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

\displaystyle \textbf{Question 1: }\text{The number of permutations of }n\text{ different things taking }r\text{ at a time when}
\displaystyle 3\text{ particular things are to be included is}
\displaystyle \text{(a) }{}^{n-3}P_{r-3}\qquad \text{(b) }{}^{n-3}P_r\qquad \text{(c) }{}^nP_{r-3}\qquad \text{(d) }r!\,{}^{n-3}C_{r-3}
\displaystyle \text{Answer: (d) }r!\,{}^{n-3}C_{r-3}
\displaystyle \text{The }3\text{ particular things must always be included.}
\displaystyle \text{Choose the remaining }r-3\text{ things from the other }n-3\text{ things.}
\displaystyle \text{Number of ways of choosing them}={}^{n-3}C_{r-3}.
\displaystyle \text{The resulting }r\text{ distinct things can be arranged in }r!\text{ ways.}
\displaystyle \therefore \text{Required number}=r!\,{}^{n-3}C_{r-3}.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The number of five-digit telephone numbers having at least one of their}
\displaystyle \text{digits repeated is}
\displaystyle \text{(a) }90000\qquad \text{(b) }100000\qquad \text{(c) }30240\qquad \text{(d) }69760
\displaystyle \text{Answer: (d) }69760
\displaystyle \text{A telephone number may begin with }0.
\displaystyle \text{Total number of five-digit telephone numbers}=10^5=100000.
\displaystyle \text{Number having no repeated digit}={}^{10}P_5.
\displaystyle =10\times9\times8\times7\times6=30240.
\displaystyle \text{Hence, number having at least one repeated digit}
\displaystyle =100000-30240=69760.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The number of words that can be formed out of the letters of the word}
\displaystyle \text{``ARTICLE'' so that vowels occupy even places is}
\displaystyle \text{(a) }574\qquad \text{(b) }36\qquad \text{(c) }754\qquad \text{(d) }144
\displaystyle \text{Answer: (d) }144
\displaystyle \text{The word ARTICLE contains }3\text{ vowels }A,I,E\text{ and }4\text{ consonants }R,T,C,L.
\displaystyle \text{The even places among }7\text{ positions are }2,4,6.
\displaystyle \text{The }3\text{ vowels can be arranged in the even places in }3!\text{ ways.}
\displaystyle \text{The }4\text{ consonants can be arranged in the remaining places in }4!\text{ ways.}
\displaystyle \text{Required number of words}=3!\times4!.
\displaystyle =6\times24=144.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{How many numbers greater than 10 lacs can be formed from }2,3,0,3,4,2,3?
\displaystyle \text{(a) }420\qquad \text{(b) }360\qquad \text{(c) }400\qquad \text{(d) }300
\displaystyle \text{Answer: (b) }360
\displaystyle \text{A number greater than }10\text{ lacs must have }7\text{ digits.}
\displaystyle \text{Hence, all the given digits }2,3,0,3,4,2,3\text{ must be used.}
\displaystyle \text{Here, }3\text{ occurs }3\text{ times and }2\text{ occurs }2\text{ times.}
\displaystyle \text{Total number of distinct arrangements}=\frac{7!}{3!\,2!}=420.
\displaystyle \text{Arrangements beginning with }0=\frac{6!}{3!\,2!}=60.
\displaystyle \text{Required number of }7\text{ digit numbers}=420-60=360.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The number of different signals which can be given from 6 flags of different}
\displaystyle \text{colours taking one or more at a time, is}
\displaystyle \text{(a) }1958\qquad \text{(b) }1956\qquad \text{(c) }16\qquad \text{(d) }64
\displaystyle \text{Answer: (b) }1956
\displaystyle \text{The }6\text{ flags are of different colours and one or more flags may be used.}
\displaystyle \text{Required number of signals} ={}^{6}P_1+{}^{6}P_2+{}^{6}P_3+{}^{6}P_4+{}^{6}P_5+{}^{6}P_6.
\displaystyle =6+30+120+360+720+720.
\displaystyle =1956.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The number of words from the letters of the word 'BHARAT' in which }
\displaystyle \text{B and H will never come together, is}
\displaystyle \text{(a) }360\qquad \text{(b) }240\qquad \text{(c) }120\qquad \text{(d) none of these}
\displaystyle \text{Answer: (b) }240
\displaystyle \text{The word BHARAT contains }6\text{ letters, of which A occurs twice.}
\displaystyle \text{Total number of distinct arrangements}=\frac{6!}{2!}=360.
\displaystyle \text{Now, consider B and H together as one unit.}
\displaystyle \text{We then have }5\text{ objects, of which A occurs twice.}
\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 \text{Number of arrangements in which B and H are together} =\frac{5!}{2!}\times2!=120.
\displaystyle \text{Required number of arrangements}=360-120=240.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The number of six letter words that can be formed using the letters }
\displaystyle \text{of the word ``ASSIST'' in which S's alternate with other letters is}
\displaystyle \text{(a) }12\qquad \text{(b) }24\qquad \text{(c) }18\qquad \text{(d) none of these}
\displaystyle \text{Answer: (a) }12
\displaystyle \text{The letters are }A,S,S,I,S,T.
\displaystyle \text{There are }3\text{ S's and }3\text{ other distinct letters }A,I,T.
\displaystyle \text{For the S's to alternate with the other letters, the possible patterns are}
\displaystyle S\,\_\,S\,\_\,S\,\_\qquad\text{or}\qquad \_\,S\,\_\,S\,\_\,S.
\displaystyle \text{For each pattern, }A,I,T\text{ can be arranged in }3!\text{ ways.}
\displaystyle \text{Required number of words}=2\times3!=2\times6=12.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The number of arrangements of the word ``DELHI'' in which E precedes I is}
\displaystyle \text{(a) }30\qquad \text{(b) }60\qquad \text{(c) }120\qquad \text{(d) }59
\displaystyle \text{Answer: (b) }60
\displaystyle \text{The word DELHI contains }5\text{ distinct letters.}
\displaystyle \text{Total number of arrangements}=5!=120.
\displaystyle \text{For every arrangement in which E precedes I, there is a corresponding arrangement}
\displaystyle \text{in which I precedes E. Hence, exactly half the arrangements have E before I.}
\displaystyle \text{Required number of arrangements}=\frac{5!}{2}.
\displaystyle =\frac{120}{2}=60.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The number of ways in which the letters of the word ``CONSTANT'' can }
\displaystyle \text{be arranged without changing the relative positions of the vowels and consonants is}
\displaystyle \text{(a) }360\qquad \text{(b) }256\qquad \text{(c) }444\qquad \text{(d) none of these}
\displaystyle \text{Answer: (a) }360
\displaystyle \text{The word CONSTANT has }2\text{ vowels }O,A\text{ and }6\text{ consonants }C,N,S,T,N,T.
