InterviewSolution
Saved Bookmarks
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.
| 151. |
There are 10 points in a plane out of these points no three are in the same straight line except 4 points which are collinear. How many (i) straight lines (ii) trian-gles (iii) quadrilateral, by joining them? |
|
Answer» (i) Required number of straight lines `=.^(10)C_(2)-.^(4)C_(2)+1=(10*9)/(1*2)-(4*3)/(1*2)+1=45-6+1=40` (ii) Required number of triangles `=.^(10)C_(3)-.^(4)C_(3)=(10*9*8)/(1*2*3)-.^(4)C_(1)=120-4=116` (iii) Required number of quadrilaterals `=.^(10)C_(4)-(.^(4)C_(3).^(4)C_(1)+.^(4)C_(4).^(6)C_(0))` `=(10*9*8*7)/(1*2*3*4)-(.^(4)C_(1)*.^(6)C_(1)+1.1)` `=210-(4xx6+1)=210-25=185` |
|
| 152. |
There are 10 points in a plane, no three of which are in the same straight line, except 4 points, which are collinear. Find the number of lines obtained from the pairs of these points,A. 40B. 39C. 45D. none of these |
|
Answer» Number of stright lines formed joining the 10 points, taking 2 at a time `""^(10)C_(2)=(10!)/(2!8!)=45`. Number of straight lines formed by joining the four points, taking 2 at a time `""^(4)C_(2)=(4!)/(2!2!)=6` But, 4 collinear points, when joined pairwise give only one line. `:.` Required number of straight lines =45-6+1=40. |
|
| 153. |
A person tries to form as many different parties as he can, out of his20 friends. Each party should consist of the same number. How many friendsshould be invited at a time? In how many of these parties would the samefriends be found?A. 5B. 10C. 8D. none of these |
|
Answer» Correct Answer - B |
|
| 154. |
A person tries to form as many different parties as he can, out of his 20 friends. Each party should consist of the same number. How many friends should be invited at a time? In how many of these parties would the same friends be found? |
|
Answer» Let the person invite r number of friends at a time. Then, the number of parties are `.^(20)C_(r)`, which is maximum, when r=10 If a particular friend will be found in p parties, then p is the number of combinations out of 20 in which this particular friend must be included. therefore, we have to select 9 more from 19 remaining friends. hence, `p=.^(19)C_(9)`. |
|
| 155. |
If `.^(n)C_(r-1)=36, .^(n)C_(r)=84 and .^(n)C_(r+1)=126`, find n and r.A. n=8,r=4B. n=9,r=3C. n=7,r=5D. none of these |
|
Answer» Correct Answer - B |
|
| 156. |
If `.^(35)C_(n+7)=.^(35)C_(4n-2)` then find the value of n.A. 28B. 3,6C. 3D. 6 |
|
Answer» Correct Answer - B |
|
| 157. |
`sum_(r=0)^m .^(n+r)C_n` is equal toA. `""^(n+m+1)C_(n+1)`B. `""^(n+m+2)C_(n)`C. `""^(n+m+3)C_(n-1)`D. none of these |
|
Answer» Correct Answer - A |
|
| 158. |
The expression `""^(n)C_(r)+4.""^(n)C_(r-1)+6.""^(n)C_(r-2)+4.""^(n)C_(r-3)+""^(n)C_(r-4)`A. `""^(n+4)C_(r)`B. `""2.^(n+4)C_(r-1)`C. `""4.^(n)C_(r)`D. `""11.^(n)C_(r)` |
|
Answer» Correct Answer - A |
|
| 159. |
