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InterviewSolution
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.
| 51. |
Find the number of ways to give 16 different things to three persons A,B, C so that B gets 1 more than A and C gets 2 more than B.A. `(16!)/(4!5!7!)`B. 4!5!7!C. `(16!)/(3!5!8!)`D. none of these |
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Answer» Suppose A gets x things, then B gets x+1 and C gets x+3 things. `:.x+x+1+x+3=16implies4`. Thus, we have to distribute 16 things to A, B and C in such a way that A gets 4 things, B gets 5 things and C gets 7 things. Required number of ways `=""^(16)C_(4)xx""^(12)C_(5)xx""^(7)C_(7)=(16!)/(4!5!7!)` |
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| 52. |
A committee of 5 is to be formed from 9 ladies and 8 men. If the committee commands a lady majority, then the number of ways this can be done isA. 2352B. 1008C. 3360D. 3486 |
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Answer» Correct Answer - D |
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| 53. |
Let A be a set of n `(>=3)` distinct elements. The number of triplets `(x, y, z)` of the A elements in which at least two coordinates is equal toA. `""^(n)C_(3)xx3!`B. `n^(3)-""^(n)C_(3)xx3!`C. `3n^(2)-2n`D. `3n^(2)(n-1)` |
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Answer» We have, Required number of triplets = Total number of triplets - Number of triplets in which no two coordinates are equal `=n^(3)-""^(n)C_(3)xx3!`. |
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| 54. |
The straight lines `I_(1),I_(2),I_(3)` are paralled and lie in the same plane. A total number of m point are taken on `I_(1),n` points on `I_(2)`, k points on `I_(3)`. The maximum number of triangles formed with vertices at these points areA. `.^(m+n+k)C_(3)`B. `.^(m+n+k)C_(3)_.^(m)C_(3)-.^(n)C_(3)-.^(k)C_(3)`C. `.^(m)C_(3)+.^(n)C_(3)+.^(k)C_(3)`D. None of these |
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Answer» Correct Answer - B Total points on all three lines`=m+n+k` `therefore`Maximum number of triangles`=.^(m+n+k)C_(3)-.^(m)C_(3)-.^(n)C_(3)-.^(k)C_(3)` |
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| 55. |
Statement-1: The total number of ways in which three distinct numbers in AP, can be selected from the set {1,2,3, . .,21}, is equal to 100. Statement-2: If a,b,c are inn AP, then a+c=2b.A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1B. Statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement-1C. Statement-1 is true, statement-2 is falseD. Statement-1 is false, statement-2 is true |
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Answer» Correct Answer - A `because a+c=2b` i.e., sum of two numbers is even, the both numbers are even or odd. In {1,2,3,4, . . ,21},11 numbers are odd and 10 numbers are even. then, total number of ways=`.^(11)C_(2)+.^(10)C_(2)=55+45=100` hence, both statements are true but statement-2 is not a correct explanation for statement-1. |
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| 56. |
The number of ways in which we can select fournumbers from 1 to 30 so as to exclude every selection of four consecutivenumbers isa. `27378`b. `27405`c. `27399`d.none of theseA. 27378B. 27405C. 27399D. none of these |
| Answer» The numbre of ways of selecting any four numbres from 1 to 30, is `""^(30)C_(4)`. Four consecutive numbers can be chosen in the following ways : (1, 2, 3, 4),(2, 3, 4, 5),…….,(27, 28, 29, 30). Thus, required number of ways `=""^(30)C_(4)-27=27378`. | |
| 57. |
Statement-1: The smallest positive integer n such that n! can be expressed as a product of n-3 consecutive integers, is 6. Statement-2: Product of three consecutive integers is divisible by 6.A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1B. Statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement-1C. Statement-1 is true, statement-2 is falseD. Statement-1 is false, statement-2 is true |
