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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
Six persons A, B, C, D, E, F, are to be seated at a circular table. Inhow many ways antis be one if A should have either Bor C on his and B must always have either C or D on his right. |
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Answer» Let the seat occupied by A be numbered as 1 and the remaining 5 seats be numbered as 2, 3,4,5,6 in anticlockwise direction. There arise two cases : Case I : B is on right of A, i.e., at number 2. Then, seat number 3 can be occupied by C or D in `.^(2)C_(1)` ways and remaining 3 persons can have remaining 3 seats in 3! ways. Hence, the number of arrangements in this case is `2xx6=12`. Case II : C is on the right of A. i.e., at number 2. Then, Bcan occupy any seat from number and 2 seats, which can be occupied in 2! ways. Hence, the number of arrangements in this case is `.^(3)C_(1)xx2!=6`. These cases are exclusive. So by the sum rule, the total number of arrangements is 12+6=18. |
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| 52. |
In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls? |
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Answer» No. of ways of selecting 3 boys out of 5 boys `= .^(5)C_(3)` No. of ways of selecting 3 girls out of 4 girls `= .^(4)C_(3)` Therefore, number of total teams `= .^(5)C_(3) xx .^(4)C_(3)` `= (5 xx 4 xx 3)/(3 xx 2 xx 1) xx (4 xx 3 xx 2)/(3 xx 2 xx 1)` `= 10 xx 4 xx 40` |
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| 53. |
Find the number of ways in which a committee of 11 members can be formed out of 6 teachers and 8 students if there are at least 4 teachers in the committee. |
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Answer» The committee can be formed in the following ways: (i) 4 teacher +7 students No. of ways of selection `= .^(6)C_(4) xx ^(8)C_(7)` (ii) 5 teachers +6 students No. of ways of selection `= .^(6)C_(5) xx ^(8)C_(6)` (iii) 6 teacher +5 students No of ways of selection `= .^(6)C_(6) xxe^(8)C_(5)` The committee can be formed in any one way, therefore from addition principle Required ays `= .^(6)C_(4) xx^(8)C_(7) +^(6)C_(5)xx^(8)C_(6) +^(6)C_(6) xx^(8)C_(5)` `= 15 xx 8 +6 xx 28 +1 xx 56` `=120 +168 +56` `= 344`. |
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| 54. |
Out of 15 balls, of which some are white and the rest are black, howmany should be white so that the number of ways in which the balls can bearranged in a row may be the greatest possible? It is assumed that the ballsof same color are alike? |
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Answer» Correct Answer - 7 or 8 Let there be r white and (15-r) black balls. Then, total number of permutations of these balls is `15!//r!(15-r)!= .^(15)C_(r )` since r white balls are alike and (15-r) black balls are alike. Therefore, the number of arrangement is `.(15)C_(r )` which is maximum, when `r=(15-1)//2 " or " (15+1)//2`, i.e., r=7 or 8. |
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| 55. |
In how many shortest ways can we reach from the point (0, 0, 0) to point(3, 7, 11) in space where the movement is possible only along het x-axis,y-axis, and z-axis or parallel to them and change of axes is permitted onlyat integral points. (An integral point is one, which has its coordinate asinteger.) |
