\displaystyle \text{MULTIPLE CHOICE QUESTIONS (MCQs)}

\displaystyle \textbf{Question 1: }\text{A coin is tossed }n\text{ times. The number of possible outcomes is}
\displaystyle \text{(a) }2n\qquad\text{(b) }{}^nC_2\qquad\text{(c) }n^2\qquad\text{(d) }2^n
\displaystyle \text{Answer:}
\displaystyle \text{For each toss of a coin, there are }2\text{ possible outcomes: Head or Tail.}
\displaystyle \text{Therefore, for }n\text{ independent tosses, the total number of outcomes is}
\displaystyle 2\times2\times\cdots\times2=2^n.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{A convex polygon has }65\text{ diagonals. The number of its sides is equal to:}
\displaystyle \text{(a) }13\qquad\text{(b) }10\qquad\text{(c) }22\qquad\text{(d) }11
\displaystyle \text{Answer:}
\displaystyle \text{If a polygon has }n\text{ sides, then the number of diagonals is }\frac{n(n-3)}{2}.
\displaystyle \therefore \frac{n(n-3)}{2}=65.
\displaystyle n(n-3)=130.
\displaystyle n^2-3n-130=0.
\displaystyle (n-13)(n+10)=0.
\displaystyle \text{Since }n\text{ is positive, }n=13.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Number of diagonals of a convex hexagon is}
\displaystyle \text{(a) }3\qquad\text{(b) }6\qquad\text{(c) }9\qquad\text{(d) }12
\displaystyle \text{Answer:}
\displaystyle \text{Number of diagonals of an }n\text{-sided polygon}=\frac{n(n-3)}{2}.
\displaystyle \text{For a hexagon, }n=6.
\displaystyle \therefore \text{Number of diagonals}=\frac{6(6-3)}{2}=\frac{18}{2}=9.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In how many ways can }10\text{ lion and }6\text{ tigers be arranged in a row so that}
\displaystyle \text{no two tigers are together?}
\displaystyle \text{(a) }10!\times{}^{11}C_6\qquad\text{(b) }10!\times{}^{10}C_6
\displaystyle \text{(c) }6!\times{}^{10}C_6\qquad\text{(d) }6!\times{}^{10}C_7
\displaystyle \text{Answer:}
\displaystyle \text{First arrange the }10\text{ lions. This can be done in }10!\text{ ways.}
\displaystyle \text{The }10\text{ lions create }11\text{ gaps in which the tigers can be placed.}
\displaystyle \_L\_L\_L\_\cdots L\_
\displaystyle \text{To ensure that no two tigers are together, choose }6\text{ of these }11\text{ gaps.}
\displaystyle \text{This can be done in }{}^{11}C_6\text{ ways.}
\displaystyle \therefore \text{Number of arrangements}=10!\times{}^{11}C_6.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The total number of ways of selecting six coins out of }20\text{ one-rupee}
\displaystyle \text{coins, }10\text{ fifty paise coins and }7\text{ twenty-five paise coins is:}
\displaystyle \text{(a) }{}^{37}C_6\qquad\text{(b) }56\qquad\text{(c) }28\qquad\text{(d) }29
\displaystyle \text{Answer:}
\displaystyle \text{Let }x,\ y,\ z\text{ be the numbers of one-rupee, fifty paise and twenty-five paise coins selected.}
\displaystyle \text{Then }x+y+z=6,\qquad x,y,z\geq0.
\displaystyle \text{Since only }6\text{ coins are selected, the available numbers }20,\ 10,\ 7\text{ impose no restriction.}
\displaystyle \text{Thus, the number of non-negative integral solutions of }x+y+z=6\text{ is}
\displaystyle {}^{6+3-1}C_{3-1}={}^{8}C_2=28.
\displaystyle \therefore \text{The total number of ways is }28.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Given }4\text{ flags of different colors, how many different signals can be generated,}
\displaystyle \text{if a signal requires the use of }2\text{ flags one below the other?}
\displaystyle \text{(a) }12\qquad\text{(b) }13\qquad\text{(c) }14\qquad\text{(d) }15
\displaystyle \text{Answer:}
\displaystyle \text{Since the two flags are placed one below the other, their order is important.}
\displaystyle \text{The first position can be filled in }4\text{ ways and the second in }3\text{ ways.}
\displaystyle \therefore \text{Number of signals}={}^{4}P_2=4\times3=12.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{In an examination, there are three multiple choice questions and each}
\displaystyle \text{question has }4\text{ choices. Number of ways in which a student can fail to get}
\displaystyle \text{all answers correct is:}
\displaystyle \text{(a) }11\qquad\text{(b) }12\qquad\text{(c) }27\qquad\text{(d) }63
\displaystyle \text{Answer:}
\displaystyle \text{Each question has }4\text{ choices.}
\displaystyle \therefore \text{Total number of ways of answering the three questions}=4^3=64.
\displaystyle \text{There is only one way in which all three answers are correct.}
\displaystyle \therefore \text{Number of ways of failing to get all answers correct}=64-1=63.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{There are }10\text{ true-false questions in an examination. Then these questions}
\displaystyle \text{can be answered in:}
\displaystyle \text{(a) }100\qquad\text{(b) }20\qquad\text{(c) }512\qquad\text{(d) }1024
\displaystyle \text{Answer:}
\displaystyle \text{Each question can be answered in }2\text{ ways: True or False.}
\displaystyle \therefore \text{Number of ways of answering }10\text{ questions}=2^{10}.
\displaystyle 2^{10}=1024.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{To fill }12\text{ vacancies, there are }25\text{ candidates of which five are from}
\displaystyle \text{scheduled caste. If }3\text{ of the vacancies are reserved for scheduled caste candidates, while}
\displaystyle \text{the rest are open to all, then the number of ways in which the selection can be made:}
\displaystyle \text{(a) }{}^5C_3\times{}^{22}C_9\qquad\text{(b) }{}^{22}C_9-{}^5C_3
\displaystyle \text{(c) }{}^5C_3+{}^{22}C_3\qquad\text{(d) None}
\displaystyle \text{Answer:}
\displaystyle \text{First select }3\text{ scheduled caste candidates out of }5\text{ for the reserved vacancies.}
\displaystyle \text{This can be done in }{}^5C_3\text{ ways.}
\displaystyle \text{After selecting these }3,\text{ there are }22\text{ candidates left.}
\displaystyle \text{The remaining }9\text{ vacancies are open to all and can be filled in }{}^{22}C_9\text{ ways.}
\displaystyle \therefore \text{Total number of selections}={}^5C_3\times{}^{22}C_9.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{A father with }8\text{ children takes them }3\text{ at a time to the zoological garden,}
\displaystyle \text{as often as he can without taking the same }3\text{ children together more than once.}
\displaystyle \text{The number of times he will go to the garden is:}
\displaystyle \text{(a) }56\qquad\text{(b) }100\qquad\text{(c) }112\qquad\text{(d) None}
\displaystyle \text{Answer:}
\displaystyle \text{Each visit consists of selecting }3\text{ children out of }8.
\displaystyle \text{Since the order of selection does not matter, the number of different groups is}
\displaystyle {}^8C_3=\frac{8!}{3!5!}.
\displaystyle =\frac{8\times7\times6}{3\times2\times1}=56.
\displaystyle \therefore \text{he will go to the garden }56\text{ times.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Number of words from the letters of the word BHARAT in which B and H}
\displaystyle \text{will never come together is:}
\displaystyle \text{(a) }210\qquad\text{(b) }240\qquad\text{(c) }422\qquad\text{(d) }400
\displaystyle \text{Answer:}
\displaystyle \text{The word BHARAT has }6\text{ letters in which A occurs twice.}
\displaystyle \text{Total number of arrangements}=\frac{6!}{2!}=360.
\displaystyle \text{Now, consider B and H together as one unit.}
\displaystyle \text{Then we have }5\text{ objects: }(BH),A,A,R,T.
\displaystyle \text{These can be arranged in }\frac{5!}{2!}\text{ ways.}
\displaystyle \text{Also, B and H can be arranged within the unit in }2!\text{ ways.}
\displaystyle \therefore \text{Number of arrangements in which B and H are together}
\displaystyle =\frac{5!}{2!}\times2!=120.
\displaystyle \therefore \text{Required number of arrangements}=360-120=240.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If all the words formed by the letters of the word HAPPY are arranged}
\displaystyle \text{according to the dictionary, then the place of HAPPY is:}
\displaystyle \text{(a) }60\qquad\text{(b) }12\qquad\text{(c) }13\qquad\text{(d) }24
\displaystyle \text{Answer:}
\displaystyle \text{The alphabetical order of the letters is }A,H,P,P,Y.
\displaystyle \text{Words beginning with A come before all words beginning with H.}
\displaystyle \text{Number of words beginning with A}=\frac{4!}{2!}=12.
\displaystyle \text{After these }12\text{ words, the first word beginning with H and followed by A is HAPPY.}
\displaystyle \therefore \text{The position of HAPPY}=12+1=13.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Everybody in a room shakes hands with everybody else. The total}
\displaystyle \text{number of handshakes is }66.\text{ The total number of persons is:}
\displaystyle \text{(a) }11\qquad\text{(b) }12\qquad\text{(c) }13\qquad\text{(d) }14
\displaystyle \text{Answer:}
\displaystyle \text{If there are }n\text{ persons, the number of handshakes is }{}^nC_2.
\displaystyle \therefore {}^nC_2=66.
\displaystyle \frac{n(n-1)}{2}=66.
\displaystyle n(n-1)=132.
\displaystyle n^2-n-132=0.
\displaystyle (n-12)(n+11)=0.
\displaystyle \text{Since }n\text{ is positive, }n=12.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{A bag contains }3\text{ black, }4\text{ white and }2\text{ red balls, all the balls being}
\displaystyle \text{different. Number of selections of at most }6\text{ balls containing balls of all the colours is:}
\displaystyle \text{(a) }1008\qquad\text{(b) }1080\qquad\text{(c) }1204\qquad\text{(d) }1130
\displaystyle \text{Answer:}
\displaystyle \text{There are }3+4+2=9\text{ different balls.}
\displaystyle \text{First count all selections containing at least one ball of each colour.}
\displaystyle (2^3-1)(2^4-1)(2^2-1)=7\times15\times3=315.
\displaystyle \text{Now exclude selections containing more than }6\text{ balls.}
\displaystyle \text{For }7\text{ balls, the number containing all three colours is }{}^9C_7-1=35.
\displaystyle \text{The subtraction of }1\text{ is for the selection containing all }3\text{ black and }4\text{ white balls}
\displaystyle \text{but no red ball.}
\displaystyle \text{For }8\text{ balls, all three colours must occur, giving }{}^9C_8=9\text{ selections.}
\displaystyle \text{For }9\text{ balls, there is }{}^9C_9=1\text{ selection.}
\displaystyle \therefore \text{Required number}=315-(35+9+1)=270.
\displaystyle \therefore \text{the mathematically correct answer is }270.
\displaystyle \text{None of the given options matches, so the question or options appear to contain a printing error.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The total number of words formed by the letters of the word}
\displaystyle \text{MISSISSIPPI such that all S's and all I's occur together, is:}
\displaystyle \text{(a) }5!\qquad\text{(b) }5!\times2\qquad\text{(c) }\frac{11!}{4!}\qquad\text{(d) None}
\displaystyle \text{Answer:}
\displaystyle \text{In MISSISSIPPI, there are }4\text{ S's, }4\text{ I's, }2\text{ P's and }1\text{ M.}
\displaystyle \text{Treat all the S's as one unit and all the I's as another unit.}
\displaystyle \text{Thus, we have the objects }(SSSS),(IIII),M,P,P.
\displaystyle \text{There are }5\text{ objects in all, of which the two P's are identical.}
\displaystyle \therefore \text{Number of arrangements}=\frac{5!}{2!}=60.
\displaystyle \therefore \text{the correct answer is }60.
\displaystyle \text{Since }60\text{ is not given among the options, the correct option is (d) None.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The number of words formed by letters of the word EQUATIONS}
\displaystyle \text{containing }2\text{ consonants and }3\text{ vowels, is:}
\displaystyle \text{(a) }72\qquad\text{(b) }120\qquad\text{(c) }7200\qquad\text{(d) }60
\displaystyle \text{Answer:}
\displaystyle \text{The word EQUATIONS contains }5\text{ vowels: }E,U,A,I,O,
\displaystyle \text{and }4\text{ consonants: }Q,T,N,S.
\displaystyle \text{Number of ways of selecting }3\text{ vowels}={}^{5}C_3.
\displaystyle \text{Number of ways of selecting }2\text{ consonants}={}^{4}C_2.
\displaystyle \text{The selected }5\text{ letters can be arranged in }5!\text{ ways.}
\displaystyle \therefore \text{Required number}={}^{5}C_3\times{}^{4}C_2\times5!.
\displaystyle =10\times6\times120=7200.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The number of ways in which a team of eleven players can be selected}
\displaystyle \text{from }22\text{ players, always including }2\text{ of them and excluding }4\text{ of them, is:}
\displaystyle \text{(a) }{}^{16}C_{11}\qquad\text{(b) }{}^{16}C_5\qquad\text{(c) }{}^{16}C_9\qquad\text{(d) }{}^{20}C_9
\displaystyle \text{Answer:}
\displaystyle \text{Out of }22\text{ players, }2\text{ must always be included and }4\text{ must always be excluded.}
\displaystyle \text{Therefore, the number of remaining eligible players}=22-2-4=16.
\displaystyle \text{Since }2\text{ players are already included, we have to select }11-2=9\text{ more players.}
\displaystyle \therefore \text{Number of ways}={}^{16}C_9.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The sum of the digits in the unit place of all the numbers formed with the}
\displaystyle \text{help of }3,4,5,6\text{ taken all at a time is:}
\displaystyle \text{(a) }432\qquad\text{(b) }108\qquad\text{(c) }36\qquad\text{(d) }18
\displaystyle \text{Answer:}
\displaystyle \text{The total number of numbers formed}=4!=24.
\displaystyle \text{Each of the digits }3,4,5,6\text{ occurs in the unit place }3!=6\text{ times.}
\displaystyle \therefore \text{Sum of all unit digits}=6(3+4+5+6).
\displaystyle =6\times18=108.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The number of ways in which }5\text{ prizes be distributed among }4\text{ boys, while}
\displaystyle \text{each boy is capable of having any number of prizes is:}
\displaystyle \text{(a) }5^4\qquad\text{(b) }4^5\qquad\text{(c) }4!\times2^4\qquad\text{(d) }4!\times5
\displaystyle \text{Answer:}
\displaystyle \text{Each of the }5\text{ prizes can be given to any one of the }4\text{ boys.}
\displaystyle \text{Thus, each prize can be distributed in }4\text{ ways.}
\displaystyle \therefore \text{Total number of ways}=4\times4\times4\times4\times4=4^5.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Total number of }n\text{-digit numbers }(n>1),\text{ having the property that no}
\displaystyle \text{two consecutive digits are same, is:}
\displaystyle \text{(a) }8^n\qquad\text{(b) }9^n\qquad\text{(c) }9\times10^{n-1}\qquad\text{(d) None}
\displaystyle \text{Answer:}
\displaystyle \text{The first digit can be chosen from }1,2,\ldots,9,\text{ giving }9\text{ choices.}
\displaystyle \text{Each succeeding digit can be any digit except the immediately preceding digit.}
\displaystyle \therefore \text{each of the remaining }n-1\text{ places has }9\text{ choices.}
\displaystyle \therefore \text{Total number}=9\times9^{n-1}=9^n.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \text{CASE BASED}

