

InterviewSolution
Saved Bookmarks
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.
101. |
If `(n-1) C_6 + (n-1)C_7 > nC_6` thenA. `ngt4`B. `ngt12`C. `nge13`D. `ngt13` |
Answer» Correct Answer - D |
|
102. |
The number of ways in which 21 objects can be grouped into three groups of 8, 7, and 6 objects,isA. `(20!)/(8!+7!+6!)`B. `(21!)/(8!7!)`C. `(21!)/(8!7!6!)`D. `(21!)/(8!+7!+6!)` |
Answer» Correct Answer - C |
|
103. |
Consider the fourteen lines in the plane given by y=x+r, y=-x+r, where `rin{0,1,2,3,4,5,6}`. The number of squares formed by these lines whose diagonals are of length 2, isA. 9B. 16C. 25D. 36 |
Answer» Correct Answer - C |
|
104. |
Let `L_1 and L_2` be two lines intersecting at P If `A_1,B_1,C_1` are points on `L_1,A_2, B_2,C_2,D_2,E_2` are points on `L_2` and if none of these coincides with P, then the number of triangles formed by these 8 points isA. 56B. 55C. 46D. 45 |
Answer» Correct Answer - D |
|
105. |
Let A and B be two finite sets having m and n elements respectively. Then the total number of mappings from A to B isA. `n^(m)`B. `""^(n)C_(m)`C. `""^(n)C_(m)xxm!`D. `m^(n)` |
Answer» In order to define a one-one function from A to B, we associate different elements of set A to distinct elements in set B. This is possible only when `ngem`. To determine the total number of one-one functions from A to B we first seletct m elements from B and then elements in A are associated to elements in B. The first part con be done in `""^(n)C_(m)` ways and the second part in m! ways. Hence, total number of one-one functions `=""^(n)C_(m)xxm!` |
|
106. |
Two different packs of cards are shuffled together. Cards dealt equally among 4 players, each getting 13 cards. The number of ways in which a player get his cards if no two cards are from the same suit with the same denomination isA. `""^(52)C_(13)`B. `2^(13)`C. `""^(52)P_(13)`D. `""^(52)C_(13)xx2^(13)` |
Answer» Here, we have 52 cards, each card being 2 in number. It is given that no two are to be of the same suit with the same denomination. So, we first draw 13 cards from 52 cards. This can be done in `""^(52)C_(13)` ways. Now each of 13 cards selected can be chosen in 2 ways either from first pack or from 2nd pack. Hence, required number of ways `=""^(52)C_(13)xx2^(13)`. |
|
107. |
How many number of 4 digits can be formed with the digits 1,2,3,4,5 if repetition of digit is allowed and not allowed? |
Answer» 1) Repetition is all allowed 4 digit number=5*5*5*=5^4=625 2) Repetition is not allowed 4 digit number=5*4*3*=120 |
|
108. |
Let n be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let m be the number in which 5 boys and 5 girls stand in such a way that exactly four girls stand consecutively in the queue. Then the value of `m/n` is ____ |
Answer» Correct Answer - 5 `n=5!xx6!` For m: 5 boys can stand in a row in 5!, creating 6 alternate space for girls. A group of 4 girls can be selected in `.^(5)C_(4)` ways. A group of 4 and single girl can be arranged at 2 places out of 6 in `.^(6)P_(2)` ways. Also, 4 girls can arrange themselves in 4! ways. `thereforem=5!xx.^(6)P_(2)xx.^(5)C_(4)xx4!=5!xx30xx5x4!=5!xx6!xx5` `implies(m)/(n)=(5!xx6!xx5)/(5!xx6!)=5` |
|
109. |
In a question paper there are two parts part A and part B each consisting of 5 questions. In how many ways a student can answer 6 questions, by selecting atleast two from each part? |
Answer» Case 1 `5C_2*5C_4=(5!)/(2!3!)=(5!)/(4!) =50` Case 2 `5C_3*5C_3=10*10=100` Case 3 `5C_4*5C_2=50` Total=100+50+50=200. |
|
110. |
The number of ways in which one or more balls can be selected out of 10 whit, 9 green and 7 blue balls, isA. 892B. 881C. 891D. 879 |
