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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.
| 1. |
In an examination, the maximum mark for each of the three papers is 50and the maximum mark for the fourth paper is 100. Find the number of ways inwhich the candidate can score 605 marks in aggregate. |
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Answer» Let the marks scored by the candidate in four papers be `x_(1),x_(2),x_(3) " and " x_(4)`. Then `x_(1)+x_(2)+x_(3)+x_(4)=150 (60% " of " 250=150)` Here, `0 le x_(1),x_(2),x_(3) le 50 " and " 0 le x_(4) le 100`. `therefore` Number of ways candidate can score 60 % marks = coefficient of `p^(150) " in " (1+p+p^(2)+..+p^(50))^(3)(1+p+p^(2)+..+p^(100))` (In each bracket series is not extended to infinite terms as upper limit of each variable is less than 150) =coefficient of `p^(150) " in " ((1-p^(51))/(1-p))^(3)((1-p^(101))/(1-p))` =coefficient of `p^(150) " in " (1-p^(51))^(3)(1-p^(101))(1-p)^(-4)` =coefficient of `p^(150) " in " (1-3p^(51)+3p^(102)-p^(101))xx(1+ ""^(4)C_(1)p+ ""^(5)C_(2)p^(2)+ ""^(6)C_(3)p^(3)++..)` `=""^(153)C_(3)-3xx""^(102)C_(3)+3xx ""^(51)C_(3)- ""^(52)C_(3)` |
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| 2. |
From 6 programmers and 4 typists, an office wants to recruit 5 people. What is the number of ways this can be done so as to recruit at least one typist ?A. 209B. 210C. 246D. 242 |
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Answer» Correct Answer - C Number of ways `.^(4)C_(1) .^(6)C_(4) + .^(4)C_(2) .^(6)C_(3) + .^(4)C_(3) .^(6)C_(2) + .^(4)C_(4) .^(6)C_(1)` `= (4) (15) + (6) (20) + (4) (15) + (1) (6)` `= 60 + 120 + 60 + 6 = 246` |
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| 3. |
In how many different ways can 3 persons A, B, C having 6 one-rupeecoin 7 one-rupee coin, 8 one-rupee coin, respectively, donate 10 one-rupee coin collectively? |
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Answer» Let A, B and C donate `x_(1),x_(2),x_(3)` number of one-rupee coins. `therefore x_(1)+x_(2)+x_(3)=10`, where `0 le x_(1) le 6, 0 le x_(2) le 7, 0 le x_(3) le 8` ` therefore` Required number of ways =coefficient to `p^(10) " in " (1+p+p^(2)+..+p^(6))xx(1+p+p^(2)+..+p^(7))xx(1+p+p^(2)+..+p^(8))` (In each bracket series is not extended to infinite terms as upper limit of each variable is less than 10) =coefficient to `p^(10) " in " ((1-p^(7))/(1-p))((1-p^(8))/(1-p))((1-p^(9))/(1-p))` =coefficient of `p^(10) " in " (1-p^(7))(1-p^(8))(1-p^(9))(1-p)^(-3)` =coefficient of `p^(10) " in " (1-p^(7)-p^(8)-p^(9))xx(1+ ""^(3)C_(1)p+ ""^(4)C_(2)p^(2)+ ""^(5)C_(3)p^(3)+..+ ""^(12)C_(10)p^(10)+..)` `=""^(12)C_(10)- ""^(5)C_(3)- ""^(4)C_(2)- ""^(3)C_(1)` =66-10-6-3 =47 |
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| 4. |
In how many ways te sum of upper faces of four distinct dices can besix. |
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Answer» Let the numbers on the upper faces of four distinct dice be `x_(1),x_(2),x_(3) " and " x_(4)`. The sum of the numbers is 6. `therefore x_(1)+x_(2)+x_(3)+x_(4)=6, " " "where" 1 le x_(1),x_(2),x_(3),x_(4) le 6` `therefore ` Number of ways the sum of the upper faces of dice is six =coefficient of `p^(6) " in" (p+p^(2)+p^(3)+p^(4)+p^(5)+p^(6))^(4)` =coefficient of `p^(6) " in" p^(4)(1+p+p^(2)+p^(3)+p^(4)+p^(5))^(4)` coefficient of `p^(2) " in" (1+p+p^(2)+p^(3)+p^(4)+p^(5))^(4)` =coefficient of `p^(2) " in" (1+p+p^(2)+.. " infinite terms")^(4)` (as terms with higher powers of p are not conisdered while calculating coefficient of `p^(2))` = coefficient of `p^(2) " in" ((1)/(1-p))^(4)` = coefficient of `p^(2) " in " (1-p)^(4)` `= .^(4+2-1)C+_(2)` `= .^(5)C_(2)=10` |
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| 5. |
There are ten points in a plane. Of these ten points, four points are in a straight line and with the exceptionof these four points, on three points are in the same straight line. Find i. the number of triangles formed, ii the number of straight lines formed iii the number of quadrilaterals formed, by joining these ten points.A. 90B. 45C. 40D. 30 |