\displaystyle \textbf{Question 21: }\text{Sunita and her friends went for a trip to Shimla. They stayed in a hotel.}
\displaystyle \text{There were }4\text{ vacant rooms A, B, C, D. Out of these }4\text{ vacant rooms, two rooms A and B}
\displaystyle \text{were double share rooms and two rooms C and D can contain one person each.}
\displaystyle \text{(Assuming Sunita and her friends are }6\text{ persons in all.)}

\displaystyle \text{(i) Find the number of ways in which room A can be filled?}
\displaystyle \text{Answer:}
\displaystyle \text{Room A can accommodate }2\text{ persons out of }6.
\displaystyle \therefore \text{Number of ways}={}^{6}C_2.
\displaystyle =\frac{6\times5}{2}=15.
\displaystyle \therefore \text{Room A can be filled in }15\text{ ways.}
\displaystyle \\

\displaystyle \text{(ii) If room A and B are already filled each, then find the number of ways in which} \\ \text{room C can be filled.}
\displaystyle \text{Answer:}
\displaystyle \text{Rooms A and B together accommodate }2+2=4\text{ persons.}
\displaystyle \text{Hence, }6-4=2\text{ persons remain.}
\displaystyle \text{Room C can be occupied by either of these }2\text{ persons.}
\displaystyle \therefore \text{Room C can be filled in }2\text{ ways.}
\displaystyle \\

\displaystyle \text{(iii) Find the total number of ways of accommodating Sunita and her friends in} \\ \text{these }4\text{ vacant rooms.}
\displaystyle \text{Answer:}
\displaystyle \text{Choose }2\text{ persons out of }6\text{ for room A in }{}^{6}C_2\text{ ways.}
\displaystyle \text{Choose }2\text{ persons out of the remaining }4\text{ for room B in }{}^{4}C_2\text{ ways.}
\displaystyle \text{The remaining }2\text{ persons can be placed in rooms C and D in }2!\text{ ways.}
\displaystyle \therefore \text{Total number of ways}={}^{6}C_2\times{}^{4}C_2\times2!.
\displaystyle =15\times6\times2=180.
\displaystyle \therefore \text{the total number of ways is }180.
\displaystyle \\

\displaystyle \text{(iv) If room A is filled with }2\text{ persons, then find the number in which room} \\ \text{C and D can be filled.}
\displaystyle \text{Answer:}
\displaystyle \text{After room A is filled, }4\text{ persons remain.}
\displaystyle \text{Room C can be filled by any one of the }4\text{ persons.}
\displaystyle \text{Room D can then be filled by any one of the remaining }3\text{ persons.}
\displaystyle \therefore \text{Number of ways}=4\times3={}^{4}P_2=12.
\displaystyle \therefore \text{rooms C and D can be filled in }12\text{ ways.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{In a certain city all telephone numbers have seven digits. City is divided}
\displaystyle \text{into six zones. Each zone is allotted a specific non-zero digit which is to be used as first}
\displaystyle \text{digit of each telephone number of that zone.}
\displaystyle \text{Based on the above information, answer the following questions:}

