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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.
| 101. |
Find the number of permutation of all the letters of the wordMATHEMATICS which starts with consonants only. |
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Answer» We have letters (M M), (A A), (T T),H,E,I,C,S Number of words starting with M to T `=(10!)/(2!xx2!)` `(because` Two pairs of identical letters are available) Number of words starting with T or C or S=`(10!)/(2!xx2!xx2!)` `(because` Three pairs of identical letters are available) Therefore, total number of words `=2xx(10!)/(2!xx2!)+3xx(10!)/(2!xx2!xx2!)` `=(7xx10!)/(2!xx2!xx2!)` |
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| 102. |
Find the total number of ways of answering five objective typequestions, each question having four choices |
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Answer» Correct Answer - `4^(5)` Since each question can be answered in 4 ways, the total number of ways of answering 5 question is `4xx4xx4xx4xx4=4^(5)`. |
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| 103. |
Eleven animals of a circus have to be placed in eleven cages (one in eachcage), if 4 of the cages are too small for 6 of the animals, then find thenumber of the ways of caging all the animals. |
| Answer» Let the 6 animals be placed in 7 of larger cages. This can be done in `.^(7)P_(6)` ways. In each of these ways, one larger cage is left vacant. The remaining five animals can be placed in the remaining five cages in 5! Ways. Hence, by the fundamental theorem, the required number of ways is `.^(7)P_(6)xx5!=604800`. | |
| 104. |
If `A={x|x`is prime number and `x |
| Answer» Here, A={2,3,5,7,11,13,17,19,23,29}. A rational number is made by taking any two numbers in any order. Therefore, the required number of rational numbers is `.^(10)P_(2)+1` (including 1). | |
| 105. |
`116` people participated in a knockout tennis tournament. The players are paired up in the first round, the winners of the first round are paired up in the second round, and so on till the final is played between two players. If after any round, there is odd number of players, one player is given a by, i.e. he skips that round and plays the next round with the winners. The total number of matches played in the tournment isA. `115`B. `53`C. `232`D. `116` |
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Answer» Correct Answer - A `(a)` Since one player emerges the winner, each of the remaining `115` players loses in some round. So `115` matches are played. Alternatively , `58+29+14+7+4+2+1=115` |
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| 106. |
The number of three-digit numbers having only two consecutive digits identical isA. `153`B. `162`C. `180`D. `161` |
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Answer» Correct Answer - B `(b)` First two digits are identical. `x x__to" Number of ways"=9xx9=81` `__x xto"Number of ways"=9xx8=72` (when last two digits are non-zero) `__0 0 to "Number of ways"=9` (when last two digits are zero) |
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| 107. |
In how many ways can 6 persons stand in a queue? |
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Answer» The number of ways in which 6 persons can stand in a queue is sam as filling 6 places with 6 persons. The number of permutations of 6 objects taken all at a time is ` .^(6)P_(6)=6!=720` |
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| 108. |
What is the total number of combination of n different things taken 1, 2, 3,...,n at a time ?A. `2^(n +1)`B. `2^(2n + 1)`C. `2^(n -1)`D. `2^(n -1)` |
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Answer» Correct Answer - D Since, combinations of taking 1, 2, 3, ..., n things at a times are `.^(n)C_(1), .^(n)C_(2), ...., .^(n)C_(n)` `:.` Total number of combinations `= .^(n)C_(1) + .^(n)C_(2) + ....+ .^(n)C_(n)` `= 1 + .^(n)C_(1) + .^(n)C_(2) +....+ .^(n)C_(n) - 1` `= 2^(n) -1` |