\displaystyle \text{The vowels can be arranged among the vowel positions in }2!\text{ ways.}
\displaystyle \text{Among the consonants, N occurs twice and T occurs twice.}
\displaystyle \text{The consonants can be arranged among the consonant positions in } \frac{6!}{2!\,2!}\text{ ways.}
\displaystyle \text{Required number of arrangements} =2!\times\frac{6!}{2!\,2!}.
\displaystyle =2\times\frac{720}{4}=360.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The number of ways to arrange the letters of the word CHEESE are}
\displaystyle \text{(a) }120\qquad \text{(b) }240\qquad \text{(c) }720\qquad \text{(d) }6
\displaystyle \text{Answer: (a) }120
\displaystyle \text{The word CHEESE contains }6\text{ letters, of which E occurs }3\text{ times.}
\displaystyle \text{The remaining letters C, H and S occur once each.}
\displaystyle \text{Therefore, the number of distinct arrangements}=\frac{6!}{3!}.
\displaystyle =\frac{720}{6}=120.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Number of all four digit numbers having different digits formed of the digits}
\displaystyle 1,2,3,4\text{ and }5\text{ and divisible by }4\text{ is}
\displaystyle \text{(a) }24\qquad \text{(b) }30\qquad \text{(c) }125\qquad \text{(d) }100
\displaystyle \text{Answer: (a) }24
\displaystyle \text{A number is divisible by }4\text{ if the number formed by its last two digits is divisible by }4.
\displaystyle \text{Using different digits from }1,2,3,4,5,\text{ the possible last two digits are}
\displaystyle 12,\ 24,\ 32,\ 52.
\displaystyle \text{Thus, there are }4\text{ choices for the last two digits.}
\displaystyle \text{For each choice, the first two places can be filled using }2\text{ of the remaining }3\text{ digits.}
\displaystyle \text{Number of ways of filling the first two places}={}^{3}P_2=3\times2=6.
\displaystyle \text{Required number of four digit numbers}=4\times6=24.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If the letters of the word KRISNA are arranged in all possible ways and }
\displaystyle \text{these words are written out in a dictionary, then the rank of the word KRISNA is}
\displaystyle \text{(a) }324\qquad \text{(b) }341\qquad \text{(c) }359\qquad \text{(d) none of these}
\displaystyle \text{Answer: (a) }324
\displaystyle \text{The alphabetical order of the letters is }A,I,K,N,R,S.
\displaystyle \text{Words beginning with A or I before K}=2\times5!=240.
\displaystyle \text{Fixing K, the letters before R are A, I, N.}
\displaystyle \text{Number of such words}=3\times4!=72.
\displaystyle \text{Fixing KR, the only letter before I is A.}
\displaystyle \text{Number of such words}=1\times3!=6.
\displaystyle \text{Fixing KRI, the letters before S are A and N.}
\displaystyle \text{Number of such words}=2\times2!=4.
\displaystyle \text{Fixing KRIS, the only letter before N is A.}
\displaystyle \text{Number of such words}=1!=1.
\displaystyle \text{Total number of words before KRISNA}=240+72+6+4+1=323.
\displaystyle \therefore \text{Rank of KRISNA}=323+1=324.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If in a group of }n\text{ distinct objects, the number of arrangements of }
\displaystyle \text{4 objects is 12 times the number of arrangements of 2 objects, then the number of objects is}
\displaystyle \text{(a) }10\qquad \text{(b) }8\qquad \text{(c) }6\qquad \text{(d) none of these}
\displaystyle \text{Answer: (c) }6
\displaystyle {}^nP_4=12\,{}^nP_2.
\displaystyle n(n-1)(n-2)(n-3)=12n(n-1).
\displaystyle (n-2)(n-3)=12.
\displaystyle n^2-5n+6=12.
\displaystyle n^2-5n-6=0.
\displaystyle (n-6)(n+1)=0.
\displaystyle n=6\text{ or }n=-1.
\displaystyle \text{Since }n\text{ is the number of objects, }n=6.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{The number of ways in which 6 men can be arranged in a row so that three}
\displaystyle \text{particular men are consecutive, is}
\displaystyle \text{(a) }4!\times3!\qquad \text{(b) }4!\qquad \text{(c) }3!\times3!\qquad \text{(d) none of these}
\displaystyle \text{Answer: (a) }4!\times3!
\displaystyle \text{Consider the three particular men as one unit.}
\displaystyle \text{Together with the other }3\text{ men, there are }4\text{ units.}
\displaystyle \text{These }4\text{ units can be arranged in }4!\text{ ways.}
\displaystyle \text{The three particular men can be arranged among themselves in }3!\text{ ways.}
\displaystyle \text{Required number of arrangements}=4!\times3!.
\displaystyle =24\times6=144.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{A 5-digit number divisible by 3 is to be formed using the digits }0,1,2,3,4
\displaystyle \text{ and }5 \ \text{without repetition. The total number of ways in which this can be done is}
\displaystyle \text{(a) }216\qquad \text{(b) }600\qquad \text{(c) }240\qquad \text{(d) }3125
\displaystyle \text{Answer: (a) }216
\displaystyle \text{The sum of all the given digits is }0+1+2+3+4+5=15.
\displaystyle \text{For divisibility by }3,\text{ the sum of the }5\text{ selected digits must be divisible by }3.
\displaystyle \text{Hence, the omitted digit must be }0\text{ or }3.
\displaystyle \text{Case I: Omit }0.
\displaystyle \text{The digits }1,2,3,4,5\text{ can be arranged in }5!=120\text{ ways.}
\displaystyle \text{Case II: Omit }3.
\displaystyle \text{Using }0,1,2,4,5,\text{ total arrangements}=5!=120.
\displaystyle \text{Arrangements beginning with }0=4!=24.
\displaystyle \text{Therefore, valid }5\text{-digit numbers}=120-24=96.
\displaystyle \text{Required number}=120+96=216.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The product of }r\text{ consecutive positive integers is divisible by}
\displaystyle \text{(a) }r!\qquad \text{(b) }r!+1\qquad \text{(c) }(r+1)!\qquad \text{(d) none of these}
\displaystyle \text{Answer: (a) }r!
\displaystyle \text{Let the }r\text{ consecutive positive integers be }n-r+1,n-r+2,\ldots,n.
\displaystyle \text{Their product}=(n-r+1)(n-r+2)\cdots n.
\displaystyle =\frac{n!}{(n-r)!}.
\displaystyle =r!\left(\frac{n!}{r!(n-r)!}\right).
\displaystyle =r!\,{}^nC_r.