\displaystyle \text{(i) How many telephone numbers are there in each zone, if digit on first place is not} \\ \text{used again?}
\displaystyle \text{(a) }9^6\qquad\text{(b) }10^6\qquad\text{(c) }{}^{10}P_6\qquad\text{(d) }{}^9P_6
\displaystyle \text{Answer:}
\displaystyle \text{The first digit is fixed for a particular zone.}
\displaystyle \text{For each of the remaining }6\text{ places, the fixed first digit cannot be used again.}
\displaystyle \therefore \text{each of the remaining places can be filled in }9\text{ ways.}
\displaystyle \therefore \text{Number of telephone numbers in each zone}=9^6.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \text{(ii) How many telephone numbers are there in the city if there is no restrictions?}
\displaystyle \text{(a) }6\times9^6\qquad\text{(b) }6\times10^6\qquad\text{(c) }6\times{}^{10}P_6\qquad\text{(d) }6\times{}^9P_6
\displaystyle \text{Answer:}
\displaystyle \text{For each zone, the first digit is fixed.}
\displaystyle \text{Each of the remaining }6\text{ digits can be chosen from }0,1,2,\ldots,9.
\displaystyle \therefore \text{Number of telephone numbers in one zone}=10^6.
\displaystyle \text{Since there are }6\text{ zones,}
\displaystyle \therefore \text{Total number of telephone numbers}=6\times10^6.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \text{(iii) How many telephone numbers are there in the city with all digits distinct?}
\displaystyle \text{(a) }9^6\qquad\text{(b) }10^6\qquad\text{(c) }{}^{10}P_6\qquad\text{(d) }{}^9P_6
\displaystyle \text{Answer:}
\displaystyle \text{For each zone, the first digit is already fixed.}
\displaystyle \text{Since all digits must be distinct, the remaining }6\text{ places are to be filled from the}
\displaystyle \text{remaining }9\text{ digits without repetition.}
\displaystyle \therefore \text{Number of telephone numbers in one zone}={}^9P_6.
\displaystyle \text{Since there are }6\text{ zones, the number in the whole city is }6\times{}^9P_6.
\displaystyle \text{Hence, as printed, none of the given options gives the total for the whole city.}
\displaystyle \text{If the question intended ``in each zone'', the correct option would be (d).}
\displaystyle \\

\displaystyle \text{(iv) How many different telephone numbers are there in the city if the first two digits of}
\displaystyle \text{different zones are }12,23,34,45,56\text{ and }67\text{?}
\displaystyle \text{(a) }6\times10^5\qquad\text{(b) }6\times8^5\qquad\text{(c) }6\times{}^{10}P_5\qquad\text{(d) }6\times{}^8P_5
\displaystyle \text{Answer:}
\displaystyle \text{For each zone, the first two digits are fixed.}
\displaystyle \text{Each of the remaining }5\text{ places can be filled by any of the }10\text{ digits.}
\displaystyle \therefore \text{Number of telephone numbers in one zone}=10^5.
\displaystyle \text{There are }6\text{ zones.}
\displaystyle \therefore \text{Total number of telephone numbers}=6\times10^5.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{The students of class XI were given a task to arrange all letters of the}
\displaystyle \text{word EQUATIONS in all possible ways.}
\displaystyle \text{Based on the above information, answer the following questions:}

\displaystyle \text{(i) In how many ways can all letters of the word EQUATIONS be arranged?}
\displaystyle \text{(a) }9!\qquad\text{(b) }10!\qquad\text{(c) }{}^9P_7\qquad\text{(d) }{}^{10}P_7
\displaystyle \text{Answer:}
\displaystyle \text{The word EQUATIONS has }9\text{ distinct letters.}
\displaystyle \therefore \text{Number of arrangements}=9!.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \text{(ii) In how many ways can all letters of the word EQUATIONS be arranged so that all} \\ \text{vowels occur together?}
\displaystyle \text{(a) }5!\times5!\qquad\text{(b) }25!\qquad\text{(c) }6!\qquad\text{(d) }5\times6!
\displaystyle \text{Answer:}
\displaystyle \text{The vowels are }E,U,A,I,O,\text{ i.e. }5\text{ vowels, and there are }4\text{ consonants.}
\displaystyle \text{Treat the }5\text{ vowels as one unit. Then there are }5\text{ objects to arrange.}
\displaystyle \text{These }5\text{ objects can be arranged in }5!\text{ ways.}
\displaystyle \text{The }5\text{ vowels within their unit can also be arranged in }5!\text{ ways.}
\displaystyle \therefore \text{Required number}=5!\times5!.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \text{(iii) In how many ways can all letters of the word EQUATIONS be arranged so that all} \\ \text{consonants occur together?}
\displaystyle \text{(a) }5!\times5!\qquad\text{(b) }2\times4!\times5!\qquad\text{(c) }6!\qquad\text{(d) }6!\times4!
\displaystyle \text{Answer:}
\displaystyle \text{The }4\text{ consonants are }Q,T,N,S.
\displaystyle \text{Treat the }4\text{ consonants as one unit. Together with }5\text{ vowels, there are }6\text{ objects.}
\displaystyle \text{These }6\text{ objects can be arranged in }6!\text{ ways.}
\displaystyle \text{The }4\text{ consonants within their unit can be arranged in }4!\text{ ways.}
\displaystyle \therefore \text{Required number}=6!\times4!.
\displaystyle =17280.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \text{(iv) In how many ways can all letters of the word EQUATIONS be arranged so}
\displaystyle \text{that the positions of vowels and consonants are unaltered?}
\displaystyle \text{(a) }5!\times5!\qquad\text{(b) }4!\times5!\qquad\text{(c) }6!\times4!\qquad\text{(d) }5!\times6!
\displaystyle \text{Answer:}
\displaystyle \text{There are }5\text{ vowel positions and }4\text{ consonant positions.}
\displaystyle \text{The }5\text{ vowels can be arranged among the vowel positions in }5!\text{ ways.}
\displaystyle \text{The }4\text{ consonants can be arranged among the consonant positions in }4!\text{ ways.}
\displaystyle \therefore \text{Required number}=5!\times4!.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{A dentist conducts a team to take surveys of people in his locality}
\displaystyle \text{about using toothpaste. A survey team has some persons and survey owner makes a team}
\displaystyle \text{out of total persons available at that time. If he has a group of }5\text{ persons available}
\displaystyle \text{at that time out of which }2\text{ are men and }3\text{ are women.}

\displaystyle \text{(i) If a committee of }3\text{ persons is to be constituted from the available group, in how}
\displaystyle \text{many ways this can be done? How many of these committees would consist of }1\text{ man}
\displaystyle \text{and }2\text{ women?}

\displaystyle \text{Answer:}
\displaystyle \text{Total number of persons}=5.
\displaystyle \therefore \text{Number of committees of }3\text{ persons}={}^{5}C_3=10.
\displaystyle \text{For a committee consisting of }1\text{ man and }2\text{ women,}
\displaystyle \text{number of ways}={}^{2}C_1\times{}^{3}C_2.
\displaystyle =2\times3=6.
\displaystyle \therefore \text{Total committees}=10,\text{ and committees with }1\text{ man and }2\text{ women}=6.
\displaystyle \\

\displaystyle \text{(ii) If }P(2n-1,n):P(2n+1,n-1)=22:7,\text{ find }n.
\displaystyle \text{Answer:}
\displaystyle \frac{P(2n-1,n)}{P(2n+1,n-1)}=\frac{22}{7}.
\displaystyle \frac{\frac{(2n-1)!}{(n-1)!}}{\frac{(2n+1)!}{(n+2)!}}=\frac{22}{7}.
\displaystyle \frac{(2n-1)!(n+2)!}{(n-1)!(2n+1)!}=\frac{22}{7}.
\displaystyle \frac{n(n+1)(n+2)}{2n(2n+1)}=\frac{22}{7}.
\displaystyle \frac{(n+1)(n+2)}{2(2n+1)}=\frac{22}{7}.
\displaystyle 7(n+1)(n+2)=44(2n+1).
\displaystyle 7n^2+21n+14=88n+44.
\displaystyle 7n^2-67n-30=0.
\displaystyle (7n+3)(n-10)=0.
\displaystyle \therefore n=-\frac{3}{7}\text{ or }n=10.
\displaystyle \text{Since }n\text{ is a positive integer, }n=10.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Two friends Swati and KOMAL are playing cards. Swati asks Komal to}
\displaystyle \text{choose any four cards from a pack of }52\text{ cards. Based on it answer the following:}

\displaystyle \text{(i) In how many ways can Komal select }4\text{ cards from the same suit and all }4\text{ cards}
\displaystyle \text{from different suits?}
\displaystyle \text{Answer:}
\displaystyle \text{There are }4\text{ suits and each suit contains }13\text{ cards.}
\displaystyle \text{For }4\text{ cards from the same suit, choose the suit in }4\text{ ways and then }4\text{ cards from }13.
\displaystyle \therefore \text{Number of ways}=4\times{}^{13}C_4.
\displaystyle =4\times715=2860.
\displaystyle \text{For }4\text{ cards from different suits, one card must be selected from each suit.}
\displaystyle \therefore \text{Number of ways}=13\times13\times13\times13=13^4.
\displaystyle =28561.
\displaystyle \\

\displaystyle \text{(ii) In how many ways can she select all face cards?}
\displaystyle \text{Answer:}
\displaystyle \text{A pack has }12\text{ face cards: }4\text{ Jacks, }4\text{ Queens and }4\text{ Kings.}
\displaystyle \text{She has to select }4\text{ cards from these }12\text{ face cards.}
\displaystyle \therefore \text{Number of ways}={}^{12}C_4.
\displaystyle =\frac{12\times11\times10\times9}{4\times3\times2\times1}=495.
\displaystyle \therefore \text{she can select }4\text{ face cards in }495\text{ ways.}
\displaystyle \\