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Answer» Correct Answer - B Statement-1 is true `therefore6!=720=8xx9xx10` i.e., product off 6-3=3 consecutive integers and statement-2 is also true, but statement-2 is not a correct explanation for statement-1. |
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| 58. |
There are 6 books on mathematics, 4 books on physics annd 5 books on chemistry in a book shop. The number of ways can a student purchase either a book on mathematics or a book on chemistry, isA. 10B. 11C. 8D. 15 |
| Answer» Correct Answer - B | |
| 59. |
lf n! ,`3xxn!` and (n+1)! are in G P, then n!, 5 x n! and (n+1)! are inA. APB. GPC. HPD. AGP |
| Answer» Correct Answer - A | |
| 60. |
Let A be a set of n `(>=3)` distinct elements. The number of triplets `(x, y, z)` of the A elements in which at least two coordinates is equal toA. `.^(n)P_(3)`B. `n^(3)-.^(n)P_(3)`C. `3n^(2)-2n`D. `3n^(2)(n-1)` |
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Answer» Correct Answer - B Required number of triplets=Total number of triplets without restrictions-number of triplets with all different coordinates `=n^()-.^(n)P_(3)` |
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| 61. |
The number of ways in which one can post 4 letters in 5 letter boxes, isA. `4^(5)`B. `5^(4)`C. 20D. 9 |
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Answer» Let `L_(1),L_(2),L_(3)andL_(4)` be four letters and `B_(1),B_(2),B_(3),B_(4)andB_(5)` be five letter boxes. The job of posting 4 letters in five letter boxes can be divided into following four sub-jobs : `J_(1)` : Posting letter `L_(1)` in a letter box. `J_(2)` : Posting letter `L_(2)` in a letter box. `J_(3)` : Posting letter `L_(3)` in a letter box. `J_(4)`: Posting letter `L_(4)` in a letter box. Since letter `L_(1)` can be posted in any one of the five letter boxes. Therefore, job `J_(1)` can be done in five ways. Similarly, each job can be done in 5 ways. So, required number of ways. `=5xx5xx5xx5=5^(4)`. |
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| 62. |
Statement-1: If `f:{a_(1),a_(2),a_(3),a_(4),a_(5)}to{a_(1),a_(2),a_(3),a_(4),a_(5)}`, f is onto and `f(x)nex` for each `xin {a_(1),a_(2),a_(3),a_(4),a_(5)}`, is equal to 44. Statement-2: The number of derangement for n objects is `n! sum_(r=0)^(n)((-1)^(r))/(r!)`.A. Statement-1 is true, statement-2 is true, statement-2 is a correct explanation for statement-1B. Statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement-1C. Statement-1 is true, statement-2 is falseD. Statement-1 is false, statement-2 is true |
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Answer» Correct Answer - a `becauseD_(n)=n!underset(r=0)overset(n)(sum)((-1)^(r))/(r!)=n!(1-(1)/(1!)+(1)/(2!)-(1)/(3!)+ . . .+((-1)^(n))/(n!))` `thereforeD_(5)=5!(1-(1)/(1!)+(1)/(2!)-(1)/(3!)+(1)/(4!)-(1)/(5!))` `=120((1)/(2)-(1)/(6)+(1)/(24)-(1)/(120))` `=6-20+5-1` `=65-21` =44 Hence, statement-1 is true, statement-2 is true and statement-2 is a correct explanation for statement-1. |
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| 63. |
If ` a,b`and `c` are three consecutive positive integers such that `1/(a!)+1/(b!)=lambda/(c!)`A. aB. bC. cD. a+b+c |
| Answer» Correct Answer - C | |
| 64. |
Let `a_(n)` denote the number of all n-digit numbers formed by the digits 0,1 or both such that no consecutive digits in them are 0. Let `b_(n)` be the number of such n-digit integers ending with digit 1 and let `c_(n)` be the number of such n-digit integers ending with digit 0. Which of the following is correct ?A. `a_(17)=a_(16)+a_(15)`B. `c_(17)nec_(16)+c_(15)`C. `b_(17)neb_(16)+c_(16)`D. `a_(17)=c_(17)+b_(16)` |
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Answer» Correct Answer - A by recurring formula `a_(17)=a_(16)+a_(15)` is correct. also, `C_(17) ne C_(16)+C_(15) impliesa_(15) ne a_(14)+a_(13)` `[becauseC_(n)=a_(n-2)]` `therefore`Incorrect, similarly, other parts are also incorrect. |
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| 65. |