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Answer» Correct Answer - B A straight line can be formed by joining 2 points `:.` Total number of straight lines `= .^(10)C_(2)` `= (10 xx 9)/(2 xx 1) = 45` |
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| 6. |
In how many ways can we get a sum of at most 17 by throwing six distinct dice ? In how many ways can we get a sum greater than 17 ? |
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Answer» Let `x_(1),x_(2),.., x_(6)` be the number that appears on the six dice. According to the questions, `x_(1)+x_(2)+x_(3)+..+x_(6) le 17, " where " 1 le x_(i) le 6` Now we can remove the inequality sign by introducing one dummy variable `x_(7)`, such that `x_(1)+x_(2)+x_(3)+..+x_(6)+x_(7)=17, " where " x_(7) ge 0` `therefore` No. of solutions of inequality (1) and equation (2) are same `therefore` Required number of solutions, = coefficient of `p^(17) " in " (p+p^(2)+..+p^(6))^(6)(1+p+p^(2)+..)` =coefficient of `p^(17) " in " p^(6) (1+p+..+p^(5))^(6)(1+p+p^(2)+..)` = coefficient of `p^(11) " in " (1+p+..+p^(5))^(6)(1+p+p^(2)+..)` = coefficient of `p^(11) " in " ((1-p^(6))/(1-p))^(6)((1)/(1-p))` = coefficient of `p^(11) " in " (1-6p^(6))(1-p)^(-7)` `= ""^(17)C_(11)-6 ""^(11)C_(5)`. Number of ways in which we can get a sum greater than 17 = total number of cases - number of cases in which the sum is at most 17 `=6^(6)-(""^(17)C_(11)-6^(11)C_(5))` |
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| 7. |
An ordinary cubical dice having six faces markedwith alphabets A, B, C, D, E, and F is thrown `n`times andht list of `n`alphabets showing p are noted. Find the total number ofways in which among the alphabets A, B, C D, E and F only three of themappear in the list. |
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Answer» Three letters from six letters can be selected in `""^(6)C_(3)` ways. Consider one such set of letters {A,B,C}. Now, if each throw of dice shows either letter A, B or C then number of cases is `3^(n)`. But this includes those cases also in which exactly one letter or exactly two letters from A, B and C appear. Number of cases in which exactly one letter appears in all throws is 3. Number of cases in which exactly two letters appear (each at least once) is `""^(3)C_(2)(2^(n)-2)`. So, required number of cases `=""^(6)C_(3)xx[3^(n)- ""^(3)C_(2)(2^(n)-2)-3]` |
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| 8. |
The members of a chess club took part in a roundrobin competition in which each player plays with other once. All membersscored the same number of points, except four juniors whose total score ere17.5. How many members were there in the club? Assume that for each win aplayer scores 1 point, 1/2 for a draw, and zero for losing. |
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Answer» Let the number of members be n. Total number of points is `""^(n)C_(2)`. Therefore, `""^(n)C_(2)-17.5=(n-4)x` (where x is the number of point scored by each player) `implies n(n-1)-35=2(n-4)x` or `2x=(n(n-1)-35)/(n-4)` (where x takes the values 0.5, 1, 1.5, etc.) `=(n^(2)-n-35)/(n-4)` `=(n(n-4)+3(n-4)-23)/(n-4)` `=(n+3)-(23)/(n-4)` `implies (23)/(n-4)` must be an integer `implies n=27` |
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| 9. |
Number of non-empty subsets of {1,2,3,..,12} having the property that sum of the largest and smallest element is 13. |
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Answer» Correct Answer - 1365 Set must contain minimum two elements, such that sum of smallest and largest element is 13. If set contains smallest number 1 and largest number 12 then we can select other elements of subset from {2,3,..,12}. So, number of subsets are `2^(10)`. If set contains smallest number 2 and largest number 11 then we can select other elements of subset from {3,4,..,10}. So, number of subsets are `2^(8)`. Similarly, we have subsets `2^(6),2^(4),2^(2),2^(0)`. So, total number of subsets `1+2^(2)+2^(4)+..+2^(10)=1365` |
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| 10. |
In a class, there are 15 boys and 10 girls. How many ways a teacher can select 1 boy and 1 girl to represent the class at a seminar. |