\displaystyle \text{Since }{}^nC_r\text{ is an integer, the product is divisible by }r!.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If }{}^{k+5}P_{k+1} =\frac{11(k-1)}{2}\,{}^{k+3}P_k,\text{ then the values of }k\text{ are}
\displaystyle \text{(a) }7\text{ and }11\qquad \text{(b) }6\text{ and }7\qquad \text{(c) }2\text{ and }11\qquad \text{(d) }2\text{ and }6
\displaystyle \text{Answer: (b) }6\text{ and }7
\displaystyle {}^{k+5}P_{k+1} =\frac{11(k-1)}{2}\,{}^{k+3}P_k.
\displaystyle \frac{(k+5)!}{4!} =\frac{11(k-1)}{2}\cdot\frac{(k+3)!}{3!}.
\displaystyle \frac{(k+5)(k+4)}{4} =\frac{11(k-1)}{2}.
\displaystyle (k+5)(k+4)=22(k-1).
\displaystyle k^2+9k+20=22k-22.
\displaystyle k^2-13k+42=0.
\displaystyle (k-6)(k-7)=0.
\displaystyle \therefore k=6\text{ or }k=7.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The number of arrangements of the letters of the word BHARAT taking 3}
\displaystyle \text{at a time is (a) }72\qquad \text{(b) }120\qquad \text{(c) }14\qquad \text{(d) none of these}
\displaystyle \text{Answer: (a) }72
\displaystyle \text{The letters of BHARAT are }B,H,A,R,A,T.
\displaystyle \text{There are }4\text{ distinct non-A letters }B,H,R,T\text{ and two identical A's.}
\displaystyle \text{Case I: No A is used.}
\displaystyle \text{Number of arrangements}={}^{4}P_3=24.
\displaystyle \text{Case II: Exactly one A is used.}
\displaystyle \text{Choose }2\text{ letters from }B,H,R,T\text{ and arrange them with A.}
\displaystyle \text{Number of arrangements}={}^{4}C_2\times3!=6\times6=36.
\displaystyle \text{Case III: Both A's are used.}
\displaystyle \text{Choose }1\text{ letter from }B,H,R,T\text{ in }4\text{ ways.}
\displaystyle \text{The letters }A,A\text{ and the chosen letter can be arranged in }\frac{3!}{2!}=3\text{ ways.}
\displaystyle \text{Number of arrangements}=4\times3=12.
\displaystyle \text{Required number}=24+36+12=72.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The number of words that can be made by re-arranging the letters of the }
\displaystyle \text{word APURBA so that vowels and consonants are alternate is}
\displaystyle \text{(a) }18\qquad \text{(b) }35\qquad \text{(c) }36\qquad \text{(d) none of these}
\displaystyle \text{Answer: (c) }36
\displaystyle \text{The vowels are }A,U,A\text{ and the consonants are }P,R,B.
\displaystyle \text{The vowels can be arranged in }\frac{3!}{2!}=3\text{ ways.}
\displaystyle \text{The consonants can be arranged in }3!=6\text{ ways.}
\displaystyle \text{There are two possible alternating patterns:}
\displaystyle VCV CVC\quad\text{or}\quad CVC VCV.
\displaystyle \text{Required number of words}=2\times\frac{3!}{2!}\times3!.
\displaystyle =2\times3\times6=36.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{The number of different ways in which 8 persons can stand in a row so that }
\displaystyle \text{between two particular persons } A\text{ and }B\text{ there are always two persons, is}
\displaystyle \text{(a) }60\times5!\qquad \text{(b) }15\times4!\times5!\qquad \text{(c) }4!\times5!\qquad \text{(d) none of these}
\displaystyle \text{Answer: (a) }60\times5!
\displaystyle \text{Since exactly two persons are between }A\text{ and }B,\text{ their positions must differ by }3.
\displaystyle \text{The possible pairs of positions are }(1,4),(2,5),(3,6),(4,7),(5,8).
\displaystyle \text{Thus, there are }5\text{ possible pairs of positions for }A\text{ and }B.
\displaystyle A\text{ and }B\text{ can interchange their positions in }2!\text{ ways.}
\displaystyle \text{The remaining }6\text{ persons can be arranged in }6!\text{ ways.}
\displaystyle \text{Required number of arrangements}=5\times2!\times6!.
\displaystyle =10\times6!=10\times6\times5!=60\times5!.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{The number of ways in which the letters of the word ARTICLE can be }
\displaystyle \text{arranged so that even places are always occupied by consonants is}
\displaystyle \text{(a) }576\qquad \text{(b) }{}^4C_3\times4!\qquad \text{(c) }2\times4!\qquad \text{(d) none of these}
\displaystyle \text{Answer: (a) }576
\displaystyle \text{The word ARTICLE has }4\text{ consonants }R,T,C,L\text{ and }3\text{ vowels }A,I,E.
\displaystyle \text{The even places are }2,4,6,\text{ and all must be occupied by consonants.}
\displaystyle \text{Choose and arrange }3\text{ of the }4\text{ consonants in these places in }{}^4P_3\text{ ways.}
\displaystyle \text{The remaining }4\text{ letters can be arranged in the remaining }4\text{ places in }4!\text{ ways.}
\displaystyle \text{Required number of arrangements}={}^4P_3\times4!.
\displaystyle =(4\times3\times2)\times24=24\times24=576.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{In a room there are 12 bulbs of the same wattage, each having a separate}
\displaystyle \text{switch. The number of ways to light the room with different amounts of illumination is}
\displaystyle \text{(a) }12^2-1\qquad \text{(b) }2^{12}\qquad \text{(c) }2^{12}-1\qquad \text{(d) none of these}
\displaystyle \text{Answer: (d) none of these}
\displaystyle \text{Since all }12\text{ bulbs have the same wattage, the amount of illumination depends}
\displaystyle \text{only on the number of bulbs which are switched on.}
\displaystyle \text{For lighting the room, the number of bulbs switched on can be }1,2,3,\ldots,12.
\displaystyle \therefore \text{There are }12\text{ different amounts of illumination.}
\displaystyle \text{Since }12\text{ is not among the given options, the correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{If }{}^{20}C_r={}^{20}C_{r-10},\text{ then }{}^{18}C_r\text{ is equal to}
\displaystyle \text{(a) }4896\qquad \text{(b) }816\qquad \text{(c) }1632\qquad \text{(d) none of these}
\displaystyle \text{Answer: (b) }816
\displaystyle \text{We know that if }{}^nC_x={}^nC_y,\text{ then }x=y\text{ or }x+y=n.
\displaystyle \text{Here, }r\ne r-10,\text{ therefore }r+(r-10)=20.
\displaystyle 2r-10=20.
\displaystyle 2r=30\Rightarrow r=15.
\displaystyle \therefore {}^{18}C_r={}^{18}C_{15}={}^{18}C_3.