In how many ways can 15 identical blankets be distributed among sixbeggars such that everyone gets at least one blanket and tow particularbeggars get equal blankets and another three particular beggars get equalblankets. |
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Answer» The number of ways of distributing blankets is equal to the number of solutions of the equations `3x+2y+z=15`. where `xge1,yge1,zge1` which is equal to coefficient of `alpha^(15)` in `(alpha^(3)+alpha^(6)+alpha^(9)+alpha^(12)+alpha^(15)+ . ..)xx(alpha^(2)+alpha^(4)+alpha^(6)+alpha^(8)+alpha^(10)+alpha^(12)+alpha^(14)+ . .)` `xx(alpha+alpha^(2)+alpha^(3)+ . .+alpha^(15)+ . . .)` =Coefficient of `alpha^(9)` in `(1+alpha^(3)+alpha^(6)+alpha^(9))xx(1+alpha^(2)+alpha^(4)+alpha^(6)+alpha^(8))` `xx(1+alpha+alpha^(2)+alpha^(3)+alpha^(4)+alpha^(5)+alpha^(6)+alpha^(7)+alpha^(8)+alpha^(9))` [negletcing higher powers] =Coefficient of `alpha^(9)` in `(1+alpha^(2)+alpha^(4)+alpha^(6)+alpha^(8)+alpha^(3)+alpha^(5)+alpha^(7)+alpha^(9)+alpha^(6)+alpha^(8)+alpha^(9))xx(1+alpha+alpha^(2)+alpha^(3)+alpha^(4)+alpha^(5)+alpha^(6)+alpha^(7)+alpha^(8)+alpha^(9))` [neglecting higher powers] `=1+1+1+1+1+1+1+1+1+1+1+1=12` Case II If the inequation `x_(1)+x_(2)+x_(3)+ . .+x_(m)len` . .. (i) [when the required sum is not fixed] In this case, we introduce a dummy variable `x_(m+1)`. so that, `x_(1)+x_(2)+x_(3)+ . ..+x_(m)+x_(m+1)=n` `x_(m+1)ge0` Here, the number of sols of eqs. (i) and (ii) will be same. |
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| 66. |
Let `A`be a set of `n`distinct elements. Then the total number of distinct function from `AtoA`is ______ and out of these, _____ are onto functions. |
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Answer» Correct Answer - `n^(n), overset(n)underset(r = 1)sum(-1)^(n - r) " "^(n)C_(r) (r)^(n)` Let `A = {x_(1), x_(2),..,x_(n)}` `therefore` Number of functions from A to A is `n^(n)` and out of these `overset(n) underset(r - 1)sum (-1)^(n-r)" "^(n)C_(r) (r)^(n)` are onto functions. |
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| 67. |
Property:Product of r consecutive number is divisible by r! |
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Answer» Correct Answer - True Let r consecutive integers be x + 1, x + 2,.., x + r. `therefore (x +1 )(x + 2) .. (x + r) = ((x + r)(x + r - 1)..(x + 1)x!)/(x!)` `= ((x + r)!)/((x)!)*(r!)/(r!) = ""^(x + r)C_(r)*(r)!` Thus, `(x + 1)(x + 2)…(x + r) = ""^(x + r)C_(r)*(r)!`, which is clearly divisible by `(r)!`. Hence, it is a true statement. |
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| 68. |
For a set of five true/false questions, no student has written allcorrect answers, and no tow students have given the same sequence of answers.What is the maximum number of students in the classes, for this to bepossible?A. 9B. 32C. 31D. 24 |
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Answer» Since a true/false type question can be answered in 2 ways either by marking it true or false. So, there are 2 ways of answering each of the 5 questions. So, Total number of different sequences of answers `=2xx2xx2xx2=2^(5)=32`. Out of these 32 sequences of answers there is only one sequence of answering all the five questions correctly. But, no student has written all the corrent answers and different stundents have given different sequecnes of answers. So, Maximum number of students in the class = Number of sequences except one sequence in which all answers are correct =32-1=31 |
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| 69. |
All possible four-digit numbers have formed using the digits 0, 1, 2, 3 so that no number has repeated digits. The number of even numbers among them, isA. 9B. 18C. 10D. none of these |
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Answer» Correct Answer - C |
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| 70. |
In a certain test, a students gave wrong answers to at least `i` questions, where `i = 1, 2, ..., k`. No student gave more than `k` wrong answers. The total number of wrong answers given is |