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Answer» There are 15 boys and 10 girls in a class. Teacher has 15 options to select one boy and 10 options to select one girl. So, using the multiplication rule of counting, Number of ways of selecting one girl and one boy `=15xx10=150` |
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| 11. |
`m`men and `n`women ae to be seated in a row so that no twowomen sit together. If `m > n`then show that the number of ways n which they fan be seated as `(m !(m+1)!)/((m-n+1)!)`. |
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Answer» Correct Answer - `((m+1)!m!)/((m-n+1)!)` m men can be seated in m! ways, creating (m+1) places for n ladies to sit. n ladies in (m+1) places can be arrnanged in `.^(m+1)P_(n)` ways. `therefore` Total ways `=m!xx .^(m+1)P_(n)` `m!xx((m+1)!)/((m+1-n)!)=((m+1)!m!)/((m-n+1)!)` |
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| 12. |
In how many ways can 3 ladies and 3 gentlemen beseated around a round table so that any two and only two of the ladies sittogether? |
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Answer» Correct Answer - 72 The number of selection of two ladies to sit together is `.^(3)C_(2)`. Let the two seats occupied by these two (selected ) ladies be numbered as 1 and 2 and remaining seats by 3,4,5, and 6. The `3^(rd)` lady cannot occupy seat number 3 and 6, so there are only two choices (viz., numbers 4 and 5) left with the `3^(rd)` lady who can make selection of seats in `.^(2)C_(1)` ways. Now, the two selected ladies (in seat numbers 1 and 2) can be permuted (i.e., change their seats ) in 2! ways and 3 gentlemen can be permuted on remaining 3 seats in 3! ways. Hence, by product rule, total number of ways is `.^(3)C_(2)xx .^(2)C_(1)xx2!xx3!=72`. |
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| 13. |
In how many ways can a student choose a programme of 5 courses if9 courses are available and 2 specific courses are compulsory for everystudent? |
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Answer» Since, 2 subjects are compulsory out of 9 subjects, so the student can choose 3 subjects out of remaining 7 subjects. So, total ways `= .^(7)C_(3)`. `= (7!)/(3!4!) = (7 xx 6xx 5 xx 4!)/(3 xx 2xx 1 xx 4!) = 35`. |
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| 14. |
A five-digit number divisible by 3 is to be formed using the digits 0,1,2,3 and 4 without repetition of digits. What is the number of ways this can be done ?A. 96B. 48C. 32D. No number can be formed |
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Answer» Correct Answer - D Since sum of digits = 10 (which is not divisible by 3) `:.` No number can be formed |
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| 15. |
If N is the number of positive integral solutions of `x_1x_2x_3x_4 = 770`, then N =A. `256`B. `729`C. `900`D. `770` |
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Answer» Correct Answer - A `(a)` `770=2xx5xx7xx11` We can assign `2` to `x_(1)`, or `x_(2)` or `x_(3)` or `x_(4)`. Thus `2` can be assigned in `4` ways similarly each of `5,7,11` can be assigned in `4` ways No. of solutions `=4xx4xx4xx4=256` |
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| 16. |
Out of 7 consonants and 4 vowels, words are to be formed by involving 3 consonants and 2 vowels. The number of such words formed is :A. 25200B. 22500C. 10080D. 5040 |
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Answer» Correct Answer - A Number of words `= 5! xx .^(7)C_(3) xx .^(4)C_(2)` `= 120 xx (7!)/(4!3!) xx (4!)/(2!2!) = 25200` |
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| 17. |
The number of positive six-digit integers which are divisible by `9` and four of its digits are `1`, `0`, `0`, `5` isA. `60`B. `120`C. `180`D. `210` |
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Answer» Correct Answer - C `(c )` The sum of digits must be divisible by `9` The pair of missing numbers must be `(1,2)` or `(0,3)` Case I : `(1,2)` `implies` when `1` is first digit the remaining digits `(1,2,0,0,5)` can be arranged in `60` days, when `5` or `2` is first digit then in `30` ways each Case II : `(0,3)` will have `3xx20=60` ways So total number of ways `=60+2xx30+3xx20=180` |