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| 109. |
In how many ways can a committee consisting of 3 men and 2 women be formed from 7 men and 5 women ?A. 45B. 350C. 700D. 4200 |
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Answer» Correct Answer - B Total no of Men = 7 Total no. of women = 5 Required number of ways `= .^(7)C_(3) xx .^(5)C_(2)` `= (7!)/(3!4!) xx (5!)/(2!3!) = (7 xx 6 xx 5)/(3 xx 2) xx (5 xx 4)/(2)` `= 7 xx 5 xx 10 = 35 xx 10 = 350` |
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| 110. |
Prove that `((n^(2))!)/((n!)^(n))` is a natural number for all n `in` N. |
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Answer» We have to prove that `(n^(2))!` is divisible by `(n!)^(n)`. We know that product of r consecutive integers is divisible by r! Now `(n^(2))!=1xx2xx3xx4xx..xx n^(2)` `=(1xx2xx3xx..xx n)xx` `[(n+1)(n+2)xx.. .. Xx(2n)]xx` `[(2n+1)(2n+2).. .. (3n)]xx` .. .. .. .. `[(n^(2)-(n-1)(n^(2)-n).. .. n^(2)]` Thus `(n^(2))!` consists n groups of product of n consecutive integers. Each group is divisible by n!. So, `(n^(2))!` is divisible by `(n!)^(n)`. |
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| 111. |
Let `y` be an element of the set `A={1,2,3,4,5,6,10,15,30}` and `x_(1)`, `x_(2)`, `x_(3)` be integers such that `x_(1)x_(2)x_(3)=y`, then the number of positive integral solutions of `x_(1)x_(2)x_(3)=y` isA. `81`B. `64`C. `72`D. `90` |
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Answer» Correct Answer - B `(b)` Number of solutions of the equation `x_(1)x_(2)x_(3)=y` is the same as the number of solutions of equation `x_(1)x_(2)x_(3)x_(4)=30=2xx3xx5` where `x_(4)` is dummy variable Now number of solutions `=` number of ways distinct integers `2`, `3` and `5` can be distributed in four boxes `x_(1)`,`x_(2)`,`x_(3)` and `x_(4)=4^(3)=64` |
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| 112. |
The are `8` events that can be schedules in a week, then The total number of ways in which the events can be scheduled isA. `8^(7)`B. `7^(8)`C. `7!`D. `8` |
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Answer» Correct Answer - B `(b)` For `1^(st)` and event there are `7` ways `2^(nd)` event there are `7` ways `8^(th)` event there are `7` ways `implies` Total no of ways `=7^(8)`. |
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| 113. |
The are `8` events that can be schedules in a week, then The total number of ways that the schedule has at least one event in each days of the week isA. `28xx5040`B. `7!8!`C. `7!xx(15!)`D. None of these |
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Answer» Correct Answer - A `(a)` `1+1+1+1+1+2=8` Total number of ways `=(8!)/(2!(1!)^(6)6!)xx7!=28xx5040` |
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| 114. |
The are `8` events that can be schedules in a week, then The total number of ways that these `8` event are scheduled on exactly `6` days of a week isA. `210xx6!`B. `7!xx266`C. `56xx7!`D. `210xx7!` |
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Answer» Correct Answer - B `(b)` `1+1+1+1+1+3=8` `1+1+1+1+2+2=8` Total number of ways `=[(8!)/(3!(1!)^(5)5!)+(8!)/((1!)^(4)(2!)^(2)2!4!)]xx7!` `=266xx7!` |
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| 115. |
Let `theta=(a_(1),a_(2),a_(3),...,a_(n))` be a given arrangement of `n` distinct objects `a_(1),a_(2),a_(3),…,a_(n)`. A derangement of `theta` is an arrangment of these `n` objects in which none of the objects occupies its original position. Let `D_(n)` be the number of derangements of the permutations `theta`. `D_(n)` is equal toA. `(n-1)D_(n-1)+D_(n-2)`B. `D_(n-1)+(n-1)D_(n-2)`C. `n(D_(n-1)+D_(n-2))`D. `(n-1)(D_(n-1)+D_(n-2))` |
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Answer» Correct Answer - D `(d)` For every choice of `r=1,2,3,….(n-1)` when the `n^(th)` object `a_(n)` goes to the `rth` place, there are `D_(n-1)+D_(n-2)` ways of the other `(n-1)` objects `a_(1)`, `a_(2)`, ….,`a_(n-1)` to be deranged. Hence `=D_(n)=(n-1)(D_(n-1)+D_(n-2))` |