\displaystyle =\frac{18\times17\times16}{3\times2\times1}=816.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{If }{}^{20}C_r={}^{20}C_{r+4},\text{ then }{}^rC_3\text{ is equal to}
\displaystyle \text{(a) }54\qquad \text{(b) }56\qquad \text{(c) }58\qquad \text{(d) none of these}
\displaystyle \text{Answer: (b) }56
\displaystyle \text{Since }r\ne r+4,\text{ we use }r+(r+4)=20.
\displaystyle 2r+4=20.
\displaystyle 2r=16\Rightarrow r=8.
\displaystyle \therefore {}^rC_3={}^8C_3.
\displaystyle =\frac{8\times7\times6}{3\times2\times1}=56.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If }{}^{15}C_{3r}={}^{15}C_{r+3},\text{ then }r\text{ is equal to}
\displaystyle \text{(a) }5\qquad \text{(b) }4\qquad \text{(c) }3\qquad \text{(d) }2
\displaystyle \text{Answer: (c) }3
\displaystyle \text{We know that if }{}^nC_x={}^nC_y,\text{ then }x=y\text{ or }x+y=n.
\displaystyle \therefore 3r=r+3\quad\text{or}\quad3r+(r+3)=15.
\displaystyle \text{From }3r=r+3,\text{ we get }2r=3\Rightarrow r=\frac{3}{2},
\displaystyle \text{which is not valid since the indices of a combination must be integers.}
\displaystyle \text{From }3r+(r+3)=15,
\displaystyle 4r+3=15\Rightarrow4r=12\Rightarrow r=3.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{If }{}^{20}C_{r+1}={}^{20}C_{r-1},\text{ then }r\text{ is equal to}
\displaystyle \text{(a) }10\qquad \text{(b) }11\qquad \text{(c) }19\qquad \text{(d) }12
\displaystyle \text{Answer: (a) }10
\displaystyle \text{Since }r+1\ne r-1,\text{ the two lower indices must be complementary.}
\displaystyle \therefore (r+1)+(r-1)=20.
\displaystyle 2r=20.
\displaystyle \therefore r=10.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{If }C(n,12)=C(n,8),\text{ then }C(22,n)\text{ is equal to}
\displaystyle \text{(a) }231\qquad \text{(b) }210\qquad \text{(c) }252\qquad \text{(d) }303
\displaystyle \text{Answer: (a) }231
\displaystyle \text{We know that if }{}^nC_x={}^nC_y,\text{ then }x=y\text{ or }x+y=n.
\displaystyle \text{Since }12\ne8,\text{ we must have }12+8=n.
\displaystyle \therefore n=20.
\displaystyle \therefore C(22,n)=C(22,20)=C(22,2).
\displaystyle =\frac{22\times21}{2\times1}=231.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{If }{}^mC_1={}^nC_2,\text{ then}
\displaystyle \text{(a) }2m=n\qquad \text{(b) }2m=n(n+1)\qquad \text{(c) }2m=n(n-1)\qquad \text{(d) }2n=m(m-1)
\displaystyle \text{Answer: (c) }2m=n(n-1)
\displaystyle {}^mC_1={}^nC_2.
\displaystyle m=\frac{n(n-1)}{2}.
\displaystyle \therefore 2m=n(n-1).
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{If }{}^nC_{12}={}^nC_8,\text{ then }n=
\displaystyle \text{(a) }20\qquad \text{(b) }12\qquad \text{(c) }6\qquad \text{(d) }30
\displaystyle \text{Answer: (a) }20
\displaystyle \text{We know that if }{}^nC_x={}^nC_y,\text{ then }x=y\text{ or }x+y=n.
\displaystyle \text{Since }12\ne8,\text{ we must have }12+8=n.
\displaystyle \therefore n=20.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{If }{}^nC_r+{}^nC_{r+1}={}^{n+1}C_x,\text{ then }x=
\displaystyle \text{(a) }r\qquad \text{(b) }r-1\qquad \text{(c) }n\qquad \text{(d) }r+1
\displaystyle \text{Answer: (d) }r+1
\displaystyle \text{Using Pascal's identity,}
\displaystyle {}^nC_r+{}^nC_{r+1}={}^{n+1}C_{r+1}.
\displaystyle \text{Comparing with }{}^nC_r+{}^nC_{r+1}={}^{n+1}C_x,
\displaystyle x=r+1.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{If }{}^{(a^2-a)}C_2={}^{(a^2-a)}C_4,\text{ then }a=
\displaystyle \text{(a) }2\qquad \text{(b) }3\qquad \text{(c) }4\qquad \text{(d) none of these}
\displaystyle \text{Answer: (b) }3
\displaystyle \text{We know that if }{}^nC_x={}^nC_y,\text{ then }x=y\text{ or }x+y=n.
\displaystyle \text{Since }2\ne4,\text{ we must have }2+4=a^2-a.
\displaystyle a^2-a=6.
\displaystyle a^2-a-6=0.
\displaystyle (a-3)(a+2)=0.
\displaystyle \therefore a=3\text{ or }a=-2.
\displaystyle \text{Among the given options, }a=3.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 32: }{}^5C_1+{}^5C_2+{}^5C_3+{}^5C_4+{}^5C_5\text{ is equal to}
\displaystyle \text{(a) }30\qquad \text{(b) }31\qquad \text{(c) }32\qquad \text{(d) }33
\displaystyle \text{Answer: (b) }31
\displaystyle \text{We know that }\sum_{r=0}^{5}{}^5C_r=2^5=32.
\displaystyle {}^5C_0+{}^5C_1+{}^5C_2+{}^5C_3+{}^5C_4+{}^5C_5=32.
\displaystyle \therefore {}^5C_1+{}^5C_2+{}^5C_3+{}^5C_4+{}^5C_5=32-{}^5C_0.
\displaystyle =32-1=31.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{Total number of words formed by 2 vowels and 3 consonants taken}
\displaystyle \text{ from 4 vowels and 5 consonants is equal to}
\displaystyle \text{(a) }60\qquad \text{(b) }120\qquad \text{(c) }7200\qquad \text{(d) none of these}
\displaystyle \text{Answer: (c) }7200
\displaystyle \text{The number of ways of selecting }2\text{ vowels from }4\text{ vowels is }{}^4C_2.
\displaystyle \text{The number of ways of selecting }3\text{ consonants from }5\text{ consonants is }{}^5C_3.
\displaystyle \text{The }5\text{ selected letters can be arranged among themselves in }5!\text{ ways.}
\displaystyle \therefore \text{Required number of words}={}^4C_2\times{}^5C_3\times5!.
\displaystyle =6\times10\times120=7200.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{There are 12 points in a plane. The number of the straight lines joining }
\displaystyle \text{any two of them when 3 of them are collinear, is}
\displaystyle \text{(a) }62\qquad \text{(b) }63\qquad \text{(c) }64\qquad \text{(d) }65
\displaystyle \text{Answer: (c) }64
\displaystyle \text{If no three points were collinear, the number of straight lines would be }{}^{12}C_2.
