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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 251. |
A man has three friends. The number of ways he caninvite one friend everyday for dinner on six successive nights so that no friend is invited morethan three times isa. `640`b. `320`c. `420`d. `510`A. 360B. 420C. 170D. 510 |
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Answer» Correct Answer - D Let x,y,z be the friends and a,b,c denote the case when x is invited a times, y is invited b times and z is invited c times. Now, we have the following possibilities (a,b,c)=(1,2,3) or (2,2,2) or (3,3,0) Hence, the total number of ways `=(6!)/(1!2!3!)xx3!+(6!)/(2!2!2!)xx(3!)/(3!)+(6!)/(3!3!0!)xx(3!)/(2!)` `=360+90+60=510` |
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| 252. |
The total number of ways in which `2n`persons canbe divided into `n`couples isa. `(2n !)/(n ! n !)`b. `(2n !)/((2!)^3)`c. `(2n !)/(n !(2!)^n)`d. none of theseA. `(2n!)/((n!)^(2))`B. `(2n!)/((2n!)^(n))`C. `(2n!)/(n!(2n!)^(2))`D. None of these |
| Answer» Correct Answer - C | |
| 253. |
Four couples (husband and wife) decide to form a committee of four members. The number of different committees that can be formed in which no couple finds a place isA. 10B. 12C. 14D. 16 |
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Answer» Correct Answer - D |
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| 254. |
A man has 10 friends, in how many ways he can invite one or more of them to a party?A. 10!B. `2^(10)`C. `10!-1`D. `2^(10)-1` |
| Answer» Correct Answer - D | |
| 255. |
The number of ways in which 12 balls can be divided between two friends, one receiving 8 and the other 4, isA. `(12!)/(8!4!)`B. `(12!12!)/(8!4!)`C. `(12!)/(8!4!2!)`D. none of these |
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Answer» Correct Answer - B |
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| 256. |
The total number of ways in which `2n`persons canbe divided into `n`couples isa. `(2n !)/(n ! n !)`b. `(2n !)/((2!)^3)`c. `(2n !)/(n !(2!)^n)`d. none of theseA. `(2n!)/(n!n!)`B. `(2n!)/(2!)^(n)`C. `(2n!)/(n!(2!)^(n))`D. none of these |
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Answer» Correct Answer - C |
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| 257. |
The number of different pairs of words (,) that can be made with the letters of the word STATICS, isA. 828B. 1260C. 396D. none of these |
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Answer» We observe that in an arrangement of the 7 letteres of the word STATICS is we put a comma, (,) after four letters, we obtain a pair of words. Hence, required number of pairs = Number of arrangements of the letters of a word STATICS `=(7!)/(2!2!)=1260` |
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| 258. |
A sequence is a termary sequence, if it contains digits 0, 1 and 2. The total numbre of teranryn sequences of length 9 which either begin with 210 or end with 210, isA. 1458B. 1431C. 729D. 707 |
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Answer» Since each digit in a ternary sequence of length 9 can be filled in 3 ways. Therefore, the number of nine-digit ternary sequence beginning with 210 is `3^(6)` and the number fo nine-digit ternary sequence ending with 210 is also `3^(6)`. The number of ternary sequence of 9 digits which being and end with 210 is `3^(3)`. Thus, the total number of sequences which either begin with 210 or end with 210 is `3^(6)+3^(6)-3^(6)=729-27=1431` |
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| 259. |
Between two junction stations A and B there are 12 intermediate stations. The number of ways in which a train can be made to stop at 4 of these stations so that no two of these halting stations are consecutive, isA. `""^(8)C_(4)`B. `""^(9)C_(4)`C. `""^(12)C_(4)-4`D. none of these |