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| 116. |
In how many ways, two different natural numbers can be selected, whichless than or equal to 100 and differ by almost 10. |
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Answer» Correct Answer - `.^(100)C_(2)-.^(90)C_(2)` Number of ways in which two numbers can be selected from 100 integers is `.^(100)C_(2)`. Let us find the number of ways by selecting two numbers, if their difference is more than 10. Let the two integers selected be P and Q. P Q x y z Here, x,y,z are number of integers before P, between P and Q, and after Q respectively. Therefore, x+y+z=98. Now, we have to find number of integral solution for `x,z ge 0` and `y ge 10`. `therefore x+y_(1)+z=88, " where" y_(1)ge 0` Then, number of solutions of above equations is `.^(88+3-1)C_(3-1)= .^(90)C_(2)`. Hence, the number of ways where two different natural numbers can be selected, which differ by almost 10 is `. ^(100)C_(2)- .^(90)C_(2)`. |
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| 117. |
The number of increasing function from `f : AtoB` where `A in {a_(1),a_(2),a_(3),a_(4),a_(5),a_(6)}`, `B in {1,2,3,….,9}` such that `a_(i+1) gt a_(i) AA I in N` and `a_(i) ne i` isA. `30`B. `28`C. `24`D. `42` |
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Answer» Correct Answer - B `(b)` If `a_(1)` is mapped with `2`, we have `.^(7)C_(5)` ways of mapping rest of the elements. If `a_(1)` is mapped with `3`, we have `.^(6)C_(5)` ways of mapping rest of the elelments. If `a_(1)` is mapped with `4`, we have `.^(5)C_(5)` ways of mapping rest of the elements. Hence total number of increasing function `=^(7)C_(5)+^(6)C_(5)+^(5)C_(5)=28` |
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| 118. |
What is the number of five-digit numbers formed with 0, 1, 2, 3, 4 without any repetition of digits ?A. 24B. 48C. 96D. 120 |
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Answer» Correct Answer - C To make a 5 digit number, 0 can not come in the bagining. So, it can be filled in 4 ways. Rest of the places can be filled in 4! Ways. So total number of digit formed `= 4 xx 4! = 4 xx 24 = 96` |
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| 119. |
On a railway there are 20 stations. The number of different tickets required in order that it may be possible to travel from every station to every station isA. 40B. 380C. 400D. 420 |
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Answer» Correct Answer - B From each railway station, there are 19 different tickets to be issued. There are 20 railway station So, total number of tickets `= 20 xx 19 = 380` |
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| 120. |
In a network of railways, a small island has 15 stations. Find thenumber of different types of tickets to be printed for each class, if every stations must have tickets for other stations. |
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Answer» For each pair of stations say A and B, two different types of tickets are required from A to B and from B to A. Now, the number of selections of 2 stations from 15 stations `= .^(15)C_(2)` `=2xx15xx7=210` |
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| 121. |
A firm of Chartered Accountants in Bombay has to send 10 clerks to 5 different companies, two clerks in each. Two of the companies are in Bombay and the others are outside. Two of the clerks prefer to work in Bombay while three others prefer to work outside. In how many ways can the assignment be made if the preferences are to be satisfied. |