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| 18. |
Find the number of ways in which one can post 5 letters in 7letterboxes.A. `7^(5)`B. `3^(5)`C. `5^(7)`D. 2520 |
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Answer» Correct Answer - A First letter can be put any 7 letters boxes = 7 ways Similarly, 2nd, 3rd, 4th and 5th letters be put in 7 ways each, respectively `:. 7 xx 7 xx 7 xx 7 xx 7 xx 7 - 7^(5)` |
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| 19. |
If `A={a , b , c , d}, B={p , q , r , s}`than which of the following are relations from `A to B ?`Give reasons for your answer.: `R_4={(a , p), (q , a), (b , s), (s , b)}`A. 4096B. 4094C. 128D. 126 |
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Answer» Correct Answer - A `A = {x, y, z}` `B{p, q, r, s}` `n(A) = 3` `n(B) = 4` `:.` Number of distinct relations `= 2^(n(A) xx n(B)) = 2^(3 xx 4) = 2^(12) = 4096` `:.` option (a) is correct. |
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| 20. |
Number of four digit positive integers if the product of their digits is divisible by `3` is.A. `2700`B. `5464`C. `6628`D. `7704` |
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Answer» Correct Answer - D `(d)` Product will be divisible by `3` if atleast one digit is , `0,3,6,9` Now total four-digit no.`=9xx10^(3)`, Number of four-digit no. without `0,3,6,9=6^(4)` Total number of four-digit integers `=9000-1296=7704` |
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| 21. |
In how many ways 11 players can be selected from 15 cricket players ? |
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Answer» No. of ways of selecting 11 players from 15 players `= .^(15)C_(11) = (15!)/(4!11!)` `= (15 xx 14 xx 13 xx 12)/(1xx 2xx 3 xx 4)` `= 1365`. |
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| 22. |
From a group of 15 cricket players, a team of 11players is to bechosen. In how many ways can this be done?A. 364B. 1001C. 1365D. 32760 |
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Answer» Correct Answer - C Number of ways that a cricket team of 11 players can be made out of 15 players `= .^(15)C_(11) = (15!)/(11!4!)` `= (15 xx 14 xx 13 xx 12 xx 11!)/(11! xx 1 xx 2 xx 3 xx 4) = 1365` |
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| 23. |
The number of ways of arranging `6` players to throw the cricket ball so that oldest player may not throw first isA. `120`B. `600`C. `720`D. `7156` |
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Answer» Correct Answer - B `(b)` For the first place `5` players (excluding the oldest) and for the remaining place `5` (including the oldest)players are available. `:.` No of ways `=5xx5xx4xx3xx2xx1=600` |
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| 24. |
Find the number of ways in which 3 distinct numbers can be selected from the set `{3^(1),3^(2),3^(3),..,3^(100),3^(101)}` so that they form a G.P. |
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Answer» Correct Answer - 2500 Let three numbers selected be `3^(a),3^(b),3^(c )` which are in G.P. `therefore (3^(b))^(2)=(3^(a))(3^(c ))` `implies 2b=a+c` `implies a,b,c ` are in A.P. Thus, selecting three number in G.P. from given set is equivalent to selecting 3 numbers from {1,2,3,..,101} which are in A.P. Now, a,b,c are in A.P. if either a and c are odd or a and c are even. Number of ways of selecting two odd numbers is `. ^(51)C_(2)` and those of selecting two even numbers is `.^(50)C_(2)`. Onece a and c are selected, b is fixed. Hence total number of ways `= .^(51)C_(2)+ .^(50)C_(2)=1275+1225=2500` |
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| 25. |
There are `10` different books in a shelf. The number of ways in which three books can be selected so that exactly two of them are consecutive isA. `60`B. `54`C. `56`D. `36` |
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Answer» Correct Answer - C `(c )` `B_(1)B_(2)B_(3)…B_(7)B_(8)B_(9)B_(10)` `(i)` When two terminal books are taken (`B_(1)B_(2)` or `B_(9)B_(10)`) then number of ways `=2xx7=14` `(ii)` When two consecutive terminal books are not taken (i.e.`B_(2)B_(3)`,`B_(3)B_(4)`,……..`B_(8)B_(9)`) Then the third book can be selected in `6` ways Then the number of ways are `7xx6=42` `:.` Total `=14+42=56` ways |