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Answer» Let `x_(1)` be the number of stations before the first halting station, `x_(2)` between first and sencond, `x_(3)` between second and third, `x_(4)` between third and fourth and `x_(5)` on the right of the 4th stations. Then, `x_(1)ge0,x_(5)ge0,x_(2),x_(3),x_(4)ge1` such that `x_(1)+x_(2)+x_(3)+x_(4)+x_(5)=8" "....(i)` The total number of ways is the number of solutions of the above equation. Let `y_(2)=x_(2)-1,y_(3)=x_(3)-1,y_(4)=x_(4)-1`. Then equation (i) reduces to `x_(1)+y_(2)+y_(3)+y_(4)+x_(5)=5`, where `y_(2),y_(3),y_(4)ge0`. The number of solutions of this equation is `""^(5+5-1)C_(5-1)=""^(9)C_(4)`. |
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| 260. |
The number of points having position vector `a hat i + b hat j + c hatk `, where `1 leq a,b,cleq10` and `a,b,c in N`, such that `2^a + 3^b+5^c` is a multiple of 4 is (A) 1000 (B) 500 (C) 250 (D) 125A. 70B. 140C. 210D. 280 |
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Answer» Correct Answer - A `because 2^(a)+3^(b)+5^(c)=2^(a)+(4-1)^(b)+(4+1)^(c)` `=2^(a)+4k+(-1)^(b)+(1)^(c)` `=2^(a)+4k+(-1)^(b)+1` I. a=1,b=even, c=any number II. `a ne 1,b=`odd, c=any number `therefore`Required number of ways`=1xx2xx5+4xx3xx5=70` [`because`even numbers=2,4, odd numbers=1,3,5 and any numbers=1,2,3,4,5] |
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| 261. |
The number of positive integers satisfying the inequality `C(n+1,n-2)-C(n+1,n-1) |
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Answer» Correct Answer - B `.^(n+1)C_(n-2)-.^(n+1)C_(n-1) le 100` `implies .^(n+1)C_(3)-.^(n+1)C_(2)le100` `implies((n+1)n(n-1))/(6)-((n+1)n)/(2)le100` ,brgt `impliesn(n+1)(n-4)le600` It is true for n=2,3,4,5,6,7,8,9. |
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| 262. |
The number of ways in which a score of 11 can be made from a throw by three persons,each throwing a single die once, isA. 45B. 18C. 27D. none of these |
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Answer» Required number of ways = Coeff. of `x^(11)" in "(x+x^(2)+......+x^(6))^(3)` = Coeff. of `x^(8)" in "(1+x+x^(2)+......+x^(5))^(3)` = Coeff. of `x^(8)" in "((1-x^(6))/(1-x))^(3)` Coeff. of `x^(8)" in "(1-x^(6))^(3)(1-x)^(-3)` = Coeff. of `x^(8)" in "(1-""^(3)C_(1)x^(6)+....)(1-x)^(-3)` = Coeff. of `x^(8)" in "(1-x)^(-3)-""^(3)C_(1)."Coff. of "x^(2)" in "(1-x)^(-3)` `=""^(8+3-1)C_(3-1)-""^(3)C_(1).""^(2+3-1)C_(3-1)=""^(10)C_(2)-3xx""^(4)C_(2)=27` |
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| 263. |
Number of positive unequal integral solutions of the equation `x+y+z=12` isA. 21B. 42C. 63D. 84 |
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Answer» Correct Answer - B We have, x+y+z=12 Assume `x lt y lt z`. Here, `x,y,z ge1` `therefore`Solutions of Eq. (i) are `(1,2,9),(1,3,8),(1,4,7),(1,5,6),(2,3,7),(2,4,6) and (3,4,5)`. Number of positive integral solutions of Eq. (i)=7 but x,y,z can be arranged in 3!=6 Hence, required number of solutions`=7xx6=42`. |
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| 264. |
In how many ways can four persons each throwing a dice, once, make a sum of 13 ?A. 220B. 180C. 140D. 80 |
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Answer» Let `x_(1),x_(2),x_(3)andx_(4)` be the numbers on the upper faces of the four dice. Then the required number of ways is equal to the dice. Then the required numbre of ways is equal to the number of solutions of `x_(1)+x_(2)+x_(3)+x_(4)=13, "where"1lex_(1),x_(2),x_(3),x_(4)le6` The number of solutions of this equation is equal to the Coefficient of `x^(13)" in "(x^(1)+x^(2)+....+x^(6))^(6)` = Coefficient of `x^(13)" in "x^(4)(x+1+x+x^(2)+....+x^(5))^(4)` = Coefficient of `x^(9)" in "((1-x^(6))/(1-x))^(4)` = Coefficient of `x^(9)" in "(1-x^(6))^(4)(1-x)^(-4)` = Coefficient of `x^(9)" in "(""^(4)C_(0)-""^(4)C_(1)x^(6)+""^(4)C_(2)x^(12)+....)(1-x)^(-4)` = Coefficient of `x^(9)" in """^(4)C_(0)(1-x)^(-4)` - Coefficient of `x^(9)" in """^(4)C_(1)x^(6)(1-x)^(-4)` `""^(4)C_(0)xx("Coefficicent of"x^(9)" in "(1-x)^(-4)` `-""^(4)C_(0)xx("Coefficicent of "x^(3)" in "(1-x)^(-4)` `=""^(4)C_(0)xx""^(9+4-1)C_(4-1)-""^(4)C_(1)xx""^(3+4-1)C_(4-1)` `=""^(12)C_(3)-4xx""^(6)C_(3)=220-80=140`. |