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Answer» Correct Answer - 5400 We have to select 4 professors for Roorkee and 6 professors for outside. Again, 2 professors prefer Roorkee and 3 outside, so we are left with 5 professors. The number of ways in which two more professors for Roorkee can be selected is ` .^(5)C_(2)=10`. And remaining 3 professors are left for outside. Now, 6 professors outside Roorkee can be allotted to 3 centres in `6!//(2!2!2!3!)xx3!` ways. Now, 4 professors for 2 centers in Roorkee can be allotted in `4!//(2!3!3!)xx2!` ways. hence, the total number of ways is `10xx(6!)/(2!2!2!3!)xx3!xx(4!)/(2!2!2!)xx2!=5400` |
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| 122. |
If ` ^(10)P_r=5040`find the value of `rdot` |
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Answer» `.^(10)P_(r )=5040` `=10xx504` `=10xx9xx8xx7` `=.^(10)P_(4)` `implies r=4` |
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| 123. |
If `""^10 P_r=5040 ,`find the value of `r.`A. 5B. 4C. 6D. None of these |
| Answer» Correct Answer - B | |
| 124. |
The number of times the digit 3 will be writtenn when listing the integers from 1 to 1000, isA. 269B. 308C. 300D. None of these |
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Answer» Correct Answer - C Before 1000 there are one digit, two digits and three digits numbers. Number of times 3 appear in one digit number `= 20 xx 9` Number of times 3 appear in two digit number `= 11 xx 9` Number of times 3 appear in three digit number = 21 Hence total number of times the digit 3 appear while writing the integers from 1 to 1000 `= 180 + 99 + 21 = 300` |
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| 125. |
Find the number of zeros at the end in product `5^6 .6^7 .7^8 .8^9 .9^(10) .30^(31)`. |
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Answer» Given product `P=5^(6).6^(7).7^(8)..(30)^(31)` `=[(5)^(6)(10)^(11)(15)^(16)(20)^(21)(25)^(26)(30)^(31)]lamda " " ("where" lamda " is an integer")` `therefore` Exponent of 5 in P=6+11+16+21+2(26)+31=137 `therefore` Number of zeros =137 |
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| 126. |
Find the sum of the series `(sum_(r=1)^(n) rxxr !)` |
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Answer» Here, the general term of the series is `T_(r )=r xxr! = (r+1-r)r!` =(r+1)r!-r! =(r+1)!-r! Hence, `T_(1)=2!-1!` `T_(2)=3!-2!` `T_(3)=4!-3!` `T_(n)=(n+1)!-n!` Adding all the above terms, we have the sum of n terms, i.e. `S_(n)=(n+1)!=1` |
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| 127. |
Find the number of zeros at the end of 130. |
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Answer» The number of zeros at the end of 130! Is equal to the exponent of 10 in 130. Now, exponent of 10 is equal to exponent of 5 as exponent of 2 is higher than exponent of 5. Now, exponent of 5 is `[(130)/(5)]+[(130)/(5^(2))]+[(130)/(5^(3))]=26+5+1=32` Also, exponent of 10 is 32, hence, there are 32 zeros at the end of 130. It should be noted that exponent of 2 is `[(130)/(2)]+[(130)/(2^(2))]+[(130)/(2^(3))]+[(130)/(2^(4))]+[(130)/(2^(5))]+[(130)/(2^(6))]+[(130)/(2^(7))]` =65+32+16+8+4+2+1=128 Hence, exponent of 10 is equal to exponent of 5. |
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| 128. |
Find the exponent of 3 in 100! |
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Answer» `100!=1xx2xx3xx..xx98xx99xx100` `=(1xx2xx4xx5xx..xx98xx100)xx(3xx6xx9xx..xx96xx99)` `=Kxx3^(33)(1xx2xx3xx.xx32xx33)` `=Kxx3^(33)(1xx2xx4xx.xx31xx32)(3xx9xx12xx.xx30xx33)` `=[K(1xx2xx4xx..xx31xx32)]xx3^(33)xx(3xx9xx12xx..xx30xx33)` `=K_(1)xx3^(33)xx3^(11)(1xx2xx3xx..xx10xx11)` `=K_(1)xx(1xx2xx4xx..xx10xx11)3^(33)xx3^(11)(3xx6xx9)` `=K_(2)xx3^(33)xx3^(11)xx3^(3)xx3` `=K_(3)xx3^(48)` Hence, exponent of 3 is 48. Alternate solution : Exponent of 3 in 100! is `[(100)/(3)]+[(100)/(3^(2))]+[(100)/(3^(2))]+[(100)/(3^(4))]=33+11+3+1=48` |
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| 129. |
What is the number of signals that can be sent by 6 flage of different colour taking one or more at a time ?A. 21B. 63C. 720D. 1956 |
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Answer» Correct Answer - B Required number of ways `6_(C_(0)) + 6_(C_(1)) + 6_(C_(2)) + ...+ .^(6)C_(6) -1 = 2^(6) -1 = 64 - 1 = 63` |