\displaystyle {}^{12}C_2=\frac{12\times11}{2}=66.
\displaystyle \text{The }3\text{ collinear points give }{}^3C_2=3\text{ pairs, but these determine only one line.}
\displaystyle \text{Hence, the same line has been counted }3\text{ times instead of once.}
\displaystyle \therefore \text{Required number of straight lines}=66-3+1.
\displaystyle =64.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{Three persons enter a railway compartment. If there are 5 seats vacant, }
\displaystyle \text{in how many ways can they take these seats?}
\displaystyle \text{(a) }60\qquad \text{(b) }20\qquad \text{(c) }15\qquad \text{(d) }125
\displaystyle \text{Answer: (a) }60
\displaystyle \text{Three persons are to occupy }3\text{ of the }5\text{ vacant seats.}
\displaystyle \text{Since the arrangement of persons in the seats matters, we use permutations.}
\displaystyle \text{Required number of ways}={}^5P_3.
\displaystyle =5\times4\times3=60.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{In how many ways can a committee of 5 be made out of 6 men and 4 women }
\displaystyle \text{containing at least one woman? (a) }246\qquad \text{(b) }222\qquad \text{(c) }186\qquad \text{(d) none of these}
\displaystyle \text{Answer: (a) }246
\displaystyle \text{There are }6+4=10\text{ persons in all.}
\displaystyle \text{Total number of committees of }5\text{ persons}={}^{10}C_5.
\displaystyle \text{Committees containing no woman consist of }5\text{ men selected from }6\text{ men.}
\displaystyle \text{Number of such committees}={}^6C_5.
\displaystyle \therefore \text{Required number of committees}={}^{10}C_5-{}^6C_5.
\displaystyle =252-6=246.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{There are 10 points in a plane and 4 of them are collinear. The number of }
\displaystyle \text{straight lines joining any two of them is}
\displaystyle \text{(a) }45\qquad \text{(b) }40\qquad \text{(c) }39\qquad \text{(d) }38
\displaystyle \text{Answer: (b) }40
\displaystyle \text{If no three points were collinear, the number of straight lines would be }{}^{10}C_2.
\displaystyle {}^{10}C_2=\frac{10\times9}{2}=45.
\displaystyle \text{The }4\text{ collinear points give }{}^4C_2=6\text{ pairs, but these determine only one line.}
\displaystyle \therefore \text{Required number of straight lines}=45-6+1.
\displaystyle =40.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{There are 13 players of cricket, out of which 4 are bowlers. In how many }
\displaystyle \text{ways a team of eleven be selected from them so as to include at least two bowlers?}
\displaystyle \text{(a) }72\qquad \text{(b) }78\qquad \text{(c) }42\qquad \text{(d) none of these}
\displaystyle \text{Answer: (b) }78
\displaystyle \text{There are }4\text{ bowlers and }9\text{ other players.}
\displaystyle \text{A team of }11\text{ players must contain at least }11-9=2\text{ bowlers.}
\displaystyle \text{Hence, every selection of }11\text{ players automatically contains at least }2\text{ bowlers.}
\displaystyle \therefore \text{Required number of teams}={}^{13}C_{11}.
\displaystyle ={}^{13}C_2=\frac{13\times12}{2}=78.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{If }C_0+C_1+C_2+\cdots+C_n=256,\text{ then }{}^{2n}C_2\text{ is equal to}
\displaystyle \text{(a) }56\qquad \text{(b) }120\qquad \text{(c) }28\qquad \text{(d) }91
\displaystyle \text{Answer: (b) }120
\displaystyle \text{We know that }{}^nC_0+{}^nC_1+{}^nC_2+\cdots+{}^nC_n=2^n.
\displaystyle \therefore 2^n=256=2^8.
\displaystyle \therefore n=8.
\displaystyle \therefore {}^{2n}C_2={}^{16}C_2.
\displaystyle =\frac{16\times15}{2}=120.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{The number of ways in which a host lady can invite for a party of 8 out of }
\displaystyle \text{12 people of whom two do not want to attend the party together is}
\displaystyle \text{(a) }2\times{}^{10}C_7+{}^{10}C_8\qquad \text{(b) }{}^{10}C_8+{}^{11}C_7
\displaystyle \text{(c) }{}^{12}C_8-{}^{10}C_6\qquad \text{(d) none of these}
\displaystyle \text{Answer: (c) }{}^{12}C_8-{}^{10}C_6
\displaystyle \text{Total number of ways of selecting }8\text{ people from }12\text{ people is }{}^{12}C_8.
\displaystyle \text{If the two particular people are both selected, the remaining }6\text{ people}
\displaystyle \text{can be selected from the other }10\text{ people in }{}^{10}C_6\text{ ways.}
\displaystyle \therefore \text{Required number of ways}={}^{12}C_8-{}^{10}C_6.
\displaystyle =495-210=285.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{Given 11 points, of which 5 lie on one circle, other than these 5, no 4 lie on }
\displaystyle \text{one circle. Then the number of circles that can be drawn so that each contains at least 3 of the }
\displaystyle \text{given points is (a) }216\qquad \text{(b) }156\qquad \text{(c) }172\qquad \text{(d) none of these}
\displaystyle \text{Answer: (b) }156
\displaystyle \text{Any }3\text{ non-collinear points determine a unique circle.}
\displaystyle \text{If no }4\text{ points were concyclic, the number of circles would be }{}^{11}C_3.
\displaystyle {}^{11}C_3=165.
\displaystyle \text{But the }5\text{ concyclic points give }{}^5C_3=10\text{ triples, all determining the same circle.}
\displaystyle \text{Thus, these }10\text{ circles must be replaced by }1\text{ circle.}
\displaystyle \therefore \text{Required number of circles}={}^{11}C_3-{}^5C_3+1.
\displaystyle =165-10+1=156.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{How many different committees of 5 can be formed from 6 men and }
\displaystyle \text{4 women on which exact 3 men and 2 women serve?}
\displaystyle \text{(a) }6\qquad \text{(b) }20\qquad \text{(c) }60\qquad \text{(d) }120
\displaystyle \text{Answer: (d) }120
\displaystyle \text{The number of ways of selecting }3\text{ men from }6\text{ men is }{}^6C_3.
\displaystyle \text{The number of ways of selecting }2\text{ women from }4\text{ women is }{}^4C_2.
\displaystyle \therefore \text{Required number of committees}={}^6C_3\times{}^4C_2.
\displaystyle =20\times6=120.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{If }{}^{43}C_{r-6}={}^{43}C_{3r+1},\text{ then the value of }r\text{ is}
\displaystyle \text{(a) }12\qquad \text{(b) }8\qquad \text{(c) }6\qquad \text{(d) }10\qquad \text{(e) }14
\displaystyle \text{Answer: (a) }12
\displaystyle \text{We know that if }{}^nC_x={}^nC_y,\text{ then }x=y\text{ or }x+y=n.