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Answer» Correct Answer - `2^(n) - 1` The number of students answering exactly `k(1 le k le n - 1)` questions wrongly is `2^(n - k) - 2^(n - k - 1)`. The number of students answering all questions wrongly is `2^(0)`. Thus, total number of wrong answers `= 1(2^(n- 1) - 2^(n - 2)) + 2(2^(n - 2) - 2^(n - 3)) + ... + (n - 1)(2^(1) - 2^(0)) + 2^(0) * n` `= 2^(n- 1) + 2^(n - 2) + 2^(n - 3) + ... + 2^(1) - 2^(0) = 2^(n)-1` |
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| 71. |
Find the number of odd proper divisors of `3^pxx6^mxx21^ndot`A. `(m+1)(n+1)(p+1)-1`B. `(m+n+p+1)(p+1)-1`C. `(m+1)(n+1)(p+1)-2`D. none of these |
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Answer» We have, `3^(m)xx6^(n)xx21^(p)=3^(m)xx2^(n)xx3^(n)xx3^(p)xx7^(p)=2^(n)xx3^(m+n+p)xx7^(p)` Clearly, Number of odd proper divisors = Number of ways of selecting any number of `3^(s)and7^(s)` from (m+n+p) identical `3^(s)andP` identical `7^(s)`. `=(m+n+p+1)(p+1)-1`. |
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| 72. |
The number of times the digit 3 will be written when listing theintegers from 1 to 1000 is:300 (b) 269(c) 271 (d) 302A. 300B. 297C. 243D. 273 |
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Answer» Since 3 does not occur in 1000. So, we have to count the number of times 3 occurs when we list the integers from 1 to 999. Any number between 1 and 999 is of the form abc where `0lea,b,cle9`. Clearly, Number of times 3 occurs =(No. of number in which 3 occurs exactly at one place) +2 (No. of numbers in which 3 occurs exactly at two places) +3(No. of numbers in which 3 occurs exactly at three places) Now No. of numbers in which 3 occurs exactly at one place: Since 3 can occur at one place in `^(3)C_(1)` ways and each of the remaining two places can be filled in 9 ways. So, number of numbers in which 3 occurs exactly at one place `=^(3)C_(1)xx9xx9`. No. of number in which 3 occurs exactly at two places : Since 3 can occru exactly at two places in `^(3)C_(2)` ways and the remaining place can be filled in 9 ways. So, number of numbers in which 3 occurs exactly at two places `=""^(3)C_(2)xx9`. No. of numbers in which 3 occurs at all the three places: Since 3 can occur in all the three digits in one way only. So, number of numbers in which 3 occurs at all the three places is one. Hence, Numbers of times 3 occurs `=""^(3)C_(1)xx9xx9+2(""^(3)C_(2)xx9)+3xx1=300`. |
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| 73. |
The number of six digit numbers that can be formed from the digits `1,2,3,4,5,6 & 7` so that digits do not repeat and terminal digits are even is:A. 144B. 72C. 288D. 720 |
| Answer» Required number `=(""^(3)C_(2)xx2!)xx(""^(5)C_(4)xx4!)=720`. | |
| 74. |
N a certain test, there are `n`question. In the test, `2^(n-i)`studentsgave wrong answers to at least `i`questions,where `i=1,,2ddot,ndot`if thetotal number of wring answers given is 2047, then `n`is equal toa. `10`b. `11`c. `12`d. `13`A. 10B. 11C. 12D. 13 |
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Answer» The number of studnets answering exactly `k(1leklen-1)` questions wrongly is `2^(n-k)-2^(n-k-1)`. The number of studnets answering all questions wrongly is `2^(0)`. Thus, the total number of wrong answers `=1.(2^(n-1)-2^(n-2))+2.(2^(n-2)-2^(n-3))+......+(n-1)(2^(1)-2^(0))+n.2^(0)` `=2^(n-1)+2^(n-2)+2^(n-3)+......+2^(0)=2^(n)-`1 Thus, `2^(n)-1=2047implies2^(n)=2048implies2^(n)=2^(11)impliesn=11` |
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| 75. |
Ten different letters of an alphabet are given. Words with five lettersare formed from these given letters. Determine the number4 of words whichhave at least one letter repeated.A. 69760B. 30240C. 99748D. none of these |
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Answer» Correct Answer - A |
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| 76. |