Answer» We have, Required number of ways `=(10+1)(9+1)(7+1)-1=879`. |
|
111. |
Find the number of groups that can be made from 5 different green balls., 4 different blue balls and 3 different red balls,if at least 1 green and 1 blue ball is to be included.A. 3700B. 3720C. 4340D. None of these |
Answer» Correct Answer - B | |
112. |
Given 5 different green dyes, four different blue dyes and three different red dyes, how many combinations of dyes can be chosen taking at least one green and one blue dye ?A. 3255B. 212C. 3720D. none of these |
Answer» Correct Answer - C |
|
113. |
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. 36B. 66C. 108D. 3 |
Answer» Correct Answer - C Total number of ways=`.^(3)C_(2)xx.^(9)C_(2)=.^(3)C_(1)xx.^(9)C_(2)=3xx(9xx8)/(1xx2)` `=3xx9xx4=108` |
|
114. |
If `""^(n)C_(r -1) = 36, ""^(n)C_(r) = 84 " and " ""^(n)C_(r +1) = 126`, then find the value of `""^(r)C_(2)`. |
Answer» Correct Answer - n = 9 and r = 3 We know that, `(""^(n)C_(r))/(""^(n)C_(r - 1)) = (n - r + 1)/(r)` `rArr " " 84/36 = 7/3 = (n - r + 1)/(r) " "` [given] `rArr " " 3n - 10r + 3 = 0 " "….(i)` Also given, `(""^(n)C_(r))/(""^(n)C_(r + 1)) = 84/126` `rArr " " (r + 1)/(n - r) = 2/3` `rArr" " 2n - 5r - 3 = 0 " "..(ii)` On solving Eqa. (i) and (ii), we get r = 3 and n = 9 |
|
115. |
Find the number of rectangles excluding squares from a rectangle of size 9 × 6. |
Answer» Here, n=6 and p=9 `therefore`Number of rectangles excluding square `=(6*9)/(4)(6+1)(9+1)-underset(r=1)overset(6)(sum)(7-r)(10-r)` `=945-underset(r=1)overset(6)(sum)(70-17r+r^(2))=945-154=791` |
|
116. |
If the letters of the word SACHIN are arranged inall possible ways and these words are written out as in dictionary, then theword SACHIN appears at serial number602 (2) 603(3) 600(4) 601A. 602B. 603C. 600D. 601 |
Answer» each question can be answered in two ways. So, total number of ways of answering 10 questions is `2xx2xx2xx....xx2` (10 times) `=2^(10)=1024` |
|
117. |
Out of 18 points in as plane, no three points are in the same straight line except five points which are collinear. The number of straight lines formed by joining them isA. 144B. 153C. 152D. 140 |
Answer» Correct Answer - A Total number of points = 18 Out of which 5 points are collinear, we get a straight line by joining any two points. `therefore" "` Number of straight line formed by joining the 18 points taking 2 at a time = `""^(18)C_(2)` and number of straight line formed by joining the 18 points taking 2 at a time `= ""^(5)C_(2)` But 5 collinear points, when joined pairwise give only one line. `therefore" "` Required number of straight line `= ""^(18)C_(2) - ""^(5)C_(2) + 1` `= 153 - 10 + 1 = 144` |
|
118. |
How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated ? |
Answer» `----- 0` total 6 digit`=5! = 120` answer |
|
119. |
At an election there are five candidates and three members to be elected, and an elector may vote for any number of candidates not greatre than the number to be elected. Then the number of ways in which an elector may vote, isA. 25B. 30C. 32D. none of these |
Answer» Correct Answer - A |
|
120. |
A bag contains 5black and 6 red balls. Determine the number of ways in which 2 black and 3red balls can be selected. |
Answer» It is given that bag contains 5 black and 6 red balls. So, 2 black balls is selected from 5 black balls in `""^(5)C_(2)` ways. and 3 red balls are selected from 6 red balls in `""^(6)C_(3)` ways. `therefore` Total number of ways in which 2 black and 3 red balls are selected `= ""^(5)C_(2) xx ""^(6)C_(3) = 10 xx 20 = 200` ways |