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| 265. |
The number of ways in which a score of 11 can be made from a throw by three persons,each throwing a single die once, isA. 45B. 18C. 27D. 68 |
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Answer» Correct Answer - C Coefficient of `x^(11)` in `(x+x^(2)+x^(3)+x^(4)+x^(5)+x^(6))^(3)` =Coefficient of `x^(8)` in `(1+x+x^(2)+x^(3)+x^(4)+x^(5))^(3)` =Coefficient of `x^(8)` in `(1-x^(6))^(3)(1-x)^(-3)` =Coefficient of `x^(8)` in `(1-3x^(6))(1+.^(3)C_(1)x+ . . )` `=.^(10)C_(2)-3xx.^(4)C_(2)=45-18=27` |
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| 266. |
In how many ways can three persons, each throwing a single dice once, make a sum of 15? |
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Answer» Number on the faces of the dice are 1,2,3,4,5,6 (least number 1, greatest number 6) here, l=1, m=6, r=3 and n=15 `therefore`Required number of ways=Coefficient of `x^(15-1xx3)` in the expansion of `(1-x^(6))^(3)(1-x)^(-3)` =Coefficient of `x^(12)` in the expansion of `(1-3x^(6)+3x^(12))(1+.^(3)C_(1)x+.^(4)C_(2)x^(2)+ . . .+.^(8)C_(6)x^(6)+ . ..+.^(14)C_(12)x^(12)+ . . .)` `=.^(14)C_(12)-3xx.^(8)C_(6)+3=.^(14)C_(2)xx.^(8)C_(2)+3` `=91-84+3=10` |
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| 267. |
In how many ways can four persons, each throwing a die once, make a sum of 6?A. `""^(9)C_(2)`B. `""^(10)C_(3)`C. `""^(8)C_(3)`D. `""^(9)C_(3)` |
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Answer» Let, `x_(1),x_(2),x_(3),x_(4)` be the number obtained by the four person on the upper faces of their respective dice. Then, the required number of ways will be equal to the number of solutions of the equation `x_(1)+x_(2)+x_(3)+x_(4)=6,"where"1lex_(1),x_(2),x_(3),x_(4)le6`. Since the upper limit is six which is equal to the sum required. So, the upper limit for each variable can be taken as infinite. Hence, required number of ways = Coefficinet of `x^(6)" in "(1+x+x^(2)+....)^(4)` = Coefficinet of `x^(6)" in "(1-x)^(4)` `=""^(6+4-1)C_(4-1)=""^(9)C_(3)=84`. |
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| 268. |
12 boys and 2 girls are to be seated in a row such that there are atleast 3 boys between the 2 girls. The number of ways this can be done is `lamdaxx12!`. The value of `lamda` isA. 55B. 110C. 20D. 45 |
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Answer» Correct Answer - B `P_(1)`=Number of ways, the girls can sit together `=(14-2+1)xx2!xx12! =26xx12!` `P_(2)=`Number of ways, one boy sits between the girls `=(14-3+1)xx2!xx12! =24xx12!` `P_(3)`=Number of ways, two boys sit between the girls `=(14-4+1)xx2!xx12! =22xx12!` `therefore`Required number of ways`=(182-26-24-22)xx12!` `thereforelamda=110" [given]"` |
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| 269. |
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:(i) exactly 3 girls ?(ii) atleast 3 girls ?(iii) atmost 3 girls ? |
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Answer» (i) `g=3,b=4` total ways = `.^4C_3.^9C_4` `=4*(9*8*7*6)/(4*3*2) = 63*8` `=504` (ii) `g= 3 or 4` total ways =`.^4C_3 .^9C_4 + .^4C_4.^9C_3` `4*(9*8*7*6*5)/(5*4*3*2) + (9*8*7)/(3*2)` `= 504+84` `=588` (iii)`g=1,2,3` total ways `= .^13C_7 - .^4C_4.^9C_3` `= (13*12*11*10*9*8)/(6*3*2*5*4) - 84` `=143*12- 84` `=1716- 84` answer |
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| 270. |
Total number of integral solutions of the system of equations `x_(1)+x_(2)+x_(3)+x_(4)+x_(5)=20and x_(1)+x_(2)+x_(3)=5" when "x_(k)gel0,"is"`A. 335B. 336C. 338D. 340 |