\displaystyle \text{If }r-6=3r+1,\text{ then }r=-\frac{7}{2},\text{ which is not admissible.}
\displaystyle \therefore (r-6)+(3r+1)=43.
\displaystyle 4r-5=43.
\displaystyle 4r=48.
\displaystyle \therefore r=12.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{The number of diagonals that can be drawn by joining the vertices of an octagon is}
\displaystyle \text{(a) }20\qquad \text{(b) }28\qquad \text{(c) }8\qquad \text{(d) }16
\displaystyle \text{Answer: (a) }20
\displaystyle \text{An octagon has }8\text{ vertices.}
\displaystyle \text{The number of line segments obtained by joining any two vertices is }{}^8C_2.
\displaystyle \text{Out of these, }8\text{ are the sides of the octagon.}
\displaystyle \therefore \text{Number of diagonals}={}^8C_2-8.
\displaystyle =\frac{8\times7}{2}-8=28-8=20.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{The value of }\left({}^7C_0+{}^7C_1\right)+\left({}^7C_1+{}^7C_2\right)+\cdots+\left({}^7C_6+{}^7C_7\right)\text{ is}
\displaystyle \text{(a) }2^7-1\qquad \text{(b) }2^8-2\qquad \text{(c) }2^8-1\qquad \text{(d) }2^8
\displaystyle \text{Answer: (b) }2^8-2
\displaystyle \text{Using Pascal's identity, }{}^7C_r+{}^7C_{r+1}={}^8C_{r+1}.
\displaystyle \therefore \text{Given expression}={}^8C_1+{}^8C_2+\cdots+{}^8C_7.
\displaystyle \text{We know that }{}^8C_0+{}^8C_1+\cdots+{}^8C_8=2^8.
\displaystyle \therefore {}^8C_1+{}^8C_2+\cdots+{}^8C_7 =2^8-{}^8C_0-{}^8C_8.
\displaystyle =2^8-1-1=2^8-2.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 46: }\text{Among 14 players, 5 are bowlers. In how many ways a team of 11 }
\displaystyle \text{may be formed with at least 4 bowlers? (a) }265\qquad \text{(b) }263\qquad \text{(c) }264\qquad \text{(d) }275
\displaystyle \text{Answer: (c) }264
\displaystyle \text{There are }5\text{ bowlers and }14-5=9\text{ other players.}
\displaystyle \text{For at least }4\text{ bowlers, the team may contain }4\text{ or }5\text{ bowlers.}
\displaystyle \text{Case I: }4\text{ bowlers and }7\text{ other players.}
\displaystyle \text{Number of ways}={}^5C_4\times{}^9C_7=5\times36=180.
\displaystyle \text{Case II: }5\text{ bowlers and }6\text{ other players.}
\displaystyle \text{Number of ways}={}^5C_5\times{}^9C_6=1\times84=84.
\displaystyle \therefore \text{Required number of teams}=180+84=264.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 47: }\text{A lady gives a dinner party for six guests. The number of ways in which }
\displaystyle \text{they may be selected from among ten friends if two of the friends will not attend the party }
\displaystyle \text{together is (a) }112\qquad \text{(b) }140\qquad \text{(c) }164\qquad \text{(d) none of these}
\displaystyle \text{Answer: (b) }140
\displaystyle \text{Total number of ways of selecting }6\text{ guests from }10\text{ friends is }{}^{10}C_6.
\displaystyle \text{If the two particular friends attend together, the remaining }4\text{ guests}
\displaystyle \text{must be selected from the other }8\text{ friends in }{}^8C_4\text{ ways.}
\displaystyle \therefore \text{Required number of ways}={}^{10}C_6-{}^8C_4.
\displaystyle =210-70=140.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 48: }\text{If }{}^{n+1}C_3=2\cdot{}^nC_2,\text{ then }n=
\displaystyle \text{(a) }3\qquad \text{(b) }4\qquad \text{(c) }5\qquad \text{(d) }6
\displaystyle \text{Answer: (c) }5
\displaystyle {}^{n+1}C_3=2\cdot{}^nC_2.
\displaystyle \frac{(n+1)n(n-1)}{3!} =2\left(\frac{n(n-1)}{2!}\right).
\displaystyle \frac{(n+1)n(n-1)}{6}=n(n-1).
\displaystyle \therefore n+1=6.
\displaystyle \therefore n=5.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 49: }\text{The number of parallelograms that can be formed from a set of four }
\displaystyle \text{parallel lines intersecting another set of three parallel lines is}
\displaystyle \text{(a) }6\qquad \text{(b) }9\qquad \text{(c) }12\qquad \text{(d) }18
\displaystyle \text{Answer: (d) }18
\displaystyle \text{To form a parallelogram, choose }2\text{ lines from the set of }4\text{ parallel lines}
\displaystyle \text{and }2\text{ lines from the set of }3\text{ parallel lines.}
\displaystyle \therefore \text{Number of parallelograms}={}^4C_2\times{}^3C_2.
\displaystyle =6\times3=18.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 1: }\text{In how many ways can 4 letters be posted in 5 letter boxes?}
\displaystyle \text{Answer:}
\displaystyle \text{Each of the 4 letters can be posted in any one of the 5 letter boxes.}
\displaystyle \text{Therefore, each letter has }5\text{ choices.}
\displaystyle \text{Required number of ways}=5\times5\times5\times5=5^4.
\displaystyle =625.
\displaystyle \therefore \text{The required number of ways is }625.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Write the number of 5 digit numbers that can be formed using digits }0,1\text{ and }2.
\displaystyle \text{Answer:}
\displaystyle \text{The first digit cannot be }0.
\displaystyle \text{Hence, the first digit can be chosen in }2\text{ ways, i.e. }1\text{ or }2.
\displaystyle \text{Each of the remaining }4\text{ places can be filled in }3\text{ ways.}
\displaystyle \text{Required number of }5\text{ digit numbers}=2\times3^4.
\displaystyle =2\times81=162.
\displaystyle \therefore \text{The required number of }5\text{ digit numbers is }162.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{In how many ways 4 women draw water from 4 taps, if no tap remains unused?}
\displaystyle \text{Answer:}
\displaystyle \text{There are }4\text{ women and }4\text{ taps, and no tap can remain unused.}
\displaystyle \text{Hence, each tap must be used by exactly one woman.}
\displaystyle \text{The }4\text{ women can be assigned to the }4\text{ taps in }4!\text{ ways.}
\displaystyle \text{Required number of ways}=4!.
\displaystyle =4\times3\times2\times1=24.
\displaystyle \therefore \text{The required number of ways is }24.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Write the total number of possible outcomes in a throw of 3 dice in which}
\displaystyle \text{at least one of the dice shows an even number.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of possible outcomes}=6^3=216.