|
121. |
A bag has contains 23 halls in which 7 are identical . Then number of ways of selecting 12 balls from bag. |
Answer» Here, `n=23,p=7,r=12 (rgtp)` `therefore` Required number of selections `=underset(r=5)overset(12)(sum).^(16)C_(r)` `=.^(16)C_(5)+.^(16)C_(6)+.^(16)C_(7)+.^(16)C_(8)+.^(16)C_(9)+.^(16)C_(10)+.^(16)C_(11)+.^(16)C_(12)` `=(.^(16)C_(5)+.^(16)C_(6))+(.^(16)C_(7)+.^(16)C_(8))+(.^(16)C_(9)+.^(16)C_(10))+(.^(16)C_(11)+.^(16)C_(12))` `=.^(17)C_(6)+.^(17)C_(8)+.^(17)C_(10)+.^(17)C_(12)" "[because .^(n)C_(r)+.^(n)C_(r-1)=.^(n+1)C_(r)]` `=.^(17)C_(11)+.^(17)C_(9)+.^(17)C_(10)+.^(17)C_(12)" "[because.^(n)C_(r)=.^(n)C_(n-r)]` `=(.^(17)C_(11)+.^(17)C_(12))+(.^(17)C_(9)+^(17)C_(10))` `=.^(18)C_(12)+.^(18)C_(10)=.^(18)C_(6)+.^(18)C_(8)` |
|
122. |
If `rgtpgtq`, the number of different selections of `p+q` things taking r at a time, where p things are identical and q other things are identical, isA. `p+q-r`B. `p+q-r+1`C. `r-p-q+1`D. none of these |
Answer» Correct Answer - B |
|
123. |
The total number of 9-digit number which have all different digits isA. `10"!"`B. `9"!"`C. `9 xx 9"!"`D. `10 xx 10"!"` |
Answer» Correct Answer - C We have to form 9-digit number with the digit 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 cannot be placed at the first place from left. So, first place from left can be filled in 9 ways. Since, repetition is not allowed, so remaining 8 places can be filled in `9"!"` ways. `therefore` Required number of ways `=9 xx 9"!"` |
|
124. |
The number of all three element subsets of the set `{a_(1),a_(2),a_(3)......a_(n)}` which contain `a_(3)` , isA. `""^(n)C_(3)`B. `""^(n-1)C_(3)`C. `""^(n-1)C_(2)`D. none of these |
Answer» The number of three element subsets contaning `a^(3)` is equal to the number ways of selecting 2 elements out of n-1 elements. So, the required number of subsets `=""^(n-1)C_(2)`. |
|
125. |
The total number of all proper factors of 75600, isA. 120B. 119C. 118D. none of these |
Answer» Correct Answer - C |
|
126. |
Find the total number of ways of answering 5 objective type questions,each question having 4 choices.A. `5^(4)`B. `4^(5)`C. 20D. 9 |
Answer» Since each question can be answered in 4 ways. So, the total number of ways of answering 5 questions `=4xx4xx4xx4xx4=4^(5)` |
|
127. |
The number of natural numbers smaller than `10^4` of which all digits are different, isA. 5274B. 5265C. 4676D. none of these |
Answer» Correct Answer - A |
|
128. |
Find the number of proper factors of the number 38808.A. 72B. 70C. 69D. 71 |
Answer» Correct Answer - B |
|
129. |
In how many ways can 30 marks be allotted to 8 questionif each question carries at least 2 marks?A. `""^(21)C_(7)`B. `""^(21)C_(8)`C. `""^(21)C_(9)`D. none of these |
Answer» Let `x_(1)` denote the marks assigned to the ith question. Then, `x_(1)+x_(2)+x_(3)+x_(4)+x_(5)+x_(6)+x_(7)+x_(8)=30` where `x_(1)ge2,i=1,2,....,8`. Let, `y_(i)=x_(i)-2,i=1,2,....8`. Then, `x_(1)+x_(2)+x_(3)+....x_(8)=30` `impliesy_(1)+y_(2)+y_(3)+......+y_(8)=14` The total number of solutions of this equation is `""^(14+8-1)C_(8-1)=""^(21)C_(7)`. |
|
130. |
If eight persons are to address a meeting, then the number of ways in which a specified speaker is to speak before another specified speaker, isA. 2520B. 20160C. 40320D. none of these |
Answer» Correct Answer - B |
|
131. |
There are five balls of different colours and five boxes of colours same as those of the balls. The number of ways in which the balls, one each in a box, could be placed such that exactly one ball goes to a box of its own colour, isA. 9B. 24C. 45D. 120 |