Write the number of words that can be formed out of the letters of theword COMMITTEE.A. `(9!)/((2!)^(3))`B. `(9!)/((2!)^(2))`C. `(9!)/(2!)`D. 9! |
|
Answer» Correct Answer - A |
|
| 160. |
The total number of numbers greater than 100 and divisible by 5, that can be formed from the digits 3, 4, 5, 6 if no digit is repeated, isA. 24B. 48C. 30D. 12 |
|
Answer» Correct Answer - D |
|
| 161. |
Number of numbers greater than 24000 can be formed by using digits 1,2,3,4,5 when no digit being repeated isA. 36B. 60C. 84D. 120 |
|
Answer» Correct Answer - C |
|
| 162. |
The number of ways of distributing `8` identical balls in 3 distinct boxes so that none of the boxes is empty isA. `""^(8)C_(3)`B. 21C. `3^(8)`D. 5 |
|
Answer» The required number of ways is same as the number of ways of distributing 8 identical items to three persons such that each person receives at lesat one item. So, required numbers of ways `=""^(8-1)C_(3-1)=""^(7)C_(2)=21` |
|
| 163. |
Four numbers are chosen at random (without replacement) from the set{1, 2, 3, ....., 20}.Statement-1:The probability that the chosen numbers whenarranged in some order will form an APIs `1/(85)`.Statement-2:If the four chosen numbers from an AP, then the setof all possible values of common difference is {1, 2, 3, 4, 5}. |
|
Answer» `a= 1 ; d= 1; {1,2,3,4,17,18,19,20,17}` `d= 2 ; {1,3,5,7, 14,16,18,20,14}` `d=3 ; {1,4,7,10,11,14,17,20}` P(4 are in AP) = `(17+2)/2 xx6/(.^20C_4) = 57/(.^20C_4)` `= (57 xx 4 xx 3 xx 2)/(20 xx 19 xx 18 xx 17) = 1/85` option 2 is correct |
|
| 164. |
No. of different garlands using 3 flowers of one kind and 12 flowers of seconds kind is |
|
Answer» a+b+c=12 `.^(12+3-1)C_(3-1)=.^14C_2.` Option C is correct. |
|
| 165. |
Different words are formed by arranging the letters of the word `SUC CESS`, thenA. 42B. 40C. 420D. 480 |
|
Answer» Correct Answer - A Total number of ways=`(7!)/(2!3!)=420` Consonants in SUCCESS are S,C,C,S,S Number of ways arranging consonants=`(5!)/(2!3!)=10` Hence, number of words in which consonants appear in alphabetic order=`(420)/(10)=42` |
|
| 166. |
The number of ways in which four letters can be selected from the word degree, isA. 7B. 6C. `(6!)/(3!)`D. none of these |
|
Answer» Correct Answer - A |
|
| 167. |
Find the number of different permutations of the letters of the wordBANANA?A. 720B. 60C. 120D. 360 |
|
Answer» Correct Answer - B |
|
| 168. |
The number of ways of painting the faces of a cube with six different colours isA. 1B. 6C. 6!D. none of these |
|
Answer» Correct Answer - A |
|
| 169. |
Let `S(n)` denotes the number of ordered pairs `(x,y)` satisfying `1/x+1/y=1/n,AA ,n in N` `S(10)` equalsA. 3B. 6C. 9D. 12 |
|
Answer» Correct Answer - C `because 10^(2)=2^(2)*5^(2)` `thereforeS(10)=3xx3=9` |
|
| 170. |
Let `S(n)` denotes the number of ordered pairs `(x,y)` satisfying `1/x+1/y=1/n,AA ,n in N` `S(10)` equalsA. S(3)+S(4)B. S(5)+S(6)C. S(8)+S(9)D. S(1)+S(11) |