\displaystyle \text{The odd numbers on a die are }1,3,5.
\displaystyle \text{Number of outcomes in which all three dice show odd numbers}=3^3=27.
\displaystyle \text{Hence, number of outcomes in which at least one die shows an even number}
\displaystyle =216-27=189.
\displaystyle \therefore \text{The required number of possible outcomes is }189.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Write the number of arrangements of the letters of the word BANANA in which}
\displaystyle \text{two N's come together.}
\displaystyle \text{Answer:}
\displaystyle \text{The letters of BANANA are }B,A,N,A,N,A.
\displaystyle \text{Treat the two N's together as one unit }(NN).
\displaystyle \text{We now have the }5\text{ objects }NN,B,A,A,A.
\displaystyle \text{Since the three A's are identical, the number of distinct arrangements is}
\displaystyle \frac{5!}{3!}=\frac{120}{6}=20.
\displaystyle \therefore \text{The required number of arrangements is }20.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Write the number of ways in which 7 men and 7 women can sit on a round}
\displaystyle \text{table such that no two women sit together.}
\displaystyle \text{Answer:}
\displaystyle \text{First, arrange the }7\text{ men around the round table.}
\displaystyle \text{Number of ways of arranging }7\text{ men}=(7-1)!=6!.
\displaystyle \text{The }7\text{ men create }7\text{ gaps between them.}
\displaystyle \text{To ensure that no two women sit together, the }7\text{ women must occupy these }7\text{ gaps.}
\displaystyle \text{The }7\text{ women can be arranged in these }7\text{ gaps in }7!\text{ ways.}
\displaystyle \text{Required number of ways}=6!\times7!.
\displaystyle =720\times5040=3628800.
\displaystyle \therefore \text{The required number of ways is }3628800.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Write the number of words that can be formed out of the letters of the word}
\displaystyle \text{'COMMITTEE'.}
\displaystyle \text{Answer:}
\displaystyle \text{The word COMMITTEE contains }9\text{ letters.}
\displaystyle M\text{ occurs }2\text{ times, }T\text{ occurs }2\text{ times and }E\text{ occurs }2\text{ times.}
\displaystyle \text{The remaining letters }C,O,I\text{ occur once each.}
\displaystyle \text{Therefore, the number of distinct words}=\frac{9!}{2!\,2!\,2!}.
\displaystyle =\frac{362880}{8}=45360.
\displaystyle \therefore \text{The required number of words is }45360.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Write the number of all possible words that can be formed using the letters}
\displaystyle \text{of the word 'MATHEMATICS'.}
\displaystyle \text{Answer:}
\displaystyle \text{The word MATHEMATICS contains }11\text{ letters.}
\displaystyle M,\ A\text{ and }T\text{ occur }2\text{ times each.}
\displaystyle \text{The remaining letters }H,E,I,C,S\text{ occur once each.}
\displaystyle \text{Therefore, the number of distinct words} =\frac{11!}{2!\,2!\,2!}.
\displaystyle =\frac{39916800}{8}=4989600.
\displaystyle \therefore \text{The required number of words is }4989600.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Write the number of ways in which 6 men and 5 women can dine at a round table}
\displaystyle \text{if no two women sit together.}
\displaystyle \text{Answer:}
\displaystyle \text{First, arrange the }6\text{ men around the round table.}
\displaystyle \text{Number of ways of arranging }6\text{ men}=(6-1)!=5!.
\displaystyle \text{The }6\text{ men create }6\text{ gaps between them.}
\displaystyle \text{The }5\text{ women must occupy }5\text{ of these }6\text{ gaps.}
\displaystyle \text{The }5\text{ women can be placed in the }6\text{ gaps in }{}^6P_5\text{ ways.}
\displaystyle \text{Required number of ways}=5!\times{}^6P_5.
\displaystyle =5!\times\frac{6!}{(6-5)!}=120\times720.
\displaystyle =86400.
\displaystyle \therefore \text{The required number of ways is }86400.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Write the number of ways in which 5 boys and 3 girls can be seated in a row}
\displaystyle \text{so that each girl is between 2 boys.}
\displaystyle \text{Answer:}
\displaystyle \text{First, arrange the }5\text{ boys in a row in }5!\text{ ways.}
\displaystyle \text{There are }4\text{ gaps between consecutive boys.}
\displaystyle \text{Each girl must occupy one of these internal gaps so that she is between two boys.}
\displaystyle \text{The }3\text{ girls can be placed in }3\text{ of these }4\text{ gaps in }{}^4P_3\text{ ways.}
\displaystyle \text{Required number of ways}=5!\times{}^4P_3.
\displaystyle =120\times(4\times3\times2).
\displaystyle =120\times24=2880.
\displaystyle \therefore \text{The required number of ways is }2880.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Write the remainder obtained when }1!+2!+3!+\cdots+200!
\displaystyle \text{is divided by }14.
\displaystyle \text{Answer:}
\displaystyle \text{For }n\geq7,\quad n!\text{ is divisible by }14.
\displaystyle \therefore 7!+8!+\cdots+200!\text{ leaves remainder }0\text{ when divided by }14.
\displaystyle \text{Hence, we need to consider only }1!+2!+3!+4!+5!+6!.
\displaystyle =1+2+6+24+120+720.
\displaystyle =873.
\displaystyle 873=14\times62+5.
\displaystyle \therefore \text{The required remainder is }5.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Write the number of numbers that can be formed using all four digits }1,2,3,4.
\displaystyle \text{Answer:}
\displaystyle \text{All four digits }1,2,3,4\text{ are distinct and all are to be used.}
\displaystyle \text{Therefore, the required number of numbers is the number of permutations of }4\text{ digits.}
\displaystyle {}^4P_4=4!.
\displaystyle =4\times3\times2\times1=24.
\displaystyle \therefore \text{The required number of numbers is }24.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Write }\sum_{r=0}^{m}{}^{n+r}C_n\text{ in the simplified form.}
\displaystyle \text{Answer:}
\displaystyle \sum_{r=0}^{m}{}^{n+r}C_n ={}^nC_n+{}^{n+1}C_n+{}^{n+2}C_n+\cdots+{}^{n+m}C_n.
\displaystyle \text{Using }{}^kC_n+{}^kC_{n+1}={}^{k+1}C_{n+1},
\displaystyle {}^nC_n+{}^{n+1}C_n={}^{n+2}C_{n+1}.
\displaystyle \text{Continuing this process, we obtain}
\displaystyle \sum_{r=0}^{m}{}^{n+r}C_n={}^{n+m+1}C_{n+1}.
\displaystyle \therefore \text{The simplified form is }{}^{n+m+1}C_{n+1}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }{}^{35}C_{n+7}={}^{35}C_{4n-2},\text{ then write the values of }n.