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| 130. |
Prove that (mn)! Is divisible by `(n!)^(m) " and" (m!)^(n)`. |
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Answer» Number of ways of distribution of (mn) distinct objects equally among n persons `=((mn)!)/((m!)^(n)n!)xxn!=((mn)!)/((m!)^(n))`. Obviously, this value is integer. So, (mn)! Is divisible by `(m!)^(n)` Similarly, number of ways of distribution of (mn) objects equally among m persons `=((mn)!)/((n!)^(m)m!)xxm!=((mn)!)/((n!)^(m))` So, (mn)! is also divisible by `(n!)^(m)`. |
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| 131. |
`5` different objects are to be distributed among `3` persons such that no two persons get the same number of objects. Number of ways this can be done is,A. `60`B. `90`C. `120`D. `150` |
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Answer» Correct Answer - B `(b)` Division of objects can be `(0,1,4)` or `(0,2,3)` So total number of distribution ways `=(5!)/(0!*1!*4!)xx3!+(5!)/(0!*2!*3!)xx3!` `=30+60=90` ways |
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| 132. |
Find the number of ways in which `n`different prizes can be distributed among `m( |
| Answer» The total number of ways is `m xx m xx`.. N times `=m^(n)`. The number of ways in which one gets all the prizes is m. Therefore the required number of ways is `m^(n)-m`. | |
| 133. |
If `r lt s le n " then prove that " ^(n)P_(s) " is divisible by "^(n)P_(r).` |
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Answer» Let s=r+k where `0 le k le s -r`. Then, `.^(n)P_(s)=(n!)/((n-s)!)` `=n(n-1)(n-2)..(n-(s-1))` `=n(n-1)(n-2)..(n-(r+k-1))` `=n(n-1)(n-2)..(n-(r-1))(n-r)(n-(r+1))..(n-(r+k-1))` `={n(n-1)(n-2)..n-(r-1)}{(n-r)(n-(r+1))..(n-(r+k-1))}` `= .^(n)P_(r ){(n-r)(n-(r+1))..(n-(r+k-1))}` `= . ^(n)P_(r )xx " Integer"` Hence, `.^(n)P_(s)` is divisible by `.^(n)P_(r )` |
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| 134. |
(a) If `.^(22)P_(r+1):^(20)P_(r+2)=11 : 52`, find r. (b) If `.^(56)P_(r+6):^(54)P_(r+3)=30800: 1`, find r. |
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Answer» Correct Answer - 41 Given, `.^(22)P_(r+1):^(20)P_(r+2)=11:52` or `(22!)/((21-r)!):(20!)/((18-r))=11:52` or `(22!)/((21-r)!)xx((18-r)!)/(20!)=(11)/(52)` or `(22xx21xx20)/((21-r)(20-r)(19-r))=(11)/(52)` or `(21-r)(20-r)(19-r)=2xx21xx52` or `(21-r)(20-r)(19-r)=2xx3xx7xx4xx13` or r=7 (b) `.^(56)P_(r+6): .^(34)P_(r+3)=30800:1` or `(56!(51-r)!)/((50-r)!54!)=30800` or `56xx55xx(51-r)=30800` or r=41 |
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| 135. |
Evaluate `.^(7)P_(4)`. |
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Answer» `.^(7)P_(4) = (7!)/((7-4)!)` `=(7!)/(3!)` `=(7.6.5.4.3!)/(3!)` `= 840`. |
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| 136. |
Prove that: `((2n)!)/(n !)={1. 3. 5 (2n-1)}2^ndot` |
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Answer» `L.H.S = ((2n)!)/(n!)` `= (1.2.3.4.5.6..(2n-1).(2n))/(n!)` `= ({2.4.6...(2n)}.{1.3.5....(2n-1)})/(n!)` `= (2^(n){1.2.3...n}{1.3.5...(2n-1)})/(n!)` `= 2^(n) {1.3.5..(2n-1)}` `= R.H.S` Hence Proved. |
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| 137. |
A fair coin is tossed `n` times. Let `a_(n)` denotes the number of cases in which no two heads occur consecutively. Then which of the following is not true ?A. `a_(1)=2`B. `a_(2)=3`C. `a_(5)=13`D. `a_(8)55` |
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Answer» Correct Answer - C `(c )` The cases for `a_(1){H,T}` i.e, `a_(1)=2` The cases for `a_(2){HT,TH,TT}`, `a_(2)=3` For `n ge 3` , if the first outcome is `H`, then next just `T` and then `a_(n-2)`. If the first out come is `T`, then `a_(n-1)` should follow. So, `a_(n)=1xx1xxa_(n-2)+1xxa_(n-1)impliesa_(n)=a_(n-2)+a_(n-1)` So, `a_(3)=a_(1)+a_(2)=5`, `a_(4)=3+5=8` and so on. |
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| 138. |
In how many ways the number 7056 can be resolved as a product of 2factors. |