Answer» Let `b_(1),b_(2),......b_(5)` be five balls and `B_(1),B_(2),......,B_(5)` be five boxes. Then, Requried number of ways `sum_(r=1)^(5)(r^(th)" ball is placed in "r^(th)" box")xx("Remaining balls are placed in wrong boxes")` `=sum_(r=1)^(5)1xx4!{-1(1)/(1!)+(1)/(2!)-(1)/(3!)+(1)/(4!)}` `=sum_(r=1)^(5)(12-4+1)=45`. |
|
132. |
Find the number of divisors of the number `N=2^3 .3^5 .5^7 .7^9`which are perfect squares. |
Answer» `becauseN=2^(3)*3^(5)*5^(7)*9^(11)` `=2^(3)*3^(5)*5^(7)*7^(9)*3^(22)` `=2^(3)*3^(27)*5^(7)*7^(9)` For perfect square of N, each prime factor must occur even number of times. 2 can be taken in 2 ways (i.e., `2^(0)` or `2^(2)`) 3 can be taken in 14 ways (i.e., `3^(0)` or `3^(2)` or `3^(4)` or `3^(6)` orr `3^(8)` or `3^(10)` orr `3^(12)` or `3^(14)` or `3^(16)` or `3^(18)` or `3^(20)` or `3^(22)` orr `3^(24)` or `3^(26)`) 5 can taken in 4 ways (i.e., `5^(0)` or `5^(2)` or `5^(4)` or `5^(6)`) and 7 can taken in 5 ways (i.e., `7^(0)` or `7^(4)` or `7^(4)` or `7^(6)` or `7^(8)`) Hence, total divisors which are perfect square `=2xx14xx4xx5=560` |
|
133. |
In how many ways the number 18900 can be split in two factors which are relatively prime or co prime |
Answer» Let N=`18900=2^(2)*3^(3)*5^(2)*7^(1)` Relatively prime or coprime Two numbers not necessarily prime are said to be relatively prime or coprime, if their HCF (highest common factor) in one as 2,3,5,7 are relatively prime numbers. `thereforen=4` [number of different primes in N] Hence, number of ways in which a composite number N can be resolved into two factors which are relativelly prime or comprime=`2^(4-1)=2^(3)=8` |
|
134. |
In how many ways the number 10800 can be resolved as a product of two factors? |
Answer» Let `N=10800=2^(4)xx3^(3)xx5^(2)` Here, N is not a perfect square [`because` power of 3 is odd] Hence, the number of ways=`(1)/(2)(4+1)(3+1)(2+1)=3`. |
|
135. |
The total number of positive integral solutions of abc=30, isA. 30B. 27C. 8D. none of these |
Answer» We have, `30=2xx3xx5=10xx3xx1` `=2xx15xx1=6xx5xx1xx30` Hence, total number of positive integral solutions of abc = 30 is `3!+3!+3!+3!+(3!)/(2!)=27` |
|
136. |
The total numbre of intergral solutins of abc =24, isA. 36B. 90C. 120D. 30 |
Answer» In example 100, we have seen that the total number of positive integral solutions of abc=24 is 30. We observe that any two of the factors in each factorization may be negative. `:.` Number of integral solutions = Number of positive integral solutions + Number of integral solutions having two negative factors `=30+""^(3)C_(2)xx30=120`. |
|
137. |
The number of ways in which 12 books can be put inthree shelves with four on each shelf isa. `(12 !)/((4!)^3)`b. `(12 !)/((3!)(4!)^3)`c. `(12 !)/((3!)^3 4!)`d. none of theseA. `(12!)/((4!)^(3))`B. `(12!)/((3!)(4!)^(3))`C. `(12!)/((3!)^(3)4!)`D. none of these |
Answer» Correct Answer - A |
|
138. |
Using the digits 1, 2, 3, 4, 5, 6,7, a number of 4 different digits is formed. Find |
Answer» (i) Total number of 4 digit formed with digit 1, 2, 3, 4, 5, 6, 7 `= 7 xx 6 xx 5 xx 4 = 840` (ii) When a number is divisible by 2. At its unit place only even numbers occurs. Total numbers `= 4 xx 5 xx 6 xx 3 = 360` (iii) Total numbers which are divisible by 25 = 40 (iv) A number is divisible 4, If its last two digit is divisible by 4. `therefore` Total such numbers = 200 |
|
139. |
No. of different matrices that can be formed with elements `0, 1, 2 or 3` each matrix having 4 elements isA. `3xx2^(4)`B. `2xx4^(4)`C. `3xx4^(4)`D. none of these |