\displaystyle \text{ASSERTION REASONING QUESTIONS: }

\displaystyle \textbf{Question 26: }\text{Assertion: The number of ways of distributing }10\text{ identical balls in }4
\displaystyle \text{distinct boxes such that no box is empty is }{}^9C_3.
\displaystyle \text{Reason: The number of ways of choosing any }3\text{ places from }9\text{ different places is }{}^9C_3.
\displaystyle \text{Mark the correct choice:}
\displaystyle \text{(a) Both A and R are true and R is correct explanation of A.}
\displaystyle \text{(b) Both A and R are true and R is not correct explanation of A.}
\displaystyle \text{(c) A is true and R is false.}
\displaystyle \text{(d) A is false and R is true.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the numbers of balls in the four boxes be }x_1,x_2,x_3,x_4.
\displaystyle x_1+x_2+x_3+x_4=10,\qquad x_1,x_2,x_3,x_4\geq1.
\displaystyle \text{By the gap method, arrange }10\text{ identical balls in a row.}
\displaystyle \text{There are }9\text{ gaps between these }10\text{ balls.}
\displaystyle \text{To divide them among }4\text{ non-empty boxes, we choose }3\text{ of these }9\text{ gaps.}
\displaystyle \therefore \text{Number of distributions}={}^9C_3.
\displaystyle \text{Thus, both Assertion and Reason are true, and Reason correctly explains the Assertion.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Assertion: The number of rectangles on a chess board is } \\ {}^8C_2\times{}^8C_2.
\displaystyle \text{Reason: To form a rectangle, we have to select any two horizontal lines and any} \\ \text{two of the vertical lines.}
\displaystyle \text{(a) Both A and R are true and R is correct explanation of A.}
\displaystyle \text{(b) Both A and R are true and R is not correct explanation of A.}
\displaystyle \text{(c) A is true and R is false.}
\displaystyle \text{(d) A is false and R is true.}
\displaystyle \text{Answer:}
\displaystyle \text{A standard chess board has }8\times8\text{ squares and hence }9\text{ horizontal and }9\text{ vertical lines.}
\displaystyle \text{To form a rectangle, we choose }2\text{ horizontal lines and }2\text{ vertical lines.}
\displaystyle \therefore \text{Number of rectangles}={}^9C_2\times{}^9C_2.
\displaystyle \text{Hence, the Assertion }{}^8C_2\times{}^8C_2\text{ is false.}
\displaystyle \text{The Reason correctly states the method of forming a rectangle and is therefore true.}
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Assertion: If }n\text{ is a positive integer, then }n(n^2-1)(n+2)\text{ is divisible by }24.
\displaystyle \text{Reason: Product of }r\text{ consecutive whole numbers is divisible by }r.
\displaystyle \text{(a) Both A and R are true and R is correct explanation of A.}
\displaystyle \text{(b) Both A and R are true and R is not correct explanation of A.}
\displaystyle \text{(c) A is true and R is false.}
\displaystyle \text{(d) A is false and R is true.}
\displaystyle \text{Answer:}
\displaystyle n(n^2-1)(n+2)=n(n-1)(n+1)(n+2).
\displaystyle =(n-1)n(n+1)(n+2).
\displaystyle \text{These are four consecutive integers.}
\displaystyle \text{Their product is divisible by }4!=24.
\displaystyle \therefore \text{the Assertion is true.}
\displaystyle \text{The Reason states only that the product of }r\text{ consecutive whole numbers is divisible by }r,
\displaystyle \text{whereas here we require the stronger result that it is divisible by }r!.
\displaystyle \text{Thus, the Reason is true but does not correctly explain the Assertion.}
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{Assertion: The product of five consecutive natural numbers is divisible by }4!
\displaystyle \text{Reason: Product of }n\text{ consecutive natural numbers is divisible by }(n+1)!
\displaystyle \text{(a) Both A and R are true and R is correct explanation of A.}
\displaystyle \text{(b) Both A and R are true and R is not correct explanation of A.}
\displaystyle \text{(c) A is true and R is false.}
\displaystyle \text{(d) A is false and R is true.}
\displaystyle \text{Answer:}
\displaystyle \text{The product of five consecutive natural numbers is always divisible by }5!.
\displaystyle \text{Since }5!=5\times4!,\text{ it is also divisible by }4!.
\displaystyle \therefore \text{the Assertion is true.}
\displaystyle \text{The Reason states that the product of }n\text{ consecutive natural numbers is divisible by }(n+1)!.
\displaystyle \text{This is not true in general. For example, for }n=2,\ 1\times2=2\text{ is not divisible by }3!=6.
\displaystyle \therefore \text{the Reason is false.}
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{Assertion: Number of lines formed by joining }n\text{ points on a circle is }\frac{n(n-1)}{2}.
\displaystyle \text{Reason: }{}^nC_3=\frac{n(n-1)}{2}.
\displaystyle \text{(a) Both A and R are true and R is correct explanation of A.}
\displaystyle \text{(b) Both A and R are true and R is not correct explanation of A.}
\displaystyle \text{(c) A is true and R is false.}
\displaystyle \text{(d) A is false and R is true.}
\displaystyle \text{Answer:}
\displaystyle \text{A line joining two of the }n\text{ points is determined by choosing any }2\text{ points.}
\displaystyle \therefore \text{Number of lines}={}^nC_2=\frac{n(n-1)}{2}.
\displaystyle \therefore \text{the Assertion is true.}
\displaystyle \text{But }{}^nC_3=\frac{n(n-1)(n-2)}{6},\text{ not }\frac{n(n-1)}{2}.
\displaystyle \therefore \text{the Reason is false.}
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER TYPE QUESTIONS}

\displaystyle \textbf{Question 31: }\text{Find }r,\text{ if }{}^nP_r=2880\text{ and }{}^nC_r=120.
\displaystyle \text{Answer:}
\displaystyle \text{We know that }{}^nP_r=r!\,{}^nC_r.
\displaystyle \therefore 2880=r!\times120.
\displaystyle \therefore r!=\frac{2880}{120}=24.
\displaystyle \therefore r!=4!.
\displaystyle \therefore r=4.
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{If there are }30\text{ students in a group, if all shake hands with one another,}
\displaystyle \text{how many handshakes are possible?}
\displaystyle \text{Answer:}
\displaystyle \text{Each handshake is formed by selecting }2\text{ students from }30.
\displaystyle \therefore \text{Number of handshakes}={}^{30}C_2.
\displaystyle =\frac{30\times29}{2}=435.
\displaystyle \therefore 435\text{ handshakes are possible.}
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{Find }r,\text{ if }{}^{10}C_{2r}={}^{10}C_{r+2}.
\displaystyle \text{Answer:}
\displaystyle \text{We know that if }{}^nC_x={}^nC_y,\text{ then }x=y\text{ or }x+y=n.
\displaystyle \text{Therefore, either }2r=r+2\text{ or }2r+(r+2)=10.
\displaystyle \text{Case I: }2r=r+2.
\displaystyle \therefore r=2.
\displaystyle \text{Case II: }3r+2=10.
\displaystyle \therefore r=\frac{8}{3}.
\displaystyle \text{Since the suffix of a combination must be a non-negative integer, }\frac{8}{3}\text{ is not admissible.}
\displaystyle \therefore r=2.
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{How many words with }2\text{ different vowels and }2\text{ different consonants can}
\displaystyle \text{be formed from the alphabets?}
\displaystyle \text{Answer:}
\displaystyle \text{There are }5\text{ vowels and }21\text{ consonants in the English alphabet.}
\displaystyle \text{Number of ways of selecting }2\text{ different vowels}={}^{5}C_2.
\displaystyle \text{Number of ways of selecting }2\text{ different consonants}={}^{21}C_2.
\displaystyle \text{The }4\text{ selected letters can be arranged in }4!\text{ ways.}
\displaystyle \therefore \text{Required number}={}^{5}C_2\times{}^{21}C_2\times4!.
\displaystyle =10\times210\times24=50400.
\displaystyle \therefore 50400\text{ words can be formed.}
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{How many committees of five persons with a chairperson can be selected}
\displaystyle \text{from }12\text{ persons?}
\displaystyle \text{Answer:}
\displaystyle \text{First select }5\text{ persons out of }12\text{ to form the committee.}
\displaystyle \text{This can be done in }{}^{12}C_5\text{ ways.}
\displaystyle \text{The chairperson can then be selected from these }5\text{ persons in }5\text{ ways.}
\displaystyle \therefore \text{Required number}={}^{12}C_5\times5.
\displaystyle =792\times5=3960.
\displaystyle \therefore 3960\text{ committees can be formed.}
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{If }{}^nC_8={}^nC_6.\text{ Find }{}^nC_2.
\displaystyle \text{Answer:}
\displaystyle \text{We know that if }{}^nC_r={}^nC_s\text{ and }r\neq s,\text{ then }r+s=n.
\displaystyle \therefore n=8+6=14.
\displaystyle \therefore {}^nC_2={}^{14}C_2.
\displaystyle =\frac{14\times13}{2}=91.
\displaystyle \therefore {}^nC_2=91.
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{How many rectangles can be formed from the chess board of any size?}
\displaystyle \text{Answer:}
\displaystyle \text{A standard chess board has }8\times8\text{ squares and hence }9\text{ horizontal and }9\text{ vertical lines.}
\displaystyle \text{A rectangle is formed by choosing }2\text{ horizontal lines and }2\text{ vertical lines.}
\displaystyle \therefore \text{Number of rectangles}={}^{9}C_2\times{}^{9}C_2.
\displaystyle =36\times36=1296.
\displaystyle \therefore 1296\text{ rectangles can be formed.}
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{Evaluate: }2\cdot5!-3\cdot4!
\displaystyle \text{Answer:}
\displaystyle 2\cdot5!-3\cdot4!=2(120)-3(24).
\displaystyle =240-72=168.
\displaystyle \therefore \text{The required value is }168.
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{Find the number of different four digit numbers that can be formed with}
\displaystyle \text{digits }2,3,4,7\text{ and using each digit once.}
\displaystyle \text{Answer:}
\displaystyle \text{There are }4\text{ distinct digits and all are to be used once.}
\displaystyle \therefore \text{Number of four digit numbers}=4!.
\displaystyle =4\times3\times2\times1=24.
\displaystyle \therefore 24\text{ different four digit numbers can be formed.}
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{Evaluate: }\frac{6}{5}\cdot5!-\frac{5}{4}\cdot4!.
\displaystyle \text{Answer:}
\displaystyle \frac{6}{5}\cdot5!-\frac{5}{4}\cdot4!
\displaystyle =\frac{6}{5}\times120-\frac{5}{4}\times24.
\displaystyle =144-30=114.
\displaystyle \therefore \text{The required value is }114.
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{Evaluate: }4!-3!.
\displaystyle \text{Answer:}
\displaystyle 4!-3!=24-6=18.
\displaystyle \therefore \text{The required value is }18.
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{Find the value of }P(7,3).
\displaystyle \text{Answer:}
\displaystyle P(7,3)={}^{7}P_3=\frac{7!}{(7-3)!}.
\displaystyle =\frac{7!}{4!}=7\times6\times5=210.
\displaystyle \therefore P(7,3)=210.
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{A college has }6\text{ good badminton players. A team of }4\text{ has to be sent for}
\displaystyle \text{inter college tournament. In how many ways can the team be selected?}
\displaystyle \text{Answer:}
\displaystyle \text{We have to select }4\text{ players out of }6.
\displaystyle \therefore \text{Number of ways}={}^{6}C_4.
\displaystyle =\frac{6!}{4!2!}=15.
\displaystyle \therefore \text{The team can be selected in }15\text{ ways.}
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{Six identical coins are arranged in a row. Determine the number of ways}
\displaystyle \text{in which }3\text{ heads and }3\text{ tails can appear.}
\displaystyle \text{Answer:}
\displaystyle \text{Out of }6\text{ positions, choose }3\text{ positions for the heads.}
\displaystyle \text{The remaining }3\text{ positions will automatically contain tails.}
\displaystyle \therefore \text{Number of ways}={}^{6}C_3.
\displaystyle =\frac{6!}{3!3!}=20.
\displaystyle \therefore 3\text{ heads and }3\text{ tails can appear in }20\text{ ways.}
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{In how many different ways can the letters of the word `HEXAGON' be}
\displaystyle \text{permuted?}
\displaystyle \text{Answer:}
\displaystyle \text{The word HEXAGON contains }7\text{ distinct letters.}
\displaystyle \therefore \text{Number of permutations}=7!.
\displaystyle =5040.
\displaystyle \therefore \text{The letters can be permuted in }5040\text{ different ways.}
\displaystyle \\