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Answer» Correct Answer - 23 `7056=2^(4)xx3^(2)xx7^(2)` Number of divisors `=(4+1)(2+1)(2+1)=5xx3xx3=45` Number of ways of resolving into 2 factors `=(45+1)/(2)=23` |
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| 139. |
Number of ways in which 7 green bottles and 8 blue bottles can be arranged in a row if exactly 1 pair of green bottles is side by side, is (Assume all bottles to be alike except for the colour).A. `84`B. `360`C. `504`D. none of these |
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Answer» Correct Answer - C `(c )` First arrange `8` blue alike bottles `to ` number of ways `=1` Now select one gap out of `9` gaps created to put two green bottles `to `number of ways `=.^(9)C_(1)` Now select `5` more gaps for other green bottles from remaining `8` gaps `to` number of ways `=.^(8)C_(5)` Hence, total number of ways `=.(9)C_(1)*^(8)C_(5)=9*(8*7*6)/(1*2*3)=504` |
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| 140. |
The interior angles of a regular polygon measure `150^@` each. The number of diagonals of the polygon isA. `35`B. `44`C. `54`D. `78` |
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Answer» Correct Answer - C `(c )` We have `pi-(2pi)/(n)=(5pi)/(6)` `implies (pi)/(6)=(2pi)/(n)` `impliesn=12` `:.` Number of diagonals `="^(10)C_(2)-12=54` |
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| 141. |
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» There will be `2I,2N,1A,1E,1X,1M, 1T,1O` in the words starting with A. No. of words starting from `A = (10!)/(2!2!) = 907200` After if, there will be words starting with `E, I,M, N,O,T,X`. Therefore, the total no. of words before the words starting with E `= 907200` |
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| 142. |
There are 3 books of mathematics, 4 of science, and 5 of literature.How many different collections can be made such that each collection consistsofone book of each subject,at least one book of each subject,at least one book of literature. |
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Answer» (i) One book of each subject is selected. So, number of ways of selections `=.^(3)C_(1)xx .^(4)C_(1)xx .^(5)C_(1)` `=3xx4xx5=60` (ii) At least one book of each subject is selected. Number of ways of ways of selection of at least one mathematics book `=3^(3)-1=7` Number of ways of selection of at least one science book `=2^(4)-1=15` Number of ways of selection of at least one literature book `=2^(5)-1=31` So, total number of ways of selection `=7xx15xx31=3255` (iii) At least one book of literature is selected. So, we can select any number of mathematics and science books including zero selection. So, number of ways of selection `=2^(3)xx2^(4)xx(2^(5)-1)` `=128xx31=3968` |
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| 143. |
A dice is rooled n times. Find the number of outcomes (i) if 6 never appear. (ii) if 6 appears at least once. (iii) if only even number appears. |
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Answer» When a dice is rolled n times, number of cases are `6xx6xx6`..n times`=6^(n)`. (i) if 6 never appear, then for each trial there are five possibilities (1,2,3,4,5). Therefore, number of outcomes for n trials will be `5xx5xx5xx`.. N times`=5^(n)` (ii) number of cases if 6 appears at least once =total number of cases-number of cases in which 6 never appear `=6^(n)-5^(n)` (iii) if only even number appear, then for each trial there are three possibilities 2,4 or 6. So, number of outcomes for n trials will be `3xx3xx3`.. times `=3^(n)` possibilities. |
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| 144. |
In how many ways 10 different balls can be put in 2 difference boxes ? |
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Answer» We have two boxes P and Q and ten balls. Now for each ball there are two options either it is put in box P or put in box Q. So, different number of ways in which balls are put in the boxes = total number of options of all balls `=2xx2xx2`.. 10 times `=2^(10)`. |