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Answer» Correct Answer - `.^(21)C_(3)xx .^(18)C_(7)` The total number of units to be covered is 3+7+11=21. A person can choose 3 units in `.^(21)C_(3)` ways. A person can choose 7 units from remaining 18 in `.^(18)C_(7)` ways. The rest 11 units can be chosen in 1 way. Therefore, the total number of ways is `.^(21)C_(3)xx .^(18)C_(7)xx1`. |
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| 56. |
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. |
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Answer» At least, one green ball can be selected out of 5 green balls in `2^(5)-1`, i.e., in 31 ways. Similarly, at least one blue ball can be selected from 4 blue balls in `2^(4)-1=15` ways. And at least one red or no red ball can be selected in `2^(3)=8` ways. Hence, the required number of ways is `31xx15xx8=3720`. |
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| 57. |
Nishi has 5 coins, each of the different denomination. Find the numberdifferent sums of money she can form. |
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Answer» Number of different sums of money she can form is equal to number of ways she select one or more coins. Therefore, Required number of ways `= .^(5)C_(1)+ .^(4)C_(2)+ .^(5)C_(3)+ .^(5)C_(4)+ .^(5)C_(5)` `=2^(5)-1=31` |
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| 58. |
A question paper on mathematics consists of 12 questions divided in to3 pars A, B and C, each containing 4 questions. In how many ways can anexaminee answerquestions selecting at least one from each part. |
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Answer» Correct Answer - `. ^(12)C_(5)-3xx .^(8)C_(5)` Required number of ways=Total number of ways in which he does not select any question from any one section `= .^(12)C_(5)-3 xx .^(8)C_(5)` |
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| 59. |
In how many ways 4 boys and 3 girls can be seated in a row so that they are alternate?A. 12B. 72C. 120D. 144 |
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Answer» Correct Answer - D BGBGBGB Required no. of ways `= 4! xx 3! = 144` |
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| 60. |
Which of the following is noth the number of ways of selecting `n` objects from `2n` objects of which `n` objects are identicalA. `2^(n)`B. `("^(2n+1)C_(0)+^(2n+1)C_(1)+...+^(2n+1)C_(n))^(1//2)`C. the number of possible subsets `{a_(1),a_(2),….,a_(n)}`D. None of these |
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Answer» Correct Answer - D `(d)` We can take `0` indentical and `n` distinct , `1` identical and `n-1` distinct , `2` identical and `n-2` distinct and so on… `:.` Total no. of solutions `=sum_(r=0)^(n).^(n)C_(r )=2^(n)` No. of possible subsets of a set containing `n` elements is `2^(n)` and `.^(2n+1)C_(0)+^(2n+1)C_(1)+^(2n+1)C_(2)+...+^(2n+1)C_(n)=2^(2n)` |
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| 61. |
How many words can be formed using all the letters of the folloiwng words ? (i) BANANA (ii) ALLAHABAD INDEPENDENCE (iv) ASSASSINATION |
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Answer» (i) We have word BANANA Letters are B, (N N), (A A A). `therefore` Number of words `=(6!)/(2!xx3!)=(6xx5xx4)/(2!)=60` (ii) We have word ALLAHABAD Letters are (A A A A), (L L), H,B,D `therefore` Number of words `=(9!)/(4!xx2!)=(9xx8xx7xx6)/(2!)=1512` (iii) We have word INDEPENDENCE Letters are I, (N N N), (D D), (E E E E),P,C. `therefore` Number of words `=(12!)/(3!xx2!xx4!)` (iv) We have word ASSANSSINATION Letters are (A A A), (S S S), (II), (N N), T, O. `therefore` Number of words `=(13!)/(3!xx4!xx2!xx2!)` |
