\displaystyle \textbf{Question 1. }\text{The number of permutations of }n\text{ different things taking }r\text{ at a time}
\displaystyle \text{when }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:}
\displaystyle \text{Since }3\text{ particular things are always included, choose remaining }r-3\text{ things}
\displaystyle \text{from the remaining }n-3\text{ things.}
\displaystyle \text{Number of selections}={}^{n-3}C_{r-3}
\displaystyle \text{Now arrange the }r\text{ selected things in }r!\text{ ways.}
\displaystyle \therefore \text{Required number}=r!\,{}^{n-3}C_{r-3}
\displaystyle \therefore \text{Correct option is (d).}
\\

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

\displaystyle \textbf{Question 3. }\text{The number of words that can be formed out of the letters of the word ARTICLE}
\displaystyle \text{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:}
\displaystyle \text{The word ARTICLE has }7\text{ letters. Vowels are }A,I,E.
\displaystyle \text{Even places are }2,4,6.
\displaystyle \text{The }3\text{ vowels can be arranged in even places in }3!\text{ ways.}
\displaystyle \text{The remaining }4\text{ consonants can be arranged in }4!\text{ ways.}
\displaystyle \therefore \text{Required number}=3!\times4!=6\times24=144
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 4. }\text{How many numbers greater than }10\text{ 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:}
\displaystyle \text{Numbers greater than }10\text{ lacs must be }7\text{-digit numbers.}
\displaystyle \text{The digits are }0,2,2,3,3,3,4.
\displaystyle \text{Total arrangements}=\frac{7!}{2!\,3!}=420
\displaystyle \text{Arrangements beginning with }0=\frac{6!}{2!\,3!}=60
\displaystyle \therefore \text{Required number}=420-60=360
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 5. }\text{The number of different signals which can be given from }6\text{ 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:}
\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+\frac{6!}{4!}+\frac{6!}{3!}+\frac{6!}{2!}+\frac{6!}{1!}+\frac{6!}{0!}
\displaystyle =6+30+120+360+720+720
\displaystyle =1956
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 6. }\text{The 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) }360\qquad \text{(b) }240\qquad \text{(c) }120\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Total arrangements of BHARAT}=\frac{6!}{2!}=360
\displaystyle \text{When B and H are together, treat BH as one unit.}
\displaystyle \text{Arrangements}=\frac{5!}{2!}\times2!=120
\displaystyle \therefore \text{Required number}=360-120=240
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 7. }\text{The number of six letter words that can be formed using the letters of the}
\displaystyle \text{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:}
\displaystyle \text{The letter S occurs }3\text{ times. Other letters are A, I and T.}
\displaystyle \text{The S's can occupy alternate places in }2\text{ ways.}
\displaystyle \text{The remaining letters A, I and T can be arranged in }3!\text{ ways.}
\displaystyle \therefore \text{Required number}=2\times3!=12
\displaystyle \therefore \text{Correct option is (a).}
\\

\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:}
\displaystyle \text{Total arrangements of DELHI}=5!=120
\displaystyle \text{In half of these arrangements, E precedes I.}
\displaystyle \therefore \text{Required number}=\frac{120}{2}=60
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 9. }\text{The number of ways in which the letters of the word CONSTANT can be}
\displaystyle \text{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:}
\displaystyle \text{Vowels are O and A. They can be arranged in }2!\text{ ways.}
\displaystyle \text{Consonants are C, N, S, T, N, T.}
\displaystyle \text{They can be arranged in }\frac{6!}{2!\,2!}\text{ ways.}
\displaystyle \therefore \text{Required number}=2!\times\frac{6!}{2!\,2!}=360
\displaystyle \therefore \text{Correct option is (a).}
\\

\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:}
\displaystyle \text{The word CHEESE has }6\text{ letters, in which E occurs }3\text{ times.}
\displaystyle \therefore \text{Number of arrangements}=\frac{6!}{3!}=120
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 11. }\text{Number of all four digit numbers having different digits formed of the}
\displaystyle \text{digits 1,2,3,4,5 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:}
\displaystyle \text{For divisibility by }4,\text{ the last two digits must be divisible by }4.
\displaystyle \text{Possible endings are }12,24,32,52.
\displaystyle \text{For each ending, the first two places can be filled in }{}^3P_2=6\text{ ways.}
\displaystyle \therefore \text{Required number}=4\times6=24
\displaystyle \therefore \text{Correct option is (a).}
\\