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| 145. |
Find the number of ways in which four distinct balls can be kept into two identical boxes so that no box remains empty. |
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Answer» Correct Answer - 7 4 distinct balls can be divided into two nonempty groups as 1,3 or 2,2 Sine boxes are identical, number of ways of division and distibution are same `therefore` Number of ways `=(4!)/(1!3!)+(4)/(2!2!2!)=4+3=7`. |
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| 146. |
Find the number of ways in which 5 distinct balls can be distributed inthree different boxes if no box remains empty.OrIf `n(A)=5a n dn(B)=3,`then find the number of ontofunctions from A to B. |
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Answer» Using above formula, the number of ways is `3^(5)- ""^(3)C_(1)(3-1)^(5)+ ""^(3)C_(2)(3-2)^(5)=243-96+3=150` |
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| 147. |
If the letters of the word BAZAR are arranged in dictionary order, then what is the 50th word ?A. ZAABRB. ZBAARC. ZBRAAD. ZAARB |
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Answer» Correct Answer - D With A at first place, rest 4 places will be arranged in 4! Ways so, Number of words begin with A = `4! = 24` Similarly with B at first place, Number of words begin with `B = (4!)/(2!) = 12` [As there are two `A_(s)`] Number of words begin with `R = (4!)/(2!) = 12` Thus, 48 words have starting letter A, B, and R. So, 49th word will be ZAABR and 50th word will be ZAARB |
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| 148. |
There are ten points in the plane, no three of which are coolinear. How many different lines can be drawn through these points ? |
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Answer» There are ten points `P_(1), P_(2),.., P_(10)`. For one line two points are required. Through point `P_(1)` there will be 9 lines when `P_(1)` is joined with any of the nine other points. Similarly, there will be nine lines passing through each point. So, number of lines is `9xx10` or 90. But there is double counting in above answer. Why ? One of the nine lines passing through point `P_(1) " is" P_(1) P_(2). "But" P_(1) P_(2)` is also one of the lines passing through point `P_(2)`. Thus, line `P_(1)P_(2)` and similarly each line is counted twice. Therefore, actual number of lines is `(90)/(2)=45`. |
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| 149. |
Find the total number of integer `n`such that `2lt=nlt=2000`and H.C.F. of `n`and 36 is 1. |
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Answer» `36=2^(2)xx3^(2)` If H.C.F. of integer n and 36 is 1, then n should not be divisible by 2 or 3. Let us first find the numbers that are divisble by 2 or 3. The number of integers in the range [2, 2000] that are divisible by 2 is 1000 (2,4,6,.., 1998, 2000). The number of integers in the range [2, 2000] that are divisible by 3 is 666 (3,6, 9, .., 1995, 1998). The number of integers in the range [2, 2000] that are divisible by 6 is 333 (6, 12, 18,.., 1992, 1998). Total number of integers divisible by 2 or 3 is 1000+666-333=1333. Thus, the total number of integers that are divisible by neither 2 nor 3 is 1999-1333=666. |
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| 150. |
There are `n`locks and `n`matching keys. If all the locks and keys are to be perfectly matched, find the maximumnumber of trails required to open a lock. |
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Answer» The maximum number of trials needed for the first key is n. For the second key, it will be n-1. Now, for the rth key, the maximum number of trials needed is n-r+1. Thus, the required answer is `n+(n-1)+..+1=(n(n+1))/(2)` |
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