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| 62. |
Find the total number of nine-digit numbers that can be formed using thedigits 2, 2, 3, 3, 5, 5, 8, 8, 8 so that the odd digitoccupy the even places. |
| Answer» Odd digits 3,3,5,5 occupy four even places in `4!//(2!2!)=6` ways. Rest five digits 2,2,8,8,8 occupy rest five places in `5!//(2!3!)=10` ways. Hence, the total number of ways is `6xx10=60`. | |
| 63. |
Three are 12 points in a plane, no of three of which are in the same straight line, except 5 points whoich are collinear. Find (i) the numbers of lines obtained from the pairs of these points. (ii) the numbers of triangles that can be formed with vertices as these points. |
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Answer» (i) No. of lines formed by joining the 12 points, taking 2 at a time `=.^(12)C_(2)` No of lines formed by joining the 5 points taking 2 at a time `= .^(5)C_(2)` But 5 collinear points when joined pairwise, give only one line `:.` The required number of lines `= .^(12)C_(2) - .^(5)C_(2) +1` `= 66 - 10 +1` `= 57` (ii) No of triangles formed by joining 12 points by taking 3 points at a time `= .^(12)C_(3)` No. of triangles formed by joining 5 points by taking 3 points at a time `= .^(5)C_(3)` But there is no triangle formed by joining 3 points out of 5 collinear points `:.` The required number of triangles formed `=.^(12)C_(3)-.^(5)C_(3)` `= 220 - 10 = 210` |
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| 64. |
The number 916238457 is an example of a nine-digit number whichcontains each of the digit 1 to 9 exactly once. Italso has the property that the digits 1 to 5 occur in their natural order,while the digits 1 to 6 do not. Find the number of such numbers. |
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Answer» Correct Answer - 2520 6 places can be selected in `.^(9)C_(6)` ways and 6 can be placed only at 5 places except the right most of the 6 selected. Remaining numbers , i.e., 7,8,9 in 3! Ways. Hence, number of numbers are `.^(9)C_(6). 5.3!=84xx30=2520` |
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| 65. |
There are `n`points in aplane in which no large no three are in a straight line except `m`which are all i straight line. Find the number of (i)different straight lines, (ii) different triangles, (iii) differentquadrilaterals that can be formed with the given points as vertices. |
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Answer» A straight line can be formed by joining any two points. So number of straight lines (i.e., selection of two points from n) is `.^(n)C_(2)` But, selection of two points from m collinear points gives no extra line. Hence, the number of distinct straight lines is `.^(n)C_(2)- (.^(m)C_(2)-1)=(1)/(2)n(n-1)-(1)/(2)m(m-1)+1` (b) Formation of traingles is equivalents to section of three points from a points. As m points are collinear, selection of three points from m collinear points from m collinear points gives no triangles. Hence, the number of triangles is `.^(n)C_(3)- .^(m)C_(3)=(1)/(6)[n(n-1)(n-2)-m(m-1)(m-2)]` (c ) Four points determine a quadrilateral. But of these four points, not more than two is to be selected from m collinear points. Now the numebr of selections of four points from all n is `.^(n)C_(4)`. The number of selections of three points from m collinear and one from rest is `.^(m)C_(3) .^(n-m)C_(1)`. The number of selections of four points from m collinear is `.^(m)C_(4)`. Hence, Number of quadrilaterals `= .^(n)C_(4)- .^(m)C_(3)xx (n-m)C_(1)- .^(m)C_(4)` |