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Answer» We have, `x_(1)+x_(2)+x_(3)+x_(4)+x_(5)=20and x_(1)+x_(2)+x_(3)=5` These two equtions reduce to `x_(4)+x_(5)=15" "(i)` and `x_(1)+x_(2)+x_(3)=5" "(ii)` Since corresponding to each solution of (i) there are solutions of equation (ii). So, total number of solution of the given system of equations = No. of solutions of (i) `xx` No. of solutions of (ii) `=""^(15+2-1)C_(2-1)xx""^(5+3-1)C_(3)=""^(16)C_(1)xx""^(7)C_(2)=336`. |
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| 271. |
Find the sum of all the numbers that can be formed with the digits 2,3, 4, 5 taken all at a time.A. 93324B. 66666C. 84844D. none of these |
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Answer» Correct Answer - A |
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| 272. |
The sum of the digits in the unit place of all numbers formed with the help of 3,4,5,6 taken all at a time isA. 432B. 108C. 36D. 18 |
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Answer» Correct Answer - B If we fixed 3 at units place. Total possible number is `3"!"` i.e., 6. Sum of the digits in unit place of all these numbers `=3"!" xx 3` Similarly, if we fixed 4, 5 and 6 at units place, in each case total possible numbers are `3"!"`. Required sum of unit digits of all such numbers `= (3 + 4+ 5 + 6) xx 3"!"` `= 18 xx 3"!" = 18 xx 6 = 108` |
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| 273. |
If the letters of the word are arranged as in dictionary, find the rank of the following words.(0) RAJU(ii) AIRTEL(ii) UMANG |
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Answer» (i) In a dictionary, the letters in alphabetical order are A,J,R,U `therefore`The first word is ARJU. Number of words beginning with A=number of ways arranging J,R,U=3!=6 The next word begin with R and it is RAJU. `therefore`Number of words before RAJU=12 `therefore`Rank of word RAJU=13. (ii) The Letters in alphabetical order are A,G,M,N,U The first word is AGMNU Number of words beginning with A=4!=24 Number of words beginning with G=4!=24 Number of words beginning with M=4!=24 Number of words beginning with N=4!=24 Number of words beginning with UA=3!=6 Number of words beginning with UG=3!=6 Number of words beginning with UMAG=1!=1 Number of words beginning with UMANG=1 `therefore` Rank of the word UMANG=24+24+24+24+6+6+1+1=110 |
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| 274. |
Find the number of integers which lie between 1 and `10^6`and which have the sum of the digits equal to 12.A. 8550B. 5382C. 6062D. 8055 |
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Answer» Correct Answer - C |
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| 275. |
If letters of the word are arranged as in dictionary, find the rank of the following words.(i) INDIA (ii) SURITI (ii) DOCOMO |
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Answer» (i) The letter in alphabetical order are A,D,I,I,N `therefore`The first word is ADIIN Number of words beginning with `A=(4!)/(2!)=12` Number of words beginning with `D=(4!)/(2!)=12` Number of words beginning with IA=3!=6 Number of words beginning with ID=3!=6 Number of words beginning with II=3!=6 Number of words beginning with INA=2!=2 Number of words beginning with INDA=1!=1 Number of words beginning with INDIA=1 `therefore`Rank of the word INDIA =12+12+6+6+6+2+1+1=46 The letters inn alphabetical order are I,I,R,S,T,U `therefore`The first word is IIRSTU Number of words beginning with I=5!=120 gt Number of words beginning with R=`(5!)/(2!)=60` Number of words beginning with SI=4!=24 Number of words beginning with SR=`(4!)/(2!)=12` Number of words beginning with ST`=(4!)/(2!)=12` Number of words beginning with SUI=3!=6 Number of words beginning with SURI=1!=1 Number of words beginning with SURITI=1 `therefore`Rank of the word SURITI =120+60+24+12+12+6+1+1=236 |
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| 276. |