\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 as 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:}
\displaystyle \text{Letters in alphabetical order are A, I, K, N, R, S.}
\displaystyle \text{Words before KRISNA}=2(5!)+3(4!)+1(3!)+2(2!)+1(1!)
\displaystyle =240+72+6+4+1=323
\displaystyle \therefore \text{Rank of KRISNA}=323+1=324
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 13. }\text{If in a group of }n\text{ distinct objects, the number of arrangements of }4\text{ objects}
\displaystyle \text{is }12\text{ times the number of arrangements of }2\text{ 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:}
\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=0
\displaystyle (n-6)(n+1)=0
\displaystyle \therefore n=6
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 14. }\text{The number of ways in which }6\text{ men can be arranged in a row so that}
\displaystyle \text{three 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:}
\displaystyle \text{Treat the }3\text{ particular men as one block.}
\displaystyle \text{Then there are }4\text{ objects to arrange.}
\displaystyle \text{These can be arranged in }4!\text{ ways.}
\displaystyle \text{The }3\text{ men inside the block can be arranged in }3!\text{ ways.}
\displaystyle \therefore \text{Required number}=4!\times3!
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 15. }\text{A }5\text{-digit number divisible by }3\text{ is to be formed using the digits }0,1,2,3,4,5
\displaystyle \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:}
\displaystyle \text{Sum of all digits }0+1+2+3+4+5=15
\displaystyle \text{For divisibility by }3,\text{ the omitted digit must be }0\text{ or }3.
\displaystyle \text{If }0\text{ is omitted, number of arrangements}=5!
\displaystyle \text{If }3\text{ is omitted, arrangements using }0,1,2,4,5=5!-4!
\displaystyle \therefore \text{Required number}=5!+(5!-4!)=120+96=216
\displaystyle \therefore \text{Correct option is (a).}
\\

\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:}
\displaystyle \text{The product of }r\text{ consecutive positive integers is always divisible by }r!.
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 17. }\text{If }{}^{k+5}P_{k+1}=\frac{11(k-1)}{2}\cdot{}^{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:}
\displaystyle {}^{k+5}P_{k+1}=\frac{11(k-1)}{2}\cdot{}^{k+3}P_k
\displaystyle \frac{(k+5)!}{4!}=\frac{11(k-1)}{2}\cdot\frac{(k+3)!}{3!}
\displaystyle (k+5)(k+4)=22(k-1)
\displaystyle k^2-13k+42=0
\displaystyle (k-6)(k-7)=0
\displaystyle \therefore k=6,7
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 18. }\text{The number of arrangements of the letters of the word BHARAT taking} \\ 3\text{ at a time is}
\displaystyle \text{(a) }72\qquad \text{(b) }120\qquad \text{(c) }14\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{The letters are B, H, A, R, A, T.}
\displaystyle \text{Words with all distinct letters}={}^{5}C_3\times3!=60
\displaystyle \text{Words with two A's and one other letter}={}^{4}C_1\times\frac{3!}{2!}=12
\displaystyle \therefore \text{Required number}=60+12=72
\displaystyle \therefore \text{Correct option is (a).}
\\

\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:}
\displaystyle \text{Vowels are A, U, A and consonants are P, R, B.}
\displaystyle \text{There are }2\text{ possible alternate patterns.}
\displaystyle \text{Vowels can be arranged in }\frac{3!}{2!}\text{ ways and consonants in }3!\text{ ways.}
\displaystyle \therefore \text{Required number}=2\times\frac{3!}{2!}\times3!=36
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 20. }\text{The number of different ways in which }8\text{ persons can stand in a row}
\displaystyle \text{so that between two particular persons A and B 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:}
\displaystyle \text{A and B can occupy positions with exactly two persons between them in }5\text{ ways.}
\displaystyle \text{A and B can be interchanged in }2!\text{ ways.}
\displaystyle \text{Remaining }6\text{ persons can be arranged in }6!\text{ ways.}
\displaystyle \therefore \text{Required number}=5\times2!\times6!=7200=60\times5!
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 21. }\text{The number of ways in which the letters of the word ARTICLE can}
\displaystyle \text{be 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:}
\displaystyle \text{The even places are }2,4,6.
\displaystyle \text{These }3\text{ places must be occupied by consonants.}
\displaystyle \text{Choose and arrange }3\text{ consonants from }4\text{ consonants in }{}^4P_3\text{ ways.}
\displaystyle \text{The remaining }4\text{ letters can be arranged in }4!\text{ ways.}
\displaystyle \therefore \text{Required number}={}^4P_3\times4!=24\times24=576
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 22. }\text{In a room there are }12\text{ 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:}
\displaystyle \text{Each bulb can be either switched on or switched off.}
\displaystyle \text{Total possible arrangements}=2^{12}
\displaystyle \text{Excluding the case when all bulbs are off, required number}=2^{12}-1
\displaystyle \therefore \text{Correct option is (c).}
\\

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