|
Answer» Correct Answer - C `because 6^(2)=2^(2)*3^(2)` `impliesS(6)=3xx3=9 and 7^(2)impliesS(7)=3` `thereforeS(6)+S(7)=12` also, `8^(2)=2^(6)` `impliesS(8)=7 and 9^(2)=3^(4) impliesS(9)=5` `thereforeS(8)+S(9)=12` `impliesS(6)+S(7)=S(8)+S(9)=12`. |
|
| 171. |
Statement-1: The number of zeros at the end of 100! Is, 24. Statement-2: The exponent of prine p in n!, is `[(n)/(p)]+[(n)/(p^(2))]+.......+[(n)/(p^(r))]` Where r is a natural number such that `P^(r)lenltP^(r+1)`.A. Statement-1 is True, Statement-2 is true, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
| Answer» Statement-2 is true (see section-2 on page 13.2) Also, Statement-1 is true (see illustrations 3 on page 13.2) | |
| 172. |
Number of ways in which RS. 18 can be distributed amongst four persons such that no body receives less than RS 4, isA. 42B. 24C. 4!D. none of these |
|
Answer» Correct Answer - D so this is the correct ans hope it helps |
|
| 173. |
The exponent of 12 in 100! isA. 24B. 25C. 47D. 48 |
| Answer» Correct Answer - C | |
| 174. |
If `2lamda` is the number of ways of selecting 3 member subset of {1,2,3, . .,29}, so that the number form of a GP with integer common ration, then the value of `lamda` is |
|
Answer» Correct Answer - 6 `2lamda`=Number of selecting 3 member subsets of {1,2,3, . . ,29} which are in GP with common raio (2 or 3 or 4 or 5). `=[29/2^(2)]+[29/3^(2)]+[29/4^(2)]+[29/5^(2)]` `=7+3+1+1=13` `thereforelamda=6` |
|
| 175. |
Let T(n) denotes the number of non-congruent triangles with integer side lengths and perimeter n. Thus `T(1)=T(2)=T(3)=T(4)=0`, while `T(5)=1`. Prove that: (i) `T(2006)ltT(2009)` ii) `T(2005)=T(2008)` |
|
Answer» a+b+c=n `T(n)=[(n+3)^2/48], n=odd` `T(n)=[n^2/48],n=even` 1)`T(2006)=[(2006)^2/48]` `T(2009)=[(2012)^2/48]` `T(2009)>T(2006)` 2)`T(2005)=[(2008)^2/48]=T(2005)`. |
|
| 176. |
Prove that 33! is divisible by `" 2^15`. what is the largest integer n such that 33! is divisible by `2^n`. |
|
Answer» For prime number `p` and `n in N`, the highest power of `p` that divides `n!` is given by, `sum_(k ge 1) [n/p^k]`. Here, `n = 33` and `p = 2`.So, highest power that divides `33! = [33/2]+[33/2^2]+[33/2^3]+[33/2^4]+[33/2^5]` `=16+8+4+2+1 = 31` So, `33!` can be divided by `2^31`. So, `33!` is divisible by `2^15` and maximun value of `n` is `31` such that `33!` is divisible by `2^31`. |
|
| 177. |
If `a_n=n/((n+1)!)` then find `sum_(n=1)^50 a_n`A. `(50!-1)/(50!)`B. `(51!-1)/(51!)`C. `(1)/(2(n-1)!)`D. none of these |
|
Answer» We have, `a_(n)+(n)/((n+1)!)` `impliesa_(n)=((n+1)-1)/((n+1)!)` `impliesa_(n)=(1)/(n!)-(1)/((n+1)!)` `:.sum_(n=1)^(50)a_(n)=sum_(n=1)^(50){(1)/(n!)-(1)/(((n+1)!))}` `impliessum_(n=1)^(50)a_(n)=((1)/(1!)-(1)/(2!))+((1)/(2!)-(1)/(3!))+((1)/(3!)-(1)/(4!))+.....+((1)/(50!)-(1)/(51!))` `impliessum_(n=1)^(50)a_(n)=1-(1)/(51!)=(51!-1)/(51!)` |