There are 10 candidates for an examination out of which 4 are appearing in Mathematics and remaining 6 are appearing indifferent subjects. In how many ways can they be seated in a row so that no two Mathematics candidates are together? |
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Answer» In this method first arrange the remaining candidates Here, remaining candidates=6 `"X0X0X0X0X0X0X"` X:Places available for Mathematics condidates 0:Places for others remining candidates can be arranged in 6! Ways. There are seven places available for mathematics candidates so that no two mathematics candidates are together. now, four candidates can be placed in these sevenn places in `.^(7)P_(4)` ways. Hence, the total number of ways`=6!xx.^(7)P_(4)=720xx840` `=604800` |
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| 277. |
The number of different four - digit numbers that can be formed with the digits 2, 3, 4, 7 and using each digit only once isA. 120B. 96C. 24D. 100 |
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Answer» Correct Answer - C Given, digits 2, 3, 4 and 7, we have to form four-digit numbers using these digits. `therefore` Required number of ways `= ""^(4)P_(4) = 4"!" = 4 xx 3 xx 2"!" = 24` |
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| 278. |
How many numbers between 5000 and 10,000 can be formed using the digits 1,2,3,4,5,6,7,8,9, each digit appearing not more than once in each number?A. `5xx""^(8)P_(3)`B. `5xx""^(8)C_(8)`C. `5!xx""^(8)P_(3)`D. `5!xx""^(8)C_(3)` |
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Answer» Correct Answer - A |
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| 279. |
Total number of 6-digit numbers in which all the odd digits appear, is |
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Answer» Among the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, clearly 1, 3, 5, 7 and 9 are odd. `therefore` Number of six-digit numbers `= 5 xx 5 xx 5 xx 5 xx 5 xx 5 = 5^(6)` |
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| 280. |
In a football championship, 153 matches were played. Every two teams played one match with each other. The number of teams, participating in the championship is …….. .A. 17B. 18C. 9D. none of these |
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Answer» Correct Answer - B |
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| 281. |
A shopkeeper sells three varieties of perfumes and he has a large number of bottles of the same size of each variety in his stock. There are 5 places in a row in his showcase. The number of different ways of displaying the three varieties of perfumes in the show case, isA. 6B. 50C. 150D. none of these |
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Answer» Correct Answer - C |
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| 282. |
The number of ways in which we can choose a committee from four men and six women, so that the committee includes atleast two men and exactly twice as many women as men isA. 94B. 126C. 128D. None of these |
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Answer» Correct Answer - A `because` Number of men = 4 and number of women = 6 It is given that committee includes two men and exactly twice as many women as men. Thus, possible selection is given in following table `{:("Men" , "Women"),(2, " "4),(3, " "6):}` Required number of committee formed `= ""^(4)C_(2) xx ""^(6)C_(4) + ""^(4)C_(3) xx ""^(6)C_(6)` `= 6 xx 15 + 4 xx 1 = 94` |
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| 283. |
If a train, five seats are vacant, the number of ways three passengers can sit, isA. 10B. 20C. 30D. 60 |
| Answer» Correct Answer - D | |
| 284. |
In a polygon, no three diagonals areconcurrent. If the total number ofpoints of intersection of diagonals interior to the polygon is 70, then thenumber of diagonals of the polygon isa. `20`b. `28`c. `8`d. none of theseA. 8B. 20C. 28D. none of these |