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| 66. |
From 7 men and 4 women a committee of 6 is to be formed such that the committee contains at least two women. What is the number of ways to do this ?A. 210B. 371C. 462D. 5544 |
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Answer» Correct Answer - B The required number of ways `= .^(11)C_(6) - (.^(7)C_(6) xx .^(4)C_(0) +.^(7)C_(5) xx .^(4)C_(1))` `= (11 xx 10 xx 9 xx 8 xx 7)/(5 xx 4 xx 3 xx 2) - (7 + (7 xx 6)/(2) xx 4)` `= 462 - (7 + 84) = 371` |
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| 67. |
If `P (32, 6) = kC(32, 6)`, then what is the value of k?A. 6B. 32C. 120D. 720 |
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Answer» Correct Answer - D Since `.^(32)P_(6) = k .^(32)C_(6)` `rArr (32!)/((32 -6)!) = k (32!)/(6!(32 -6)!)` `rArr k = 6! = 720` |
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| 68. |
If `.^(22)P_(r+1) :.^(20)P_(r+2) = 11:52`, find the value of r. |
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Answer» `.^(22)P_(r+1): .^(20)P_(r+1) = 11:52` `rArr (22!)/((22-r-1)!) : (20!)/((20-r-2)!) = 11:52` `rArr (22!)/((21-r)!): (20!)/((18-r)!) = 11:52` `rArr (22!)/((21-r)!) xx ((18-r)!)/(20!) = (11)/(52)` `rArr (22 xx 21 xx 20!)/((21-r) xx (20-r) xx (19-r) xx(18-r)!) xx ((18-r)!)/(20!) = (11)/(52)` `rArr (22 xx 21)/((21-r)(20-r)(19-r)) = (11)/(52)` `rArr (21-r) (20-r) (19-r) = (22 xx 21 xx 52)/(11)` `= 14 xx 13 xx 12` On comparing: `rArr 21 - r = 14` `rArr r = 7`. |
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| 69. |
If `C(n,12)=C(n,8)` then find the values of `C(n,17)` and `C(22,n)`A. 131B. 231C. 256D. 292 |
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Answer» Correct Answer - B Give `C(n, 12) = C(n, 8)` `rArr .^(n)C_(12) = .^(n)C_(8)` `rArr (n!)/((n-12)!2!) = (n!)/((n-8)!8!)` `rArr (1)/((n-12)!(12 xx 11 xx 10 xx 9 xx 8!))` `= (1)/((n-8)(n-9) (n-10)(n-1)(n-12)!8!)` `rArr (1)/(12 xx 11 xx 10 xx 9) = (1)/((n-8)(n-9)(n-10)(n-11))` `rArr (n-8) (n-9) (n-10)(n-11)` `= 12 xx 11 xx 10 xx 9` `rArr n-8 = 12, n-9 = 11, n - 10 = 10 and n-11 = 9` `rArr n = 20` `rArr C(22, n) = .^(22)C_(20)` `= (22!)/(2!20!) = (22 xx 21)/(2) = 231` |
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| 70. |
In a football championship, 153 matches were played. Every two teams played one match with each other. The number of teams, participating in the championship is …….. .A. 21B. 18C. 17D. 15 |
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Answer» Correct Answer - B Let total no. of team participated in a championship ben. Since, every team played one match with each other team. `:. .^(n)C_(2) = 153 rArr (n!)/(2!(n -2)!) = 153` `rArr (n(n-1)(n-2)!)/(2!(n-2)!) = 153 rArr (n(n-1))/(2) = 153` `rArr n(n-1) = 306` `rArr n^(2) - n - 306 = 0` `rArr n (n -18) + 17 (n -18) = 0` `rArr n = 18, -17` n cannot be negative `:. n != - 17` `rArr n = 18` |
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| 71. |
Prove that `.^(n)C_(r )+.^(n-1)C_(r )+..+.^(r )C_(r )=.^(n+1)C_(r+1)` |