|
| 178. |
How many unordered pairs (a b) of positive integers a and b are there such that lcm (a, b) = 1,26,000? (Note: An unordered pair fa, b) means a, b) {b, a |
|
Answer» LCM(126000)=`10^3*2*7*9` `=2^4*3^2*5^3*7` Patterns=(1*(2)*(1*4)*(1*3)*(1*5) =2*4*3*5=120. |
|
| 179. |
The number of zeros at the end of 100!, isA. 16B. 5C. 7D. 70 |
|
Answer» We have, `70!""=2^(a)xx3^(b)xx5^(c)xx7^(d)xx.....` Now, `c=E_(2)(70!)=[(70)/(2)]+[(70)/(2^(2))]+[(70)/(2^(4))]+[(70)/(2^(5))]+[(70)/(2^(6))]` `impliesa=35+17+8+4+2+1=67` and, `c=E_(5)(70!)=[(70)/(2)]+[(70)/(5^(2))]=14+2=16` `:.70!""=2^(67)xx5^(16)xx3^(b)xx7^(d)xx......` `implies70!""=(2xx5)^(16)xx2^(51)xx3^(b)xx7^(d)xx......` `implies70!""=10^(16)xx2^(51)xx3^(b)xx7^(d)xx......` Thus, the number of zeros at the end of 70! is 16. |
|
| 180. |
Find the remainder when `sum_(r=1)^(n)r!` is divided by 15, if `n ge5`. |
|
Answer» Let `N=underset(r=1)overset(n)(sum)r!=1!+2!+3!+4!+5!+6!+7!+ . . .+n!` `=(1!+2!+3!+4!)+(5!+6!+7!+ . . .+n!)` `=33+(5!+6!+7!+ . . .+n!)` `implies(N)/(15)=(33)/(15)+((5!+6!+7!+ . . .+n!))/(15)` `=2+(3)/(15)+` interger [as `5!,6!`, . . . are divisible by 15] `=(3)/(15)+`Integer Hence, remainder is 3. |
|
| 181. |
Prove that 33! Is divisible by `2^(19)` and what is the largest integer n such that 33! Is divisible by `2^(n)`? |
|
Answer» In terms of prime factors, 33! Can be written as `2^(a)*3^(b)*7^(d)` . . . Now, `E_(2)(33!)=[(33)/(2)]+[(33)/(2^(2))]+[(33)/(2^(3))]=[(33)/(2^(4))]+[(33)/(2^(5))]` + . . . `=16+8+4+2+1+0` . . =31 Hence, the exponent of 2 in 33!. now , 33! is divisible by `2^(31)` which is also divisible by `2^(19)` `therefore`Largest value of n in 31. |
|
| 182. |
Find the exponent of 3 in 100!. |
|
Answer» In terms of prime factors 100! Can be written as `2^(a)*3^(b)*5^(c)*7^(d)` Now, `b=E_(3)(100!)` `=[(100)/(3)]+[(100)/(3^(4))]+[(100)/(3^(3))]+[(100)/(3^(4))]` + . . . `=33+11+3+1+0+ . . .=48` Hence, exponent of 3 is 48. |
|
| 183. |
If `a_n = n (n!)`, then `sum_(r=1)^100 a_r` is equal toA. 101!B. 100!-1C. 101!-1D. 101!+1 |
|
Answer» We have, `sum_(r=1)^(100)a_(r)=sum_(r=1)^(100)r(r!)=sum_(r=1)^(100){(r+1)!-1}r!` `impliessum_(r=1)^(100)a_(r)=sum_(r=1)^(100){(r+1)!-r!}` `impliessum_(r=1)^(100)a_(r)=(2!-1!)+(3!-2!)+(4!-3!)+......+(101!-100!)` `impliessum_(r=1)^(100)a_(r)=101!-1` |
|
| 184. |
If `l=`LCM of `8!, 10!` and `12!` and `h =` HCF of of `8!, 10!` and `12!` then `l/h` is equal toA. 132B. 11800C. 11880D. None of these |
|
Answer» Correct Answer - C `l=LCM` of 8!,10! And 12!=12! and h=HCF of 8!,10! And 12!=8! `therefore(l)/(h)=(12!)/(8!)=12*11*10*9=11880` |
|
| 185. |
1.1!+2.2!+3.3!+.......n.n!` is equal toA. `(n+1)!`B. `(n+1)!+1`C. `(n+1)!-1`D. none of these |
|
Answer» Correct Answer - C We have, `1.1!+2.2!+3.3!+....+n.n!` `=sum_(r=1)^(n)r(r!)` `=sum_(r=1)^(n)[(r+1)r!-r!]` `=sum_(r=1)^(n)[(r+1)!-r!]` `=(2!-1!)+(3!-2!)+....+[(n+1)!-n!]` `(n+1)!-1!""(n+1)!-1` |