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| 26. |
The number of ways of distributing `3` identical physics books and `3` identical methematics books among three students such that each student gets at least one books isA. `45`B. `55`C. `64`D. `72` |
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Answer» Correct Answer - B `(b)` `n(A)=^(3+2-1)C_(2-1)xx^(3+2-1)C_(2-1)=16` `n(B)=n(C )=16` `n(AnnB)=^(3+2-1)C_(1-1)xx^(3+2-1)C_(1-1)` `=1=n(BnnC)=n(AnnC)` `n(AnnBnnC)=0` Required no. of ways `=100-(16+16+16-1-1-1+0)=55` |
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| 27. |
A shelf contains 20 books of which 4 are single volume and the otherform sets of 8, 5, and 3 volumes. Find the number of ways in which the booksmay be arranged on the shelf so thatvolumes of each set will not be separated.volumes of each set remain in their due order. |
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Answer» (i) Considering each set as single unit, permutations of 7 units is 7!. Permutations of books of the set of 8 volumes among themselves is 8!. Respective permutations of books of the set of 5 volumes is 5! And that of books of 3 volumes is 3!. By the product rule, total number of permutations is 7! 8! 5! 3!. (ii) Since the books in a set of books containing any number of volumes can be arranged in due order in 2 ways, the total number of permutations is `7!xx2xx2xx2=8xx7!=8`. |
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| 28. |
A positive integer `n` is of the form `n=2^(alpha)3^(beta)`, where `alpha ge 1`, `beta ge 1`. If `n` has `12` positive divisors and `2n` has `15` positive divisors, then the number of positive divisors of `3n ` isA. `15`B. `16`C. `18`D. `20` |
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Answer» Correct Answer - B `(b)` `n=2^(alpha)*3^(beta)` No. of divisors `=(alpha+1)(beta+1)=12` `2n=2^(alpha+1)3^(beta)` No.of divisors `=(alpha+2)(beta+1)=15` `implies(alpha+2)/(alpha+1)=(5)/(4)` `implies4alpha+8=5alpha+5impliesalpha=3` `impliesbeta=2implies3n=2^(3)3^(3)` No. of divisors `=(3+1)(3+1)=16` |
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| 29. |
There are two sets of parallel lines, their equations being `x cos alpha+y sin alpha=p` and `x sin alpha- y cos alpha=p` , `p=1,2,3,….n` and `alpha in (0,pi//2)`. If the number of rectangles formed by these two sets of lines is `225`, then the value of `n` is equals toA. `4`B. `5`C. `6`D. `7` |
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Answer» Correct Answer - C `(c )` `"^(n)C_(2)*^(n)C_(2)=225impliesn=6` |
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| 30. |
`m` equi spaced horizontal lines are inersected by `n` equi spaced vertical lines. If the distance between two successive horizontal lines is same as that between two successive vertical lines, then find the number of squares formed by the lines if `(m < n)` |
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Answer» Let set A of m equi spaced horizontal lines be `l_(1),l_(2), l_(3),.., l_(m)` Let set B of n equi spaced vertical lines be `k_(1),k_(2),k_(3),.., k_(n)` Also let the distance between two consecutive lines be 1 unit. For a square of area 1 sq. units, we must select two consecutive horizontal lines and two consecutive vertical lines which can be done in `(m-1)xx(n-1)` ways. For a square of area 4 sq. units, we must select two lines from set A as `(l_(1),l_(3),(l_(2),l_(4)),.., (l_(m-2),l_(m))` and two from set B as `(k_(1),k_(3)),(k_(2),k_(4)),..,(k_(n-2),k_(n))` `therefore` Number of squares `=(m-2)xx(n-2)` Similarly number of squares of area 9 sq. units `=(m-3)xx(n-3)` and so on. Number of squares of area `(m-1)^(2)` sq. units `=(m-(m-1))xx(n-(m-1))` `therefore` Total number of squares `= underset(r=1)overset(m-1) sum (m-r)(n-r)` |
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| 31. |