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Answer» `.^(r )C_(r ) + .^(r+1)C_(r ) + .^(r+2)C_(r) +.. + .^(n-1)C_(r )+ .^(n)C_(r )` `.^(r+1 )C_(r+1 )+ .^(r+1)C_(r )+ .^(r+2)C_(r ) +.. + .^(n-1)C_(r )+ .^(n)C_(r )` `= .^(r+1)C_(r+1) + .^(r+1)C_(r ) +.. + .^(n+1)C_(r )+ .^(n)C_(r )` `= .^(r+3)C_(r+1) +..+ .^(n-1)C_(r ) + .^(n)C_(r )` On adding similar way, we get L.H.S. `= .^(n-1)C_(r+1)+ .^(n-1)C_(r ) + .^(n)C_(r )` `= .^(n)C_(r+1) + .^(n)C_(r )` `= .^(n+1)C_(r+1)=R.H.S.` |
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| 72. |
Find the value (s) of r satisfying the equation `.^(69)C_(3r-1)-.^(69)C_(r^(2)-1)-.^(69)C_(3r)` |
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Answer» `.^(69)C_(3r-1)+ .^(69)C_(r^(2)-1)+ .^(69)C_(r^(2))` `implies .^(70)C_(3r)= .^(70)C_(r^(2))` `implies 3r=r^(2) " or" r=0,3` or `3r+r^(2)-70=0` or r=7, -10 Hence, possible values of r and 3 and 7 as for these values all the terms in equation are defined. |
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| 73. |
If `.^(n)C_(8)=.^(n)C_(6)`, then find `.^(n)C_(2)`. |
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Answer» If `.^(n)C_(x)= .^(n)C_(y) " and" x ne y`, then x+y=n. Hence, `.^(n)C_(8)= .^(n)C_(6)` `implies n=(8+6)=14` Now, `.^(n)C_(2)= .^(14)C_(2)=(14xx13)/(2)=91` |
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| 74. |
If `.^(n)C_(8) = .^(n)C_(2)`, find `.^(n)C_(2)`. |
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Answer» `.^(n)C_(x) = .^(n)C_(Y)` `rArr n = x +y` `:. .^(n)C_(8) = .^(n)C_(2)` `rArr n = 8 +2 = 10` Therefore, `.^(n)C_(2) = .^(10)C_(2) = (10!)/(2!8!) = (10 xx 9 xx 8!)/(2xx 1 xx 8!) = 45`. |
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| 75. |
Determine n if (i) `.^(2n)C_(3) : .^(n)C_(3) = 12:1` (ii) `.^(2n)C_(3): .^(n)C_(3)= 11:1` |
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Answer» (i) `.^(2n)C_(3):^(n)C_(2) = 12:1` `rArr ((2n)!)/(3!(2n-3)!): (n!)/(2!(n-2)!) = 12:1` `rArr ((2n).(2n-1)(2n-2))/(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` (ii) `.^(2n)C_(3):^(n)C_(3) = 11:1` `rArr .^(2n)C_(3)xx1 = 11 xx .^(n)C_(3)` `rArr (2n(2n-1)(2n-2))/(3xx 2xx1) = 11 xx (n(m-1)(n-2))/(3 xx 2xx 1)` `rArr (2n(2n-1).2(n-1))/(6) xx (6)/(n(n-1)(n-2) = 11` `rArr (4(2n-1))/(n-2) =11` `rArr 8 n - 4 = 11 n - 22` `rArr 22 - 4 = 11n - 8n` `rArr 18 = 3n rArr n = 6`. |
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| 76. |
If `^0P_5+5^9P_4=^(10)P_r ,`find the value of `rdot` |
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Answer» `.^(10)P_(r )=.^(9)P_(5)+5^(9)P_(4)` `=(9!)/((9-5)!)+5xx(9!)/((9-4)!)` `=(9!)/(4!)+(9!)/(4!)` `=2xx(9!)/(4!)` `=(5xx2xx9!)/(5xx4!)` `=(10xx9!)/(5!)` `=(10!)/(5!)` `=.^(10)P_(5)` `implies r=5` |
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| 77. |
If `^n P_r=^n P_(r+1)a n d^n C_r=^n C_(r-1,)`then thevalue of `n+r`is. |
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Answer» `.^(n)C_(r) = .^(n)C_(r-1)` `rArr n = r +(r-1)` `rArr n = 2r - 1` ...(1) and `.^(n)P_(r)- .^(n)P_(r+1)` `rArr (n!)/((n-r)!) =(n!)/((n-r-1)!)` `rArr (n-r)! = (n-r-1)!` `rArr (n-r)(n-r-1)! = (n-r-1)!` `rArr n- r = 1` `rARr 2r - 1 - r = 1` [From eq. (1)] `rArr r = 2` `rArr` From eq. (1) `n = 2r - 1` `=2(2) - 1 = 3` `n = 3, r = 2` |
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| 78. |
A class has tree teachers, Mr. `X`, Ms.`Y` and Mrs.`Z` and six students `A`, B`, `C`, `D`, `E`, `F`. Number of ways in which they can be seated in a line of `9` chairs,if between any two teachers there are exactly two students isA. `18xx6!`B. `12xx6!`C. `24xx6!`D. `6xx6!` |