\displaystyle \textbf{Question 46: }\text{There are }3\text{ different rings to be worn in four fingers with at most one in}
\displaystyle \text{each finger. In how many ways can this be done?}
\displaystyle \text{Answer:}
\displaystyle \text{We have to place }3\text{ different rings on }3\text{ of the }4\text{ fingers.}
\displaystyle \therefore \text{Number of ways}={}^{4}P_3.
\displaystyle =4\times3\times2=24.
\displaystyle \therefore \text{The rings can be worn in }24\text{ ways.}
\displaystyle \\

\displaystyle \textbf{Question 47: }\text{In how many ways }2\text{ different prizes be awarded to }15\text{ students, without}
\displaystyle \text{giving both to the same student?}
\displaystyle \text{Answer:}
\displaystyle \text{The first prize can be awarded to any one of }15\text{ students.}
\displaystyle \text{The second prize can then be awarded to any one of the remaining }14\text{ students.}
\displaystyle \therefore \text{Number of ways}={}^{15}P_2.
\displaystyle =15\times14=210.
\displaystyle \therefore \text{The prizes can be awarded in }210\text{ ways.}
\displaystyle \\

\displaystyle \textbf{Question 48: }\text{How many words of three distinct English alphabets are there?}
\displaystyle \text{Answer:}
\displaystyle \text{There are }26\text{ letters in the English alphabet.}
\displaystyle \text{We have to arrange }3\text{ distinct letters chosen from }26.
\displaystyle \therefore \text{Number of words}={}^{26}P_3.
\displaystyle =26\times25\times24=15600.
\displaystyle \therefore 15600\text{ words can be formed.}
\displaystyle \\

\displaystyle \textbf{Question 49: }\text{There are }12\text{ buses running between Jammu and Delhi. In how many}
\displaystyle \text{ways can a man go from Jammu to Delhi and return by the same bus?}
\displaystyle \text{Answer:}
\displaystyle \text{The man can choose any one of the }12\text{ buses for going to Delhi.}
\displaystyle \text{Since he has to return by the same bus, there is only one choice for the return journey.}
\displaystyle \therefore \text{Number of ways}=12\times1=12.
\displaystyle \therefore \text{The required number of ways is }12.
\displaystyle \\

\displaystyle \textbf{Question 50: }\text{Find }r\text{ if }P(11,r)=P(12,r-1).
\displaystyle \text{Answer:}
\displaystyle \frac{11!}{(11-r)!}=\frac{12!}{(12-(r-1))!}.
\displaystyle \frac{11!}{(11-r)!}=\frac{12!}{(13-r)!}.
\displaystyle \frac{(13-r)!}{(11-r)!}=\frac{12!}{11!}=12.
\displaystyle (13-r)(12-r)=12.
\displaystyle r^2-25r+144=0.
\displaystyle (r-9)(r-16)=0.
\displaystyle \therefore r=9\text{ or }r=16.
\displaystyle \text{But }r\leq11\text{ for }P(11,r).
\displaystyle \therefore r=9.
\displaystyle \\

\displaystyle \text{SHORT ANSWER TYPE QUESTIONS }

\displaystyle \textbf{Question 51: }\text{Out of }18\text{ points in a plane, no three points are in the same line except}
\displaystyle \text{five points which are collinear. Find the number of lines that can be formed joining the points.}
\displaystyle \text{Answer:}
\displaystyle \text{If no three points were collinear, the number of lines would be }{}^{18}C_2.
\displaystyle {}^{18}C_2=\frac{18\times17}{2}=153.
\displaystyle \text{But the }5\text{ collinear points give }{}^5C_2=10\text{ pairs, all representing the same line.}
\displaystyle \text{Therefore, these }10\text{ lines must be replaced by }1\text{ line.}
\displaystyle \therefore \text{Required number of lines}=153-10+1=144.
\displaystyle \therefore \text{The number of lines is }144.
\displaystyle \\

\displaystyle \textbf{Question 52: }\text{There are two identical red, two identical black and two identical white balls. In}
\displaystyle \text{how many ways can the balls be placed in the cells such that balls of the same color do not}
\displaystyle \text{occupy any two consecutive cells?}
\displaystyle \text{Answer:}
\displaystyle \text{Total arrangements of }R,R,B,B,W,W=\frac{6!}{2!2!2!}=90.
\displaystyle \text{Let }A_R,A_B,A_W\text{ denote arrangements in which the two balls of a colour are together.}
\displaystyle n(A_R)=n(A_B)=n(A_W)=\frac{5!}{2!2!}=30.
\displaystyle \therefore n(A_R)+n(A_B)+n(A_W)=90.
\displaystyle n(A_R\cap A_B)=n(A_B\cap A_W)=n(A_W\cap A_R)=\frac{4!}{2!}=12.
\displaystyle \therefore \text{sum of pairwise intersections}=3\times12=36.
\displaystyle n(A_R\cap A_B\cap A_W)=3!=6.
\displaystyle \text{By the principle of inclusion and exclusion,}
\displaystyle \text{Required number}=90-90+36-6=30.
\displaystyle \therefore \text{the balls can be arranged in }30\text{ ways.}
\displaystyle \\

\displaystyle \textbf{Question 53: }\text{Determine }n\text{ if }{}^{2n}C_3:{}^nC_3=11:1.
\displaystyle \text{Answer:}
\displaystyle \frac{{}^{2n}C_3}{{}^nC_3}=11.
\displaystyle \frac{\frac{2n(2n-1)(2n-2)}{6}}{\frac{n(n-1)(n-2)}{6}}=11.
\displaystyle \frac{2n(2n-1)\,2(n-1)}{n(n-1)(n-2)}=11.
\displaystyle \frac{4(2n-1)}{n-2}=11.
\displaystyle 8n-4=11n-22.
\displaystyle 3n=18.
\displaystyle \therefore n=6.
\displaystyle \\

\displaystyle \textbf{Question 54: }\text{A boy has }4\text{ movie tickets and }9\text{ movies of his interest in the theatre. Of}
\displaystyle \text{these }9,\text{ he does not want to see Marvels part }2,\text{ unless part }1\text{ is seen.}
\displaystyle \text{In how many ways can he choose }3\text{ movies to be seen?}
\displaystyle \text{Answer:}
\displaystyle \text{Without restriction, }3\text{ movies can be selected from }9\text{ in }{}^9C_3\text{ ways.}
\displaystyle \text{Invalid selections contain part }2\text{ but not part }1.
\displaystyle \text{After selecting part }2,\text{ the other }2\text{ movies are selected from the remaining }7\text{ movies.}
\displaystyle \therefore \text{Number of invalid selections}={}^7C_2=21.
\displaystyle \therefore \text{Required number}={}^9C_3-{}^7C_2.
\displaystyle =84-21=63.
\displaystyle \therefore \text{He can choose }3\text{ movies in }63\text{ ways.}
\displaystyle \\

\displaystyle \textbf{Question 55: }\text{How many }4\text{-letter codes can be formed using the first }10\text{ letters of the}
\displaystyle \text{English alphabet, if no letter can be repeated?}
\displaystyle \text{Answer:}
\displaystyle \text{We have to select and arrange }4\text{ different letters from }10\text{ letters.}
\displaystyle \therefore \text{Number of codes}={}^{10}P_4.
\displaystyle =10\times9\times8\times7=5040.
\displaystyle \therefore 5040\text{ different codes can be formed.}
\displaystyle \\