In ;a plane, there are 5 straight lines which will pass through a givenpoint, 6 others which all pass through another given point, and 7 otherswhich all as through a third given point. Supposing no three lines intersectat any point and no two are parallel, find the number of triangles formed bythe intersection of the straight line. |
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Answer» Let 5 straight lines be passing through A, 6 passing through B, and 7 passing through C. In all, there are 18 straight lines. To find the number of triangles, we have to find the number of selection of 3 lines from these 18 lines, keeping in mind that selection of 3 lines from the lines passing through A, B, or C will not give any triangle. Hence, the required number of triangles is `.^(18)C_(3)-(.^(5)C_(3)+ .^(6)C_(3) + .^(7)C_(3))=751`. |
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| 32. |
If there are six straight lines in a plane, no two of which are parallel and no three of which pass through the same point, then find the number of points in which these lines intersect. |
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Answer» Correct Answer - 15 There are six straight lines `L_(1),L_(2),L_(3),L_(4),L_(5),L_(6)` in a plane. Since no two lines are parallel and no three of which pass through the same point, we have maximum points of intersection. Line `L_(1)` intersect five lines, so there will be five points of intersection. Similarly, for each of the lines `L_(2),L_(3),L_(4),L_(5),L_(6)`, there will be five points of intersection. But there will be double counting . So, total number of points of intersection is `(5xx6)/(2)=15` |
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| 33. |
If 7 points out of 12 are in the same straight line, then the number of triangles formed isA. 84B. 175C. 185D. 201 |
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Answer» Correct Answer - C Number of triangle formed from 12 point `= .^(12)C_(3)` Since 7 parts are collinear, then `.^(7)C_(3)` triangles will not be formed so. ltbr. `= .^(12)C_(3) - .^(7)C_(3)` `= (12!)/(3!9!) - (7!)/(3!4!) = (12.11.10)/(3.2.1) - (7.6.5)/(3.2.1)` `= 220 - 35 = 185` |
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| 34. |
A, B, C, D and E are coplanar points and three of them lie in a straight line. What is the maximum number of triangles that can be drawn with these points as their vertices ?A. 5B. 9C. 10D. 12 |
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Answer» Correct Answer - B Number of triangles using 5 points out of three are on a straight line `= .^(5)C_(3) - .^(3)C_(3) = (5!)/(3!2!) - 1 = (5 xx 4)/(2) - 1` `= 10 - 1 = 9` |
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| 35. |
If 7 points out of 12 are in the same straight line, then the number of triangles formed isA. 185B. 158C. 172D. None of these |
| Answer» Correct Answer - A | |
| 36. |
How many 10-digit numbers can be formed by using digits 1 and 2A. `10^(2)`B. `2^(10)`C. `.^(10)C_(2)`D. None of these |
| Answer» Correct Answer - B | |
| 37. |
If `""^(n)C_(r -1) = 36, ""^(n)C_(r) = 84 " and " ""^(n)C_(r +1) = 126`, then find the value of `""^(r)C_(2)`.A. (8,4)B. (9,3)C. (7,5)D. (6,5) |
| Answer» Correct Answer - B | |
| 38. |
No. of diagonals of a polygon are 170. No. of sides in this polygon are:A. 18B. 20C. 17D. None of these |
| Answer» Correct Answer - B | |
| 39. |
Solve: (i) 8! (ii) 4! - 3! |
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Answer» (i) `8! = 1 xx 2 xx 3 xx 4 xx 5 xx 6 xx 7xx 8` `= 40320` (ii) `4! - 3! = 1 xx 2 xx 3 xx 4 - 1 xx 2 xx 3` `=24 - 6 = 18`. |
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| 40. |
Convert 1.3.5.7.9 into factorial |
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Answer» `1.3.5.7.9 = (1.2.3.4.5.6.7.8.9.10)/(2.4.6.8.10)` `=(10!)/(2^(5).1.2.3.4.5) = (10!)/(2^(5).5!)` |
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| 41. |
Convert 5.6.7.8 into factorial. |
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Answer» 5.6.7.8 `=(1.2.3.4.5.6.7.8)/(1.2.3.4) = (8!)/(4!)` |
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| 42. |
If `(1)/(6!) +(1)/(7!) = (x)/(8!)`, find x. |