Answer» A matrix having 4 elements can of the order `1xx4or4xx1or,2xx2`. Since, each element can take 4 values viz. 0, 1, 3. Similarly number of matrices of each order is `4^(4)`. Hence, required number of matrices `=4^(4)+4^(4)+4^(4)=3xx4^(4)` |
|
140. |
There are 10 professors and 20 lecturers, out of whom a committee of 2 professors and 3 lecturers is to be formed. Find |
Answer» (i) We have to select 2 professors out of 10 and 3 lecturers out of `20 = ""^(10)C_(2) xx ""^(20)C_(3)` (ii) When a particular professor included `= ""^(10 -1)C_(1) xx ""^(20)C_(3) = ""^(9)C_(1) xx ""^(20)C_(3)` (iii) When a particular lecturer included `= ""^(10)C_(2) xx ""^(19)C_(2)` (iv) When a particular lecturer excluded `= ""^(10)C_(2) xx ""^(19)C_(3)` |
|
141. |
`""^(n)C_(r)+2""^(n)C_(r-1)+C_(r-2)` is equal toA. `""^(n+1)C_(r)`B. `""^(n)C_(r+1)`C. `""^(n-1)C_(r+1)`D. none of these |
Answer» Correct Answer - D |
|
142. |
Seven different lecturers are to deliver lectures in seven perday. A, B and C are three oftne lectures. The number of ways in which a routine for theday canr be made such that A delivers his lecture before B and B before C, isA. 420B. 120C. 210D. none of these |
Answer» Let us consider the arrangements of 7 objects out of which three are identical and remaining distinct. In each arrangement of these 7 objects, let us mark the first identical item as P, second as Q and third as R. We find that in all the `(7!)/(3!)` arrangements P will occur prior to Q and Q will occur prior to R. In the given sum also the required number of ways `=(7!)/(3!)=840` |
|
143. |
The number of ways in which 9 persons can be divided into three equal groups isA. 1680B. 840C. 560D. 280 |
Answer» Correct Answer - D |
|
144. |
In how any ways can 8 different books be distributed among 3 studentsif each receives at least 2 books?A. 490B. 980C. 2940D. 5880 |
Answer» Correct Answer - C | |
145. |
Find the number of distinct rational numbers `x`such that `oA. 15B. 13C. 12D. 11 |
Answer» Correct Answer - D As `0 lt x lt 1`, we have `p lt q` The number of rational numbers=5+4+3+2+1=15 When p,q have a common factor, we get some rational numbers which are not different from those already counted. There are 4 such numbers `(2)/(4),(2)/(6),(3)/(6),(4)/(6)`. `therefore` The required of rational numbers=15-4=11 |
|
146. |
Let A be the set of 4-digit numbers `a_1 a_2 a_3 a_4` where `a_1 > a_2 > a_3 > a_4`, then `n(A)` is equal toA. 126B. 84C. 210D. None of these |
Answer» Correct Answer - C Any selection of four digits from the ten digits 0,1,2,3,. . . ,9 gives one number. So, the required number of numbers`=^(10)C_(4)=120` |
|
147. |
If `a_1,a_2,a_3,.....,a_(n+1)` be `(n+1)` different prime numbers, then the number of different factors (other than1) of `a_1^m.a_2.a_3...a_(n+1)`, isA. m+1B. (m+1)2^(n)`C. `m*2^(n)+1`D. None of these |
Answer» Correct Answer - D | |
148. |
The number of committees of 5 persons consisting of at least one female number, that can be formed from 6 males and 4 females, isA. 246B. 252C. 6D. none of these |
Answer» Correct Answer - A |
|
149. |
`""^(15)C_(8) + ""^(15)C_(9) - ""^(15)C_(6) - ""^(15)C_(7)` is equal to ………. . |
Answer» `""^(15)C_(8) + ""^(15)C_(9) - ""^(15)C_(6) - ""^(15)C_(7) = ""^(15)C_(15 -8) + ""^(15)C_(15 - 9) - ""^(15)C_(6) - ""^(15)C_(7) " " [because ""^(n)C_(r) = ""^(n)C_(n - r) ]` `= ""^(15)C_(7) + ""^(15)C_(6) - ""^(15)C_(6) - ""^(15)C_(7)` = 0 |
|
150. |
12 persons are to be arranged to a round table. If two particular persons among them are not to be side by side, the total number of arrangements, isA. `9(10!)`B. `2(10!)`C. `45(8!)`D. `10!` |
Answer» Correct Answer - A |
|