|
| 186. |
If `102!=2^(alpha)*3^(beta)*5^(gamma)*7^(delta)`…, thenA. `alpha=98`B. `beta=2gamma+1`C. `alpha=2beta`D. `2gamma=3delta` |
|
Answer» Correct Answer - A::B::C::D `becauseE_(2)(102!)=98,E_(3)(102!)=49`, `E_(5)(102!)=24 and E_(7)(102!)=16` `therefore alpha=98,beta=49,gamma=24 and delta=16` |
|
| 187. |
The number of ways of choosing triplet `(x , y ,z)`such that `zgeqmax{x, y}a n dx ,y ,z in {1,2, n ,n+1}`isa. `^n+1C_3+^(n+2)C_3`b. `n(n+1)(2n+1)//6`c. `1^2+2^2++n^2`d. `2((^(n+2)C_3))_(-^(n+2))C_2`A. `.^(n+1)C_(3)+.^(n+2)C_(3)`B. `(n(n+1)(2n+1))/(6)`C. `1^(2)+2^(2)+3^(2)+ . . .+n^(2)`D. `2(.^(n+2)C_(3))-.^(n+1)C_(2)` |
|
Answer» Correct Answer - A::B::C::D Triplets with (i) x=y`ltz` (ii) `x lt y lt z` (iii) `y lt x lt z` can be chosen in `.^(n+1)C_(2),.^(n+1)C_(3),.^(n+1)C_(3)` ways. `therefore.^(n+1)C_(2)+.^(n+1)C_(3)+.^(n+1)C_(3)=.^(n+2)C_(3)+.^(n+1)C_(3)` `=2(.^(n+2)C_(3))-.^(n+1)C_(2)` `=(n(n-1)(2n+1))/(6)` |
|
| 188. |
Sum of the series `sum_(r=1)^n (r^2+1)r!,` isA. `(n+1)!`B. `(n+2)!-1`C. `n(n+1)!`D. none of these |
|
Answer» We have, `sum_(r=1)^(n)(r^(2)+1)r!` `sum_(r=1)^(n){(r+2)(r+1)-3(r+1)+2}r!` `sum_(r=1)^(n){(r+2)!-(r+1)!}-2sum_(r=1)^(n){(r+1)!-r!}` `=(n+2)!-2!-2{(n +1)!-1}=n(n+1)!` |
|
| 189. |
Find the exponent of 3 in 100!A. 47B. 48C. 49D. 46 |
|
Answer» Let `E_(p)(n)` denote the exponent of p in n. Then, `E_(p)(n!)=[(n)/(p)]+[(n)/(p^(2))]+....+[(n)/(p^(s))]` where s is the largest positive integer such that `p^(s)lenltp^(s+1)` Here, n=100, p=3 `:.3^(4)lt100lt3^(5)` `:.s=4` So, `E_(3)(100!)=[(100)/(3)]+[(100)/(3^(2))]+[(100)/(3^(3))]+[(100)/(3^(4))]` `=33+11+3+1=48` Hence, the exponent of 3 in 100! is 48. |
|
| 190. |
Find the numbers of positive integers from 1 to 1000, which aredivisible by at least 2, 3, or 5. |
|
Answer» Let `A_(k)` be the set of positive integers from 1 to 1000, which is divisible by k. obviously, we have to find `n(A_(2) cup A_(3) cup A_(5))`. If `[*]` denotes the greatest integer function, then `n(A_(2))=[1000/2]=[500]=500` `n(A_(3))=[1000/3]=[333.33]=333` `n(A_(5))=[1000/5]=[200]=200` `n(A_(2) cap A_(3))=[1000/6]=[166.67]=166` `n(A_(3)capA_(5))][1000/15]=[66.67]=66` `n(A_(2)capA_(5))][1000/10][100]=100` and `n(A_(2)capA_(3)capA_(5))=[1000/30]=[33.33]=33` From principal of inclusion and exclusion `n(A_(2)cupA_(3)cupA_(5))=n(A_(2))+n(A_(3))+n(A_(5))-n(A_(2)capA_(3))` `-n(A_(3)capA_(5))-n(A_(2)capA_(5))+n(A_(2)capA_(3)capA_(5))` `=500+333+200-166-66-100+33=734` Hence, the number of positive integers from 1 to 1000, which are divisible by atleast 2,3 or 5 734. |
|
| 191. |
If m and n are positive integers more than or equal to 2, `mgtn`, then (mn)! is divisible byA. `(m!)^(n),(n!)^(m)and(m+n)!` but not by (m-n)!B. `(m+n)!,(m-n)!,(m!)^(n)` but `(n!)^(m)`C. `(m!)^(n),(n!)^(m),(m+n)!and(m-n)!`D. `(m!)^(n)and(n!)^(m)` but not by (m+n)!and(m-n)! |
|