Ten different letters of an alphabet are given. Words with five lettersare formed from these given letters. Determine the number4 of words whichhave at least one letter repeated.A. `10^(5)`B. 30240C. 69760D. none of these |
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Answer» The total number of 5-letter words which can be formed from 10 letters when one or more of its letters is repeated `=10xx10xx10xx10xx10=10^(5)`. The total number of 5-letter words when none of their letters is repeated `=^(10)C_(5)xx5!""=30240`. Hence, the number of 5-letter words which have at least one of their letters repeated `=10^(5)-30240=69760`. |
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| 77. |
In a party15 people shake their hands with each other. How many times did thehand-shakes take place?(a)105 (b) 120 (c) 135 (d) 165A. `""^(15)P_(2)`B. `""^(15)C_(2)`C. 15!D. `2xx15!` |
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Answer» Correct Answer - B |
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| 78. |
Every two persons shakes hands with each other in a party and the total number of handis 66. The number of guests in the party isA. 11B. 12C. 13D. 14 |
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Answer» Correct Answer - B |
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| 79. |
Find the total number of positive integral solutions for `(x ,y ,z)`such that `x y z=24.`Also find out the total number of integral solutions.A. 36B. 90C. 120D. None of these |
| Answer» Correct Answer - C | |
| 80. |
The number of integral solutions of `x+y+z=0`with `xgeq-5,ygeq-5,zgeq-5`isA. 135B. 136C. 455D. 105 |
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Answer» Let `u=x+5,v=y+5andw=z+5`. Then, `u,v,wge0andx+y+z=0impliesu+v+w=15`. The number of solutions of this equation is `""^(15+3-1)C_(3-1)=""^(17)C_(2)=136` |
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| 81. |
Number of non-negative integral solutions of the equation a+b+c=6 isA. 28B. 32C. 36D. 56 |
| Answer» Correct Answer - A | |
| 82. |
If a,b and c are integers and `age1,bge2 and c ge 3`. If `a+b+c=15`, the number of possible solutions of the equation isA. 55B. 66C. 45D. None of these |
| Answer» Correct Answer - A | |
| 83. |
The number of integral solutions of `x+y+z=0`with `xgeq-5,ygeq-5,zgeq-5`isa. `134`b. `136`c. `138`d. `140`A. 272B. 136C. 240D. 120 |
| Answer» Correct Answer - B | |
| 84. |
The number of words which can be formed out of the letters `a`, `b`, `c`, `d`, `e` `f` taken 3 together, each word containing one vowel at least isA. 72B. 48C. 96D. none of these |
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Answer» Correct Answer - C |
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| 85. |
The smallest value of x satisfying the inequality `""^(10)C_(x-1)gt2.""^(10)C_(x)` is.A. 7B. 10C. 9D. 8 |
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Answer» Correct Answer - D |
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| 86. |
There are n seats round a table numbered `1,2, 3,..., n`. The number of ways in which `m( |
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Answer» Correct Answer - B |
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| 87. |
The total number of ways in which 30 mangoes can be distributed among 5 persons isA. `""^(30)C_(5)`B. `30^(5)`C. `5^(30)`D. `""^(34)C_(4)` |
| Answer» Required number of ways is same as the number of ways of dividing 30 identical items among 5 persons i.e., `""^(30+5-1)C_(5)=""^(34)C_(4)` | |
| 88. |
The number of divisor of 4200, isA. 42B. 48C. 54D. none of these |
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Answer» We have, `4200=2^(3)xx3^(1)xx5^(2)xx7^(1)` `:.` Number of divisors `=(3+1)(1+1)(2+1)(1+1)=48` |
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| 89. |
Numbers greater than 1000 but not greater than4000 which can be formed with the digits 0, 1, 2, 3, 4 (repetition of digitsis allowed) area. `350`b. `375`c. `450`d. `576`A. 375B. 374C. 376D. none of these |