\displaystyle \textbf{Question 56: }\text{Evaluate }\frac{n!}{(n-r)!}\text{ when}
\displaystyle \text{(i) }n=6,\ r=2\qquad\text{(ii) }n=9,\ r=5.
\displaystyle \text{Answer:}
\displaystyle \frac{n!}{(n-r)!}={}^nP_r.
\displaystyle \text{(i) For }n=6,\ r=2,
\displaystyle \frac{6!}{(6-2)!}=\frac{6!}{4!}=6\times5=30.
\displaystyle \therefore \text{the required value is }30.
\displaystyle \text{(ii) For }n=9,\ r=5,
\displaystyle \frac{9!}{(9-5)!}=\frac{9!}{4!}=9\times8\times7\times6\times5.
\displaystyle =15120.
\displaystyle \therefore \text{the required value is }15120.
\displaystyle \\

\displaystyle \textbf{Question 57: }\text{Find the number of }4\text{-digit numbers that can be formed using the digits}
\displaystyle 1,2,3,4,5\text{ if no digit is repeated. How many of these will be even?}
\displaystyle \text{Answer:}
\displaystyle \text{To form a }4\text{-digit number, we select and arrange }4\text{ digits out of }5.
\displaystyle \therefore \text{Total number of }4\text{-digit numbers}={}^{5}P_4.
\displaystyle =5\times4\times3\times2=120.
\displaystyle \text{For an even number, the unit digit must be }2\text{ or }4.
\displaystyle \therefore \text{the unit place can be filled in }2\text{ ways.}
\displaystyle \text{The remaining }3\text{ places can be filled from the remaining }4\text{ digits in }{}^4P_3\text{ ways.}
\displaystyle \therefore \text{Number of even numbers}=2\times{}^4P_3.
\displaystyle =2\times24=48.
\displaystyle \therefore \text{Total numbers}=120\text{ and even numbers}=48.
\displaystyle \\

\displaystyle \textbf{Question 58: }\text{In how many ways can a team of }3\text{ boys and }3\text{ girls be selected from }5
\displaystyle \text{boys and }4\text{ girls?}
\displaystyle \text{Answer:}
\displaystyle \text{Number of ways of selecting }3\text{ boys from }5={}^{5}C_3.
\displaystyle \text{Number of ways of selecting }3\text{ girls from }4={}^{4}C_3.
\displaystyle \therefore \text{Required number of ways}={}^{5}C_3\times{}^{4}C_3.
\displaystyle =10\times4=40.
\displaystyle \therefore \text{The team can be selected in }40\text{ ways.}
\displaystyle \\

\displaystyle \textbf{Question 59: }\text{Determine the number of }5\text{ card combinations out of a deck of }52\text{ cards if}
\displaystyle \text{there is exactly one ace in each combination.}
\displaystyle \text{Answer:}
\displaystyle \text{There are }4\text{ aces and }48\text{ non-ace cards in a deck.}
\displaystyle \text{Choose exactly }1\text{ ace from }4\text{ in }{}^4C_1\text{ ways.}
\displaystyle \text{Choose the remaining }4\text{ cards from the }48\text{ non-ace cards in }{}^{48}C_4\text{ ways.}
\displaystyle \therefore \text{Required number of combinations}={}^4C_1\times{}^{48}C_4.
\displaystyle =4\times194580=778320.
\displaystyle \therefore \text{The number of combinations is }778320.
\displaystyle \\

\displaystyle \textbf{Question 60: }\text{How many words, with or without meaning, each of }2\text{ vowels and }3
\displaystyle \text{consonants can be formed from the letters of the word DAUGHTER?}
\displaystyle \text{Answer:}
\displaystyle \text{The vowels in DAUGHTER are }A,U,E,\text{ i.e. }3\text{ vowels.}
\displaystyle \text{The consonants are }D,G,H,T,R,\text{ i.e. }5\text{ consonants.}
\displaystyle \text{Choose }2\text{ vowels from }3\text{ in }{}^3C_2\text{ ways.}
\displaystyle \text{Choose }3\text{ consonants from }5\text{ in }{}^5C_3\text{ ways.}
\displaystyle \text{The selected }5\text{ letters can be arranged in }5!\text{ ways.}
\displaystyle \therefore \text{Required number}={}^3C_2\times{}^5C_3\times5!.
\displaystyle =3\times10\times120=3600.
\displaystyle \therefore 3600\text{ words can be formed.}
\displaystyle \\

\displaystyle \textbf{Question 61: }\text{It is required to seat }5\text{ men and }4\text{ women in a row so that the women}
\displaystyle \text{occupy the even places. How many such arrangements are possible?}
\displaystyle \text{Answer:}
\displaystyle \text{There are }9\text{ positions, of which }2,4,6,8\text{ are even positions.}
\displaystyle \text{The }4\text{ women can be arranged in these }4\text{ even positions in }4!\text{ ways.}
\displaystyle \text{The }5\text{ men can be arranged in the remaining }5\text{ positions in }5!\text{ ways.}
\displaystyle \therefore \text{Required number of arrangements}=4!\times5!.
\displaystyle =24\times120=2880.
\displaystyle \therefore 2880\text{ arrangements are possible.}
\displaystyle \\

\displaystyle \textbf{Question 62: }\text{In how many ways can the letters of the word ASSASSINATION be}
\displaystyle \text{arranged so that all the S's are together?}
\displaystyle \text{Answer:}
\displaystyle \text{In ASSASSINATION, A occurs }3\text{ times, S occurs }4\text{ times, I occurs }2\text{ times,}
\displaystyle \text{N occurs }2\text{ times, and T and O occur once each.}
\displaystyle \text{Treat all }4\text{ S's as one unit.}
\displaystyle \text{We then have }10\text{ objects, with }3\text{ A's, }2\text{ I's and }2\text{ N's identical.}
\displaystyle \therefore \text{Required number of arrangements}=\frac{10!}{3!2!2!}.
\displaystyle =151200.
\displaystyle \therefore 151200\text{ arrangements are possible.}
\displaystyle \\

\displaystyle \textbf{Question 63: }\text{We wish to select }6\text{ persons from }8,\text{ but if the person A is chosen, then B}
\displaystyle \text{must be chosen. In how many ways can selections be made?}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of ways of selecting }6\text{ persons from }8={}^{8}C_6=28.
\displaystyle \text{Invalid selections are those in which A is chosen but B is not chosen.}
\displaystyle \text{With A selected and B excluded, choose }5\text{ persons from the remaining }6.
\displaystyle \therefore \text{Number of invalid selections}={}^{6}C_5=6.
\displaystyle \therefore \text{Required number of selections}=28-6=22.
\displaystyle \therefore 22\text{ selections can be made.}
\displaystyle \\

\displaystyle \textbf{Question 64: }\text{If all letters of the word RACHIT are arranged in all possible ways as}
\displaystyle \text{listed in dictionary, then what is the rank of the word RACHIT?}
\displaystyle \text{Answer:}
\displaystyle \text{The alphabetical order of the letters is }A,C,H,I,R,T.
\displaystyle \text{Before R, the first letter can be }A,C,H\text{ or }I.
\displaystyle \therefore \text{Number of words before those beginning with R}=4\times5!=480.
\displaystyle \text{Among the words beginning with R, RACHIT is the first in dictionary order.}
\displaystyle \therefore \text{Rank of RACHIT}=480+1=481.
\displaystyle \therefore \text{The rank of RACHIT is }481.
\displaystyle \\

\displaystyle \textbf{Question 65: }\text{How many automobile licence plates can be made if each plate contains}
\displaystyle \text{two different letters followed by three different digits?}
\displaystyle \text{Answer:}
\displaystyle \text{The two different letters can be selected and arranged from }26\text{ letters in }{}^{26}P_2\text{ ways.}
\displaystyle \text{The three different digits can be selected and arranged from }10\text{ digits in }{}^{10}P_3\text{ ways.}
\displaystyle \therefore \text{Number of licence plates}={}^{26}P_2\times{}^{10}P_3.
\displaystyle =(26\times25)(10\times9\times8).
\displaystyle =650\times720=468000.
\displaystyle \therefore 468000\text{ licence plates can be made.}
\displaystyle \\

\displaystyle \textbf{Question 66: }\text{If }{}^{12}P_{x+1}>2\,{}^{12}P_x,\text{ then find the set of values of }x.
\displaystyle \text{Answer:}
\displaystyle {}^{12}P_{x+1}>2\,{}^{12}P_x.
\displaystyle \frac{{}^{12}P_{x+1}}{{}^{12}P_x}>2.
\displaystyle \frac{\frac{12!}{(11-x)!}}{\frac{12!}{(12-x)!}}>2.
\displaystyle 12-x>2.
\displaystyle \therefore x<10.
\displaystyle \text{Since }x\text{ is a positive integer,}
\displaystyle \therefore x\in\{1,2,3,4,5,6,7,8,9\}.
\displaystyle \\

\displaystyle \textbf{Question 67: }\text{Find the number of positive integers greater than }6000\text{ and less than}
\displaystyle 7000\text{ which are divisible by }5,\text{ provided that no digit is to be repeated.}
\displaystyle \text{Answer:}
\displaystyle \text{Since the number lies between }6000\text{ and }7000,\text{ the thousands digit must be }6.
\displaystyle \text{For divisibility by }5,\text{ the unit digit must be }0\text{ or }5.
\displaystyle \text{Case I: Unit digit is }0.
\displaystyle \text{The hundreds digit can be chosen in }8\text{ ways and the tens digit in }7\text{ ways.}
\displaystyle \therefore \text{Number of such numbers}=8\times7=56.
\displaystyle \text{Case II: Unit digit is }5.
\displaystyle \text{Again, the hundreds digit can be chosen in }8\text{ ways and the tens digit in }7\text{ ways.}
\displaystyle \therefore \text{Number of such numbers}=8\times7=56.
\displaystyle \therefore \text{Total number}=56+56=112.
\displaystyle \therefore \text{The required number of integers is }112.
\displaystyle \\