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Answer» `(1)/(6!) + (1)/(7!) = (x)/(8!)` `rArr x = 8! ((1)/(6!)+(1)/(7!))= (8!)/(6!) +(8!)/(7!)` `=(8 xx 7 xx 6!)/(6!) +(8 xx 7!)/(7!)` ` = 56 +8 = 64` |
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| 43. |
How many 3-digit numbers can be formed from the digits 1,2,3,4 and 5 assuming that (i) repetition of the digits is allowed? (ii) repetition of the digits is not allowed? |
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Answer» (i) When repetition of digits is allowed: No. of ways of choosing firsy digits = 5 No. of ways of choosing second digit = 5 No. of ways of choosing third digit = 5 Therefore, total possible numbers `= 5 xx 5 xx 5 = 125` (ii) When repetition of digits is not allowed: No. of ways of choosing first digit = 5 No. of ways of choosing second digit = 4 No. of ways of choosing thrid digit = 3 Total possible numbers `= 5 xx 4 xx 3 = 60`. |
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| 44. |
Determine n if(i) `^2n C_2:^n C_2=12 :1`(ii) `^2n C_3:^n C_3=11 :1` |
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Answer» `.^(2n)C_(3): .^(n)C_(2) = 12:1` `rArr ((2n)!)/(3!(2n-3)!) : (n!)/(2!(n-2)) = 12:1` `rArr ((2n)(2n-1)(2n-1))/(6): (n(n-1))/(2) = 12:1` `rArr ((2n)(2n-1)(2n-2))/(6) xx (2)/(n(n-1)) = (12)/(1)` `rArr (2(2n-1).2)/(3) = 12` `rArr 2n - 1 = 9` `rArr n = 5`. |
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| 45. |
If `^n+2C_8:^(n-2)P_4: 57 : 16 ,`find `ndot` |
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Answer» `.^(n+2)C_(8):^(n-2)P_(4) = 57:16` `(.^(n+2)C_(8))/(.^(n-2)P_(4)) = (57)/(16)` `rArr ((n+1)!)/(8!(n-6)!).((n-6)!)/((n-2)!) = (57)/(16)` `rArr ((n+2)(n+1)n(n-1)(n-2)!)/((n-2)!) =(57)/(16) xx 8!` `rArr (n+2)(n+1)n(n-1)` `= (57 xx 1 xx 2 xx 3xx 4xx 5xx 6 xx7xx8)/(16)` `rArr (n+2) (n+1).n(n-1) = 21 xx 20 xx 19 xx 18` `rArr n = 19` (on comparison). |
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| 46. |
How many numbers of 5 digits can be formed by the digits 1,2,2,1,3? |
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Answer» In the given 5 digits, the digits 1,2 occur twice. `:.` Total 5 digit numbers formed `= (5!)/(2!.2!)` `= (120)/(2 xx 2) = 30`. |
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| 47. |
In how many ways can 5 boys and 3 girls sit in a row so that no two girls are together ? |
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Answer» Correct Answer - `.^(6)P_(3)xx5!` Five boys can sit in 5! Ways, in this case, there are 6 vacant places where the girls can sit in `.^(6)P_(3)xx5!` |
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| 48. |
A person invites a group of 10 friends at dinner and sits5 on a round table and 5 more on another round table,4 on one round table and 6 on the other round table.Find the number of ways in each case in which he can arrange the guest. |
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Answer» (i) The number of ways of selection of 5 friends for the first table is `.^(10)C_(5)`. Remaining 5 friends are left for the second table. The total number of permutations of 5 guests on each table is 4!. Hence, the total number of arrangements is `.^(10)C_(5)xx4!xx4!=((10)!)/(5!xx5!)xx4!xx4!=10!//25`. (ii) The number of ways of selection of 6 guests is `.^(10)C_(6)` The number of ways of permutations of 6 guests on round table is 5!. The number of permutation of 4 guests on another round table is 3!. Therefore, total number of arrangements is `.^(10)C_(6)xx5!xx3!=((10)!)/(6!xx4!)5!xx3!=((10)!)/(24)` |
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| 49. |
Find the number of ways in which 10 different diamonds can be arrangedto make a necklace. |
| Answer» Since diamonds do not have natural order of left and right so clockwise and anticlockwise arrangements are taken as identical. Therefore, the number of arrangements of 10 different diamonds to make a necklace is `9!//2=181440`. | |
| 50. |
In how many ways can 6 persons be seated at a round table? |
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Answer» No. of arrangement in which 6 persons can be seated at a round table `= (6-1)! = 5! = 120`. |
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