Answer» Correct Answer - C |
|
| 192. |
the exponent of `15` in `100!,` isA. 48B. 24C. 25D. 23 |
|
Answer» We have, `15=3xx5` Now, `E_(3)(100!)=48` [See illustration 1] and `E_(5)(100!)=[(100)/(5)]+[(100)/(5^(2))]=20+4=24` `:.` Exponent of 15 in 100!=min (24,48) =24. |
|
| 193. |
The number `24!` is divisible byA. `6^(24)`B. `24^(6)`C. `12^(12)`D. `48^(5)` |
| Answer» Correct Answer - B | |
| 194. |
If `15! =2^oo* 3^beta*5^gamma*7^delta*11^theta * 13^phi`, then the value of `oo-beta+gamma-delta+theta-phi`, isA. 4B. 6C. 8D. 10 |
| Answer» Correct Answer - B | |
| 195. |
Number of zeros in the expansion of `100!` isA. 22B. 23C. 24D. 25 |
|
Answer» In terms of prime factors 100! Can be written as `2^(a)3^(b)5^(c)7^(d).......` Now, `E_(2)(100!)=[(100)/(2)]+[(100)/(2^(2))]+[(100)/2^(3)]+[(100)/(2^(4))]+[(100)/(2^(5))]+[(100)/(2^(6))]` `impliesE_(2)(100!)=50+25+12+6+3+1=97` and, `E_(5)(100!)=[(100)/(5)]+[(100)/(5^(2))]=20+4=24` `:.100!""=2^(97)xx3^(6)xx5^(24)xx7^(d)xx......` `implies100!""=2^(73)xx(2xx5)^(24)xx7^(d)xx....` `implies100!""=10^(24)xx2^(73)xx3^(b)xx7^(d)xx....` Thus, the number of zeros at the end of 100! is 24. |
|
| 196. |
The last non-zero number at the end of 20! is ?A. 2B. 4C. 6D. 8 |
| Answer» Correct Answer - B | |
| 197. |
The number of naughts standing at the end of 125! IsA. 29B. 30C. 31D. 32 |
| Answer» Correct Answer - C | |
| 198. |
There are `(n+1)`white and `(n+1)`blackballs, each set numbered `1ton+1.`The numberof ways in which the balls can be arranged in a row so that the adjacentballs are of different colors isa. `(2n+2)!`b. `(2n+2)!xx2`c. `(n+1)!xx2`d. `2{(n+1)!}^2`A. `(2n+2)!`B. `(2n+2)!xx2`C. `(n+1)!xx2`D. `2{(n+1)!}^(2)` |
|
Answer» Correct Answer - D |
|
| 199. |
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. 3600B. 3720C. 3800D. 3600 |
|
Answer» Correct Answer - B Possible number of choosing green dyes `= 2^(5)` Possible number of choosing blue dyes `= 2^(4)` Possible number of choosing red dyes `= 2^(3)` If atleast one blue and one green dyes are selected. Then, total number of selection `= (2^(5) - 1) (2^(4) - 1) xx 2^(3) = 3720` |
|
| 200. |
There are 2 identical white balls, 3 identical red balls and 4 green balls of different shades. The number of ways in which they can be arranged in a row so that atleast one ball is separated from the balls of the same colour, is : |
|
Answer» Number of ways balls are arranged in a row so that at least one ball is separated from the balls of same colors, = Total Number of ways - All identical balls together Total number of ways balls can be arranged `n(S) = (9!)/(2!3!)` `n(S) = (9!)/12 = (9**8**7!)/12 = 6 (7!)` The number of ways balls are arranged so that identical balls are together `n(T) = 3!4!` So, the number of ways balls are arranged so that at least one ball is separated ` =n(S) - n(T) = 6 (7!) -3!4! = 6 (7!) -64! = 6(7!-4!)` |
|