\displaystyle \textbf{Question 69: }\text{A student is allowed to select at least one piece of fruit from }6\text{ bananas, }5
\displaystyle \text{oranges and }4\text{ apples. In how many ways can he make a selection?}
\displaystyle \text{Answer:}
\displaystyle \text{The number of bananas selected can be }0,1,2,\ldots,6,\text{ giving }7\text{ choices.}
\displaystyle \text{The number of oranges selected can be }0,1,2,\ldots,5,\text{ giving }6\text{ choices.}
\displaystyle \text{The number of apples selected can be }0,1,2,\ldots,4,\text{ giving }5\text{ choices.}
\displaystyle \therefore \text{Total number of selections}=7\times6\times5=210.
\displaystyle \text{The selection in which no fruit is selected must be excluded.}
\displaystyle \therefore \text{Required number of selections}=210-1=209.
\displaystyle \therefore \text{The student can make the selection in }209\text{ ways.}
\displaystyle \\

\displaystyle \textbf{Question 70: }\text{Find the number of ways in which }13\text{ different books can be packed into}
\displaystyle \text{five parcels if four parcels contain }3\text{ books each and the fifth only one.}
\displaystyle \text{Answer:}
\displaystyle \text{First choose }1\text{ book for the parcel containing only one book in }{}^{13}C_1\text{ ways.}
\displaystyle \text{The remaining }12\text{ books are to be divided into }4\text{ groups of }3\text{ books each.}
\displaystyle \text{Number of ways}=\frac{12!}{(3!)^4\,4!}.
\displaystyle \text{The factor }4!\text{ is used because the four parcels containing }3\text{ books each are interchangeable.}
\displaystyle \therefore \text{Required number}={}^{13}C_1\times\frac{12!}{(3!)^4\,4!}.
\displaystyle =\frac{13!}{(3!)^4\,4!}.
\displaystyle =200200.
\displaystyle \therefore \text{The books can be packed in }200200\text{ ways.}
\displaystyle \\

\displaystyle \text{LONG QUESTIONS}

\displaystyle \textbf{Question 71: }\text{How many words, with or without meaning can be made from the letters}
\displaystyle \text{of the word MONDAY, assuming that no letter is repeated, if}
\displaystyle \text{(i) }4\text{ letters are used at a time,}
\displaystyle \text{(ii) All letters are used at a time,}
\displaystyle \text{(iii) all letters are used but first letter is a vowel?}
\displaystyle \text{Answer:}
\displaystyle \text{The word MONDAY has }6\text{ distinct letters: }M,O,N,D,A,Y.
\displaystyle \text{The vowels are }O\text{ and }A.

\displaystyle \text{(i) Number of words formed using }4\text{ letters at a time}
\displaystyle ={}^{6}P_4.
\displaystyle =6\times5\times4\times3=360.
\displaystyle \therefore 360\text{ words can be formed.}

\displaystyle \text{(ii) When all }6\text{ letters are used, the number of arrangements is}
\displaystyle 6!=720.
\displaystyle \therefore 720\text{ words can be formed.}

\displaystyle \text{(iii) The first letter must be a vowel.}
\displaystyle \text{The first position can be filled by }O\text{ or }A,\text{ giving }2\text{ choices.}
\displaystyle \text{The remaining }5\text{ letters can be arranged in }5!\text{ ways.}
\displaystyle \therefore \text{Required number}=2\times5!.
\displaystyle =2\times120=240.
\displaystyle \therefore 240\text{ words can be formed with a vowel in the first place.}
\displaystyle \\

\displaystyle \textbf{Question 72: }\text{In how many ways can the letters of the word PERMUTATIONS be}
\displaystyle \text{arranged if the}
\displaystyle \text{(i) Words start with P and end with S,}
\displaystyle \text{(ii) Vowels are all together,}
\displaystyle \text{(iii) There are always }4\text{ letters between P and S?}
\displaystyle \text{Answer:}
\displaystyle \text{The word PERMUTATIONS has }12\text{ letters, of which T occurs twice.}
\displaystyle \text{The vowels are }E,U,A,I,O,\text{ i.e. }5\text{ distinct vowels.}

\displaystyle \text{(i) P is fixed at the first position and S at the last position.}
\displaystyle \text{The remaining }10\text{ letters are to be arranged, with T repeated twice.}
\displaystyle \therefore \text{Number of arrangements}=\frac{10!}{2!}.
\displaystyle =1814400.
\displaystyle \therefore 1814400\text{ words start with P and end with S.}

\displaystyle \text{(ii) Treat all }5\text{ vowels as one unit.}
\displaystyle \text{Along with the }7\text{ consonants, we have }8\text{ objects, with T repeated twice.}
\displaystyle \text{These }8\text{ objects can be arranged in }\frac{8!}{2!}\text{ ways.}
\displaystyle \text{The }5\text{ vowels within their unit can be arranged in }5!\text{ ways.}
\displaystyle \therefore \text{Required number}=\frac{8!}{2!}\times5!.
\displaystyle =20160\times120=2419200.
\displaystyle \therefore 2419200\text{ arrangements are possible.}

\displaystyle \text{(iii) If there are }4\text{ letters between P and S, their positions differ by }5.
\displaystyle \text{The possible pairs of positions are }(1,6),(2,7),\ldots,(7,12).
\displaystyle \text{Thus, there are }7\text{ pairs of positions.}
\displaystyle \text{For each pair, P and S can interchange their positions in }2!\text{ ways.}
\displaystyle \therefore \text{Number of ways of placing P and S}=7\times2=14.
\displaystyle \text{The remaining }10\text{ letters can be arranged in }\frac{10!}{2!}\text{ ways.}
\displaystyle \therefore \text{Required number}=14\times\frac{10!}{2!}.
\displaystyle =7\times10!=25401600.
\displaystyle \therefore 25401600\text{ arrangements have exactly }4\text{ letters between P and S.}
\displaystyle \\

\displaystyle \textbf{Question 73: }\text{A committee of }7\text{ has to be formed from }9\text{ boys and }4\text{ girls. In how many}
\displaystyle \text{ways can this be done when the committee consists of:}
\displaystyle \text{(i) Exactly }3\text{ girls?}
\displaystyle \text{(ii) At least }3\text{ girls?}
\displaystyle \text{(iii) At most }3\text{ girls?}
\displaystyle \text{Answer:}

\displaystyle \text{(i) Exactly }3\text{ girls means }3\text{ girls and }4\text{ boys.}
\displaystyle \text{Number of ways}={}^{4}C_3\times{}^{9}C_4.
\displaystyle =4\times126=504.
\displaystyle \therefore 504\text{ committees can be formed.}

\displaystyle \text{(ii) At least }3\text{ girls means either }3\text{ girls or }4\text{ girls.}
\displaystyle \text{For }3\text{ girls and }4\text{ boys, number of ways}={}^{4}C_3\times{}^{9}C_4.
\displaystyle =504.
\displaystyle \text{For }4\text{ girls and }3\text{ boys, number of ways}={}^{4}C_4\times{}^{9}C_3.
\displaystyle =1\times84=84.
\displaystyle \therefore \text{Required number}=504+84=588.
\displaystyle \therefore 588\text{ committees can be formed.}

\displaystyle \text{(iii) At most }3\text{ girls means }0,1,2\text{ or }3\text{ girls.}
\displaystyle \text{For }0\text{ girls, number of ways}={}^{4}C_0\times{}^{9}C_7=36.
\displaystyle \text{For }1\text{ girl, number of ways}={}^{4}C_1\times{}^{9}C_6=4\times84=336.
\displaystyle \text{For }2\text{ girls, number of ways}={}^{4}C_2\times{}^{9}C_5=6\times126=756.
\displaystyle \text{For }3\text{ girls, number of ways}={}^{4}C_3\times{}^{9}C_4=504.
\displaystyle \therefore \text{Required number}=36+336+756+504=1632.
\displaystyle \therefore 1632\text{ committees can be formed.}
\displaystyle \\

\displaystyle \textbf{Question 74: }\text{A bag contains six white marbles and five red marbles. Find the number}
\displaystyle \text{of ways in which four marbles can be drawn from the bag if:}
\displaystyle \text{(i) They can be of any color}
\displaystyle \text{(ii) Two must be white and two red}
\displaystyle \text{(iii) They must be all of same color.}
\displaystyle \text{Answer:}

\displaystyle \text{There are }6\text{ white and }5\text{ red marbles, making a total of }11\text{ marbles.}

\displaystyle \text{(i) If the four marbles can be of any color, choose any }4\text{ from }11.
\displaystyle \text{Number of ways}={}^{11}C_4.
\displaystyle =\frac{11\times10\times9\times8}{4\times3\times2\times1}=330.
\displaystyle \therefore 330\text{ selections are possible.}

\displaystyle \text{(ii) We have to select }2\text{ white marbles and }2\text{ red marbles.}
\displaystyle \text{Number of ways}={}^{6}C_2\times{}^{5}C_2.
\displaystyle =15\times10=150.
\displaystyle \therefore 150\text{ selections are possible.}

\displaystyle \text{(iii) All four marbles must be white or all four must be red.}
\displaystyle \text{Number of ways of selecting }4\text{ white marbles}={}^{6}C_4=15.
\displaystyle \text{Number of ways of selecting }4\text{ red marbles}={}^{5}C_4=5.
\displaystyle \therefore \text{Required number}=15+5=20.
\displaystyle \therefore 20\text{ selections are possible.}
\displaystyle \\

\displaystyle \textbf{Question 75: }\text{From a class of }25\text{ students, }10\text{ are to be chosen for an excursion party.}
\displaystyle \text{There are }3\text{ students who decide that either all of them will join or none}
\displaystyle \text{of them will join. In how many ways can the excursion party be chosen?}
\displaystyle \text{Answer:}

\displaystyle \text{There are two possible cases.}

\displaystyle \text{Case I: All the }3\text{ particular students join the excursion party.}
\displaystyle \text{Then the remaining }7\text{ students are to be selected from the other }22\text{ students.}
\displaystyle \text{Number of ways}={}^{22}C_7.