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Answer» Correct Answer - A `(a)` The possible arrangements of teachers and students can be as follows : `(i) T S S T S S T S S` `(ii) S T S S T S S T S` `(iii) S S T S S T S S T` Hence, total number of ways `=3*(3!)6!=(18)6!` |
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| 79. |
Find the number of ways of arranging 15 students `A_1,A_2,........A_15` in a row such that (i) `A_2`, must be seated after `A_1 and A_2`, must come after `A_2` (ii) neither `A_2` nor `A_3` seated brfore `A_1`A. `(2!xx15!)/(3!)`B. `(15!)/(3!)`C. `2!15!`D. None of these |
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Answer» Correct Answer - A `(a)` First arrange `12` persons `A_(4),A_(5),….A_(15)` in `"^(15)P_(12)` ways There remains `3` places. Keep `A_(1)` in the first place and arrange `A_(2)`, `A_(3)` in the remaining two places in `2!` ways. `:.` Required no. of arrangements `="^(15)P_(12).1.2!` |
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| 80. |
The number of words that can be formed using all the letters of the word REGULATIONS such that `G` must come after `R`, `L` must come after `A`, and `S` must come after `N` areA. `11!//8`B. `11!`C. `"^(11)P_(6)`D. none of these |
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Answer» Correct Answer - A `(a)` The other five letters (other than `G`, `R`, `L`, `A`, `S`, `N`) in `11` places can be arranged in `"^(11)P_(5)` ways. Then three remains six places. Choose two of them in `"^(6)C_(2)` ways. In these places we can put `G`, `R` only `(i)` one way. Similarly others. `:.` Required no. of ways `="^(11)P_(5)xx^(6)C_(2)xx^(4)C_(2)xx^(2)C_(2)=11!//8` |
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| 81. |
There are 10 stations on a circular path. A train has to stop at 3 stations such that no two stations are adjacent. The number of such selections must be: (A) `50` (B) `84` (C) `126` (D) `70`A. `50`B. `60`C. `70`D. `80` |
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Answer» Correct Answer - A `(a)` Total selections `="^(10)C_(3)=120` Number of selections in which `3` stations are adjacent `=10` Number of selections in which `2` stations are adjacent `=6` But there are `10` such pairs. `implies` Total invalid selections `=10+6xx10=70` |
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| 82. |
A guard of `12` men is formed from a group of `n` soldiers. It is found that `2` particular soldiers `A` and `B` are `3` times as often together on guard as `3` particular soldiers `C, D` & `E`. Then `n` is equal toA. `28`B. `27`C. `32`D. `36` |
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Answer» Correct Answer - C `(c )` `.^(n-2)C_(12-2)=3.^(n-3)C_(12-3)` `implies.^(n-2)C_(10)=3.^(n-3)C_(9)` `implies(n-2)/(10)=3impliesn=32` |
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| 83. |
Find the number of integers between 1 and 1000 having the sum of thedigits 18. |
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Answer» Correct Answer - `.^(33)C_(18)-6xx.^(23)C_(8)` Any number between 1 and 100000 must be of less than six digits. Therefore, it must be of the form `a_(1)a_(2)a_(3)a_(4)a_(5)a_(6)` where `a_(1),a_(2),a_(3),a_(4),a_(5),a_(6) in {0,1,2,..,9}` Now given that `a_(1)+a_(2)+a_(3)+a_(4)+a_(5)+a_(6)=18` where `0 le a_(i) le9, i=1,2,3,..,9` `therefore` Required number of integers =coefficient of `p^(18) " in" (1+p+p^(2)+..+ p^(9))^(6)` = coefficient of `p^(18) " in" (1-p^(10))/(1-p))^(6)` = coefficient of `p^(18) " in" [(1-p^(10))^(6)(1-p)^(-6)]` =coefficient of `p^(18) " in" [(1- .^(6)C_(1)p^(10))(1-p)^(-16)]` `= .^(33)C_(18)-6xx .^(23)C_(8)` |