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Answer» Let there be n-sides of the polygon. Then, number of vertivces is also n. We know that in a convex polygon every group of four vertices determines two diagonals and hence one point of intersection So. Total numbers of points of interesction of diagonals is `""^(n)C_(4)`. `:.""^(n)C_(4)=70` `impliesn(n-1)(n-2)(n-3)=70xx24` `impliesn(n-1)(n-2)(n-3)=5xx6xx7xx8` `impliesn=8` Hence, number of diaonals `=""^(n)C_(2)-n=""^(8)C_(2)-8=20`. |
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| 285. |
The interior angles of a regular polygon measure `150^@` each. The number of diagonals of the polygon is |
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Answer» Each of exterior angle`=30^(@)` `therefore`Number of sides`=(360^(@))/(30^(@))=(360xx(pi)/(180))/(30xx(pi)/(180))=12` `therefore`Number of diagonals`=.^(12)C_(2)-12=66-12=54` |
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| 286. |
We wish to select 6 persons from 8, but if the person A is chosen, thenB must be chosen. In how many ways can the selections be made?A. 20B. 21C. 22D. 30 |
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Answer» Correct Answer - C Total number of person = 8 Number of person to be selected = 6 It is given that, if A is chosen then, B must be chosen. Therefore, following cases arise. Case I When A is chosen, B must be chosen. Number of ways `= ""^(8 -2)C_(6-2) = ""^(6)C_(4)` Case II When A is not chosen. `therefore" "` Number of ways `= ""^(8 -1)C_(6) = ""^(7)C_(6)` Hence, required number of ways `= ""^(6)C_(4) + ""^(7)C_(6)` `= 15 + 7 = 22` |
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| 287. |
If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary; how many words are there in this list before the first word starting with E? |
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Answer» `A= (10!)/(2!2!)` `= 90*56*30*6` `= 100*27*56*6` `= 162*56*100` `= 907200` Answer |
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| 288. |
If the letters of the word LATE be permutedand the words so performed be arranged as in a dictionary, find rank of theword LATE.A. 12B. 13C. 14D. 15 |
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Answer» Correct Answer - C |
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| 289. |
The letters of the word COCHIN are permuted andall the permutations are arranged in an alphabetical order s in an Englishdictionary. The number of words that appear before the word COCHIN isa.`360`b. `192`c. `96`d.`48`A. 360B. 192C. 96D. 48 |
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Answer» The number of words starting with CC=4! The number of words starting with CH=4! The number of words starting with CI=4! The number of words starting with CN=4! COCHIN is the first word in the list of words beginning with CO. `:.` No. of words that appear befor the word COCHIN =96 |
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| 290. |
There are two urns. Urn A has 3 distinct red balls and urn B has 9distinct blue balls. From each urn two balls are taken out at random and thentransferred to the other. The number of ways in which this can be done is(1) 36 (2) 66 (3) 108 (4) 3A. 66B. 108C. 3D. 36 |
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Answer» Clearly, Required number of ways `=""^(3)C_(2)xx""^(9)C_(2)=108` |
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| 291. |
Eight chairs are numbered 1 to 8. Two women andthree men wish to occupy one chair each. First, the women choose the chairsfrom amongst the chairs marked 1 to 4, and then the men select th chairs fromamongst the remaining. The number of possible arrangements isa.`^6C_3xx^4C_2`b. `^4P_2xx^4P_3`c. `^4C_2xx^4P_3`d. none of theseA. `""^(4)C_(3)xx""^(4)C_(2)`B. `""^(4)C_(2)xx""^(4)P_(3)`C. `""^(4)P_(2)xx""^(4)P_(3)`D. none of these |
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Answer» Correct Answer - D |
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| 292. |