\displaystyle \text{Case II: None of the }3\text{ particular students joins the excursion party.}
\displaystyle \text{Then all }10\text{ students are to be selected from the other }22\text{ students.}
\displaystyle \text{Number of ways}={}^{22}C_{10}.

\displaystyle \therefore \text{Required number of ways}={}^{22}C_7+{}^{22}C_{10}.
\displaystyle =170544+646646=817190.
\displaystyle \therefore \text{The excursion party can be chosen in }817190\text{ ways.}
\displaystyle \\

\displaystyle \textbf{Question 76: }\text{4 cards are chosen from the deck of }52\text{ cards. Find the number of ways to}
\displaystyle \text{choose}
\displaystyle \text{(i) four cards of same suit}
\displaystyle \text{(ii) two are red and two are black cards}
\displaystyle \text{(iii) four cards belong to four different suits}
\displaystyle \text{(iv) cards are of same color}
\displaystyle \text{(v) are face cards}
\displaystyle \text{(vi) at least two of them are red cards?}
\displaystyle \text{Answer:}

\displaystyle \text{(i) There are }4\text{ suits, each containing }13\text{ cards.}
\displaystyle \text{Choose one suit in }4\text{ ways and then choose }4\text{ cards from its }13\text{ cards.}
\displaystyle \therefore \text{Number of ways}=4\times{}^{13}C_4.
\displaystyle =4\times715=2860.
\displaystyle \therefore 2860\text{ selections are possible.}

\displaystyle \text{(ii) There are }26\text{ red cards and }26\text{ black cards.}
\displaystyle \text{Choose }2\text{ red cards and }2\text{ black cards.}
\displaystyle \therefore \text{Number of ways}={}^{26}C_2\times{}^{26}C_2.
\displaystyle =325\times325=105625.
\displaystyle \therefore 105625\text{ selections are possible.}

\displaystyle \text{(iii) To get four different suits, select one card from each of the }4\text{ suits.}
\displaystyle \therefore \text{Number of ways}=13\times13\times13\times13=13^4.
\displaystyle =28561.
\displaystyle \therefore 28561\text{ selections are possible.}

\displaystyle \text{(iv) All }4\text{ cards must either be red or all must be black.}
\displaystyle \text{Number of ways of choosing }4\text{ red cards}={}^{26}C_4.
\displaystyle \text{Number of ways of choosing }4\text{ black cards}={}^{26}C_4.
\displaystyle \therefore \text{Required number}=2\times{}^{26}C_4.
\displaystyle =2\times14950=29900.
\displaystyle \therefore 29900\text{ selections are possible.}

\displaystyle \text{(v) There are }12\text{ face cards in a deck: }4\text{ Jacks, }4\text{ Queens and }4\text{ Kings.}
\displaystyle \therefore \text{Number of ways of choosing }4\text{ face cards}={}^{12}C_4.
\displaystyle =495.
\displaystyle \therefore 495\text{ selections are possible.}

\displaystyle \text{(vi) At least }2\text{ red cards means exactly }2,\ 3\text{ or }4\text{ red cards.}
\displaystyle \text{For exactly }2\text{ red cards, number of ways}={}^{26}C_2\times{}^{26}C_2.
\displaystyle =105625.
\displaystyle \text{For exactly }3\text{ red cards, number of ways}={}^{26}C_3\times{}^{26}C_1.
\displaystyle =2600\times26=67600.
\displaystyle \text{For exactly }4\text{ red cards, number of ways}={}^{26}C_4=14950.
\displaystyle \therefore \text{Required number}=105625+67600+14950.
\displaystyle =188175.
\displaystyle \therefore 188175\text{ selections are possible.}
\displaystyle \\

\displaystyle \textbf{Question 77: }\text{A tea party is arranged for }18\text{ persons among two sides of a long table}
\displaystyle \text{with }9\text{ chairs on each side. Four guests wish to sit on one particular side}
\displaystyle \text{and three on the other side. In how many ways can they be seated?}
\displaystyle \text{Answer:}

\displaystyle \text{Four specified guests must sit on the first side and three specified guests on the second side.}
\displaystyle \text{Thus, }18-4-3=11\text{ guests remain to be allotted between the two sides.}

\displaystyle \text{The first side has }9-4=5\text{ vacant seats.}
\displaystyle \text{Choose }5\text{ of the remaining }11\text{ guests for this side in }{}^{11}C_5\text{ ways.}
\displaystyle \text{The remaining }6\text{ guests automatically go to the second side.}

\displaystyle \text{The }9\text{ persons on the first side can be arranged in }9!\text{ ways.}
\displaystyle \text{The }9\text{ persons on the second side can also be arranged in }9!\text{ ways.}

\displaystyle \therefore \text{Required number of arrangements}={}^{11}C_5\times9!\times9!.
\displaystyle =462(9!)^2.
\displaystyle \therefore \text{The required number of ways is }462(9!)^2.
\displaystyle \\

\displaystyle \textbf{Question 78: }\text{If }C(n,r-1)=36,\ C(n,r)=84\text{ and }C(n,r+1)=126,\text{ then find }C(r,2).
\displaystyle \text{Answer:}

\displaystyle \frac{C(n,r)}{C(n,r-1)}=\frac{n-r+1}{r}.
\displaystyle \therefore \frac{84}{36}=\frac{n-r+1}{r}.
\displaystyle \therefore \frac{7}{3}=\frac{n-r+1}{r}.
\displaystyle \therefore 3n+3=10r.\qquad ...(1)

\displaystyle \text{Also, }\frac{C(n,r+1)}{C(n,r)}=\frac{n-r}{r+1}.
\displaystyle \therefore \frac{126}{84}=\frac{n-r}{r+1}.
\displaystyle \therefore \frac{3}{2}=\frac{n-r}{r+1}.
\displaystyle \therefore 2n-2r=3r+3.
\displaystyle \therefore 2n=5r+3.\qquad ...(2)

\displaystyle \text{From (1), }3n=10r-3.
\displaystyle \text{From (2), }6n=15r+9.
\displaystyle \therefore 20r-6=15r+9.
\displaystyle \therefore 5r=15.
\displaystyle \therefore r=3.

\displaystyle \therefore C(r,2)=C(3,2)=3.
\displaystyle \therefore \text{The required value is }3.
\displaystyle \\

\displaystyle \textbf{Question 79: }\text{In a village, there are }87\text{ families of which }52\text{ families have at most }2
\displaystyle \text{children. In a rural development program, }20\text{ families are to be helped and chosen for}
\displaystyle \text{assistance, of which at least }18\text{ families must have at most }2\text{ children. In how many ways}
\displaystyle \text{can the choice be made?}
\displaystyle \text{Answer:}

\displaystyle \text{Number of families having at most }2\text{ children}=52.
\displaystyle \text{Number of remaining families}=87-52=35.

\displaystyle \text{At least }18\text{ of the }20\text{ selected families must have at most }2\text{ children.}
\displaystyle \text{Therefore, there are three possible cases.}

\displaystyle \text{Case I: }18\text{ families from }52\text{ and }2\text{ families from }35.
\displaystyle \text{Number of ways}={}^{52}C_{18}\times{}^{35}C_2.

\displaystyle \text{Case II: }19\text{ families from }52\text{ and }1\text{ family from }35.
\displaystyle \text{Number of ways}={}^{52}C_{19}\times{}^{35}C_1.

\displaystyle \text{Case III: All }20\text{ families are selected from the }52\text{ families.}
\displaystyle \text{Number of ways}={}^{52}C_{20}.

\displaystyle \therefore \text{Required number of ways}
\displaystyle ={}^{52}C_{18}\times{}^{35}C_2+{}^{52}C_{19}\times{}^{35}C_1+{}^{52}C_{20} \times {}^{35}C_{0}
\displaystyle =28188434477042385.
\displaystyle \therefore \text{The choice can be made in }28188434477042385\text{ ways.}
\displaystyle \\

\displaystyle \textbf{Question 80: }\text{Eight chairs are numbered from }1\text{ to }8.\text{ Two women and }3\text{ men wish to}
\displaystyle \text{occupy one chair each. First the women choose the chairs amongst the chairs }1\text{ to }4
\displaystyle \text{and then men select from the remaining chairs. Find the possible arrangements.}
\displaystyle \text{Answer:}

\displaystyle \text{The }2\text{ women have to occupy }2\text{ of the first }4\text{ chairs.}
\displaystyle \text{Number of ways of seating the women}={}^{4}P_2.
\displaystyle =4\times3=12.

\displaystyle \text{After seating the women, }8-2=6\text{ chairs remain vacant.}
\displaystyle \text{The }3\text{ men can occupy }3\text{ of these }6\text{ chairs in }{}^{6}P_3\text{ ways.}
\displaystyle {}^{6}P_3=6\times5\times4=120.

\displaystyle \therefore \text{Total number of arrangements}={}^{4}P_2\times{}^{6}P_3.
\displaystyle =12\times120=1440.
\displaystyle \therefore \text{The number of possible arrangements is }1440.
\displaystyle \\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.