50+20+10) = 100-80 = 20   Q.5 In a group of persons, 70% of the persons are male and 30% of the persons are married. If two-sevenths of the males are married, what fraction of the females is single?   A) 2/7   B) 1/3   C) 3/7   D) 2/3 Ans. D Let total no. of persons =100 Therefore total no. of males= 70 Total no. of females= 30 Given that, no. of unmarried persons =30 So, Number of married males= 2/7 * 70=20 Therefore, No. of married females =30-20 =10 Therefore, No. of unmarried females =30-10 = 20 Therefore, Required fraction of single females= 20/30 = 2/3