Find the total number of selections of at least one red ball from a bag containing 4 red balls and 5 black balls, balls of the same colour being identical. |
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Answer» Total number of selection `= [(4 + 1) (5 + 1) - 1]-5` `= (5 xx 6 - 1)-5` `=(30 - 1)-5=24` |
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| 293. |
A code word consists of three letters of the English alphabet followed by two digits of the decimal system. If neither letter nor digit is repeated in any code word, then the total number of code words, isA. 1404000B. 16848000C. 2808000D. none of these |
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Answer» Correct Answer - A |
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| 294. |
How many words (with or without dictionary meaning) can be made from the letters of the word MONDAY, assuming that no letter is repeated, if |
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Answer» (i) 4 letters are used at a time `= ""^(6)P_(4) = (6"!")/(2"!") = 6 xx 5 xx 4 xx 3 = 360` (ii) All letters used at a time `= 6"!" = 6 xx 5 xx 4 xx 3 xx 2 xx 1 = 720` (iii) All letters used but first is vowel `= 2xx 5"!" = 2 xx 5 xx 4 xx 3 xx 2 xx 1 = 240` |
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| 295. |
If the 13 letter words (need no be meaningful) are to be formed using the letters from the word "MEDITERRANEAN" such that the first letter is R and the fourth letter is, E, then the total number of such words, isA. `(11!)/((2!)^(3))`B. 59C. 110D. 56 |
| Answer» There are 1 M, 3E, 1D, 1I, 1T, 2R, 2A, and 2N in the word "MEDITERRNEAN". After placing R at the first place and E at the fourth place, 11 letters are left which can be arranged in `(11!)/((2!)^(3))` ways. Hence, total number of words in `(11!)/((2!)^(3))`. | |
| 296. |
Total number of divisors of `n = 3^5. 5^7. 7^9` that are in the form of `4lambda + 1; lamda >=0` is equal toA. 15B. 30C. 120D. 240 |
| Answer» Correct Answer - D | |
| 297. |
The number of all four digit numbers which are divisible by 4 that can be formed from the digits 1, 2, 3, 4, and 5, isA. 125B. 30C. 95D. none of these |
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Answer» Correct Answer - A |
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| 298. |
The number of ways of arranging `m`positiveand `n(A. `""^(m+1)+P_(n)`B. `""^(n+1)+P_(m)`C. `""^(m+1)+C_(n)`D. `""^(n+1)+C_(m)` |
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Answer» Correct Answer - C |
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| 299. |
If all the words (with or without meaning) having five letters, formedusing the letters of the word SMALL and arranged as in a dictionary; then theposition of the word SMALL is :(1) 46th(2) 59th(3) 52nd(4) 58thA. 46thB. 59thC. 52ndD. 58th |
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Answer» Correct Answer - D Clearly, number of words start with `A = (4!)/(2!) = 12` Number of words start with `L = 4! = 24` Number of words start with `M = (4!)/(2!) = 12` Number of words start with `SA = (3!)/(2!) = 3` Number of words start with `SL = 3! = 6` Note that, next word will be "SMALL". Hence, the position of word "SMALL" is 58th. |
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| 300. |
A five digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5, without repetition. The total number of ways this can be done, is |
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Answer» A number will be divisible by 3, if sum of the digits in number obe divisible by 3. Here, 0+1+2+3+4+5=15, which is divisible by 3. therefore, the digit that can be left out, while the sum still is multiple of 3, is either 0 or 3. If 0 left out Then, possible numbers=`.^(5)P_(5)=5!=120` If 3 left out Then, possible numbers=`.^(5)P_(5)-.^(4)P_(4)=5!-4!=120-24=96` Hence, required total numbers=120+96=216 |
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