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Answer» Correct Answer - A |
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| 90. |
How many numbers lying between 100 and 1000 can be formed with the digits 0, 1, 2, 3, 4, 5, if the repetition of the digits is not allowed? |
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Answer» total ways are `.^6P_3 - .^5P_2` `= (6!)/((6-3)!)- (5!)/((5-2)!)` `=6*5*4- 5*4` `= 5*5*4= 100` answer |
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| 91. |
How many 4-digit numbers can be formed by using the digits 1 to 9 if repetition of digits is not allowed? |
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Answer» For the first digit place there are 9 options , second digit place there are 8 options, third digit place there are 7 options and fourth digit place there are 6 options. so number of 4 digit numbers=`9 xx 8 xx 7xx6=3024` |
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| 92. |
Write the number of all possible words that can be formed using theletters of the word MATHEMATICS.A. `(11!)/(2!2!2!)`B. 11!C. `""^(11)C_(1)`D. none of these |
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Answer» Correct Answer - A |
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| 93. |
Nishi has 5 coins, each of the different denomination. Find the numberdifferent sums of money she can form.A. 32B. 25C. 31D. none of these |
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Answer» Correct Answer - C |
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| 94. |
In a group of boys, two boys are brothers and sixmore boys are present in the group. In how many ways can they sit if thebrothers are not to sit along with each other?a.`2xx6!`b. `^7P_2xx6!`c. `^7C_2xx6!`d. none of theseA. 4820B. 1410C. 2830D. none of these |
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Answer» Correct Answer - D |
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| 95. |
There are three coplanar parallel lines. If any `p`points aretaken on each of the lines, the maximum number of triangles with vertices onthese points isa. `3p^2(p-1)+1`b. `3p^2(p-1)`c. `p^2(4p-3)`d. none of theseA. `3n^(2)(n-1)`B. `3n^(2)(n-1)+1`C. `n^(2)(4n-3)`D. none of these |
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Answer» The maximum number of triangles = Number of triangles with vertices on different lines + Number of trigangles with 2 vertices on one line and third vertex on any one of the remaining two lines `=""^(n)C_(1)xx""^(n)C_(1)xx""^(n)C_(1)+""^(3)C_(1)xx(""^(n)C_(2)xx""^(2n)C_(1))=3n^(2)(n-1)`. |
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| 96. |
Find the number of ways in which 8 different flowered can be strung toform a garland so that four particular flowers are never separated.A. `4!.4!`B. `(8!)/(4!)`C. 288D. none of these |
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Answer» Correct Answer - A |
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| 97. |
We are required to form different words with the help of the word INTEGER. Let `m_1` be the number of words in which I and N are never together and `m_2` be the number or words which begin with and I and end with R, then `m_1/m_2`A. 42B. 30C. 6D. `1//30` |
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Answer» Correct Answer - B |
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| 98. |
The number of ways four boys can be seated around a round talble in four chairs of ditterent colours, isA. 24B. 12C. 23D. 64 |
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Answer» Correct Answer - A |
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| 99. |
If `.^(n)P_(5)=20 .^(n)P_(3)`, find the value of n.A. 4B. 8C. 6D. 7 |
| Answer» Correct Answer - B | |
| 100. |
The number of prime numbers among the numbers `105! + 2, 105! +3, 105! +4,...., 105! + 104, 105!+105,` isA. 31B. 32C. 33D. None of these |
| Answer» Correct Answer - D | |