\displaystyle \text{Answer:}
\displaystyle \text{We know that if }{}^nC_x={}^nC_y,\text{ then }x=y\text{ or }x+y=n.
\displaystyle \therefore n+7=4n-2\quad\text{or}\quad(n+7)+(4n-2)=35.
\displaystyle \text{From }n+7=4n-2,
\displaystyle 3n=9\Rightarrow n=3.
\displaystyle \text{From }(n+7)+(4n-2)=35,
\displaystyle 5n+5=35.
\displaystyle 5n=30\Rightarrow n=6.
\displaystyle \therefore \text{The values of }n\text{ are }3\text{ and }6.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Write the number of diagonals of an }n\text{-sided polygon.}
\displaystyle \text{Answer:}
\displaystyle \text{A line segment is obtained by choosing any }2\text{ vertices from }n\text{ vertices.}
\displaystyle \text{Number of such line segments}={}^nC_2.
\displaystyle \text{Out of these, }n\text{ line segments are the sides of the polygon.}
\displaystyle \therefore \text{Number of diagonals}={}^nC_2-n.
\displaystyle =\frac{n(n-1)}{2}-n.
\displaystyle =\frac{n(n-3)}{2}.
\displaystyle \therefore \text{The number of diagonals is }\frac{n(n-3)}{2}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Write the expression }{}^nC_{r+1}+{}^nC_{r-1}+2\times{}^nC_r\text{ in the simplest form.}
\displaystyle \text{Answer:}
\displaystyle {}^nC_{r+1}+{}^nC_{r-1}+2{}^nC_r
\displaystyle =\left({}^nC_{r+1}+{}^nC_r\right) +\left({}^nC_r+{}^nC_{r-1}\right).
\displaystyle ={}^{n+1}C_{r+1}+{}^{n+1}C_r.
\displaystyle ={}^{n+2}C_{r+1}.
\displaystyle \therefore \text{The simplest form is }{}^{n+2}C_{r+1}.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Write the value of }\sum_{r=1}^{6}{}^{56-r}C_3+{}^{50}C_4.
\displaystyle \text{Answer:}
\displaystyle \sum_{r=1}^{6}{}^{56-r}C_3+{}^{50}C_4
\displaystyle ={}^{55}C_3+{}^{54}C_3+{}^{53}C_3+{}^{52}C_3+{}^{51}C_3+{}^{50}C_3+{}^{50}C_4.
\displaystyle \text{Using }{}^nC_r+{}^nC_{r-1}={}^{n+1}C_r,
\displaystyle {}^{50}C_4+{}^{50}C_3={}^{51}C_4.
\displaystyle {}^{51}C_4+{}^{51}C_3={}^{52}C_4.
\displaystyle \text{Continuing in the same manner, we get}
\displaystyle \sum_{r=1}^{6}{}^{56-r}C_3+{}^{50}C_4={}^{56}C_4.
\displaystyle =\frac{56\times55\times54\times53}{4\times3\times2\times1}.
\displaystyle =367290.
\displaystyle \therefore \text{The required value is }367290.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{There are 3 letters and 3 directed envelopes. Write the number of ways in which}
\displaystyle \text{no letter is put in the correct envelope.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the letters be }A,B,C\text{ and their correct envelopes be }A,B,C.
\displaystyle \text{The arrangements in which no letter is put in its correct envelope are}
\displaystyle (B,C,A)\quad\text{and}\quad(C,A,B).
\displaystyle \therefore \text{The required number of ways is }2.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Write the maximum number of points of intersection of 8 straight lines in a plane.}
\displaystyle \text{Answer:}
\displaystyle \text{For the maximum number of points of intersection, no two lines should be parallel}
\displaystyle \text{and no three lines should pass through the same point.}
\displaystyle \text{Each pair of straight lines gives one distinct point of intersection.}
\displaystyle \therefore \text{Maximum number of points of intersection}={}^8C_2.
\displaystyle =\frac{8\times7}{2}=28.
\displaystyle \therefore \text{The maximum number of points of intersection is }28.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Write the number of parallelograms that can be formed from a set of four parallel lines}
\displaystyle \text{intersecting another set of three parallel lines.}
\displaystyle \text{Answer:}
\displaystyle \text{To form a parallelogram, choose }2\text{ lines from the set of }4\text{ parallel lines}
\displaystyle \text{and }2\text{ lines from the set of }3\text{ parallel lines.}
\displaystyle \therefore \text{Number of parallelograms}={}^4C_2\times{}^3C_2.
\displaystyle =6\times3=18.
\displaystyle \therefore \text{The required number of parallelograms is }18.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Write the number of ways in which 5 red and 4 white balls can be drawn from a bag containing 10 red and 8}
\displaystyle \text{white balls.}
\displaystyle \text{Answer:}
\displaystyle \text{The number of ways of selecting }5\text{ red balls from }10\text{ red balls is }{}^{10}C_5.
\displaystyle \text{The number of ways of selecting }4\text{ white balls from }8\text{ white balls is }{}^8C_4.
\displaystyle \therefore \text{Required number of ways}={}^{10}C_5\times{}^8C_4.
\displaystyle =252\times70=17640.
\displaystyle \therefore \text{The required number of ways is }17640.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Write the number of ways in which 12 boys may be divided into three groups of 4 boys each.}
\displaystyle \text{Answer:}
\displaystyle \text{The first group of }4\text{ boys can be selected in }{}^{12}C_4\text{ ways.}
\displaystyle \text{The second group of }4\text{ boys can be selected from the remaining }8\text{ boys in }{}^8C_4\text{ ways.}
\displaystyle \text{The remaining }4\text{ boys form the third group in }{}^4C_4\text{ way.}
\displaystyle \text{Since the three groups are identical, each division has been counted }3!\text{ times.}
\displaystyle \therefore \text{Required number of ways}=\frac{{}^{12}C_4\times{}^8C_4\times{}^4C_4}{3!}.
\displaystyle =\frac{495\times70\times1}{6}=5775.
\displaystyle \therefore \text{The required number of ways is }5775.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Write the total number of words formed by 2 vowels and 3 consonants}
\displaystyle \text{taken from 4 vowels and 5 consonants.}
\displaystyle \text{Answer:}
\displaystyle \text{The number of ways of selecting }2\text{ vowels from }4\text{ vowels is }{}^4C_2.
\displaystyle \text{The number of ways of selecting }3\text{ consonants from }5\text{ consonants is }{}^5C_3.
\displaystyle \text{The }5\text{ selected letters can be arranged among themselves in }5!\text{ ways.}
\displaystyle \therefore \text{Required number of words}={}^4C_2\times{}^5C_3\times5!.
\displaystyle =6\times10\times120=7200.
\displaystyle \therefore \text{The total number of words is }7200.
\displaystyle \\

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