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| 84. |
Find the number of `n`digit numbers, which contain the digits 2 and 7, but not the digits 0, 1, 8,9. |
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Answer» Correct Answer - `6^(n)-2xx5^(n)+4^(n)` Number of n-digit number formed by using the digits 2,3,4,5,6,7 is `6^(n)`. Number of n-digit numbers formed by using the digits 3,4,5,6,7 is `n(A)=5^(n)`. Number of n-digit numbers formed by using the digits 2,3,4,5,6 is `n(B)=5^(n)`. Number of n-digit numbers formed by using the digits 3,4,5,6 is `n(A cap B)=4^(n)`. So, total number of numbers which contains digits 2 and 7 is `6^(n)-5^(n)-5^(n)+4^(n)`. |
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| 85. |
`P=n(n^(2)-1)(n^(2)-4)(n^(2)-9)…(n^(2)-100)` is always divisible by , `(n in I)`A. `2!3!4!5!6!`B. `(5!)^(4)`C. `(10!)^(2)`D. `10!11!` |
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Answer» Correct Answer - A::B::C::D `(a,b,c,d)` We know that product of `r` consecutive integer is divisible by `r!` We have `P=(n-10)(n-9)(n-8)…(n+10)` is produt of `21` consecutive integers Which is divisible by `21!` `[(n-10)(n-9)(n-8)...n][(n+1)(n+2)(n+3)...(n+10)]` Which is divisible by `11! 10!` `[(n-10)...(n-6)]xx[(n-5)...(n-1)]xx[n(n+1)...(n+4)]xx[(n+5)...(n+9)]xx(n+10)` which is divisible by `(5!)^(4)`. Also it can be shown that `P` is divisible by `2!3!4!5!6!` |
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| 86. |
The number of `n` digit number formed by using digits `{1,2,3}` such that if `1` appears, it appears even number of times, isA. `2^(n)+1`B. `(1)/(2)(3^(n)+1)`C. `(1)/(2)(3^(n)-1)`D. `(1)/(2)(2^(n)-1)` |
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Answer» Correct Answer - B `(b)` Required number of numbers `="(n)C_(0)(2)^(n)+"(n)C_(2)(2)^(n-2)+…` We know that `(1+x)^(n)+(x-1)^(n)=2sum_(k=0)^(n)"(n)C_(k)x^(k)` Hence, put `x=2`, get the result `2["^(n)C_(0)(2)^(n)+^(n)C_(2)(2)^(n-2)+^(n)C_(2)(2)^(n-4)+^(n)C_(6)(2)^(n-6)+...]` |
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| 87. |
Let `f:A->A` be an invertible function where `A= {1,2,3,4,5,6}` The number of these functions in which at least three elements have self image is |
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Answer» Correct Answer - 56 If exactly r element have self image then number of functions =(Number of ways of selecting r elements)`xx` (Derangement of remaining elements) `=""^(n)C_(r ) r!(1-(1)/(1!)+(1)/(2!)-..)` `therefore` Required number of functions `= ""^(6)C_(3)xx3!xx(1-(1)/(1!)+(1)/(2!)-(1)/(3!))+ ""^(6)C_(4)xx2!(1-(1)/(1!)+(1)/(2!))+""^(6)C_(5)xx1!xx(1-(1)/(1!))+""^(6)C_(6)` =40+15+0+1 =56 |
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| 88. |
A contest consists of ranking `10` songs of which `6` are Indian classic and `4` are westorn songs. Number of ways of ranking so that (mention correct statements)A. There are exactly `3` indian classic songs in top `5` is `(5!)^(3)`.B. Top rank goes to Indian classic song is `6(9!)`C. The ranks of all western songs are consecutive is `4!7!`D. The `6` Indian classic songs are in a specified order is `"^(10)P_(4)`. |
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Answer» Correct Answer - A::B::C::D `(a,b,c,d)` `(a) ^(6)C_(3)^(4)C_(2)5!5! =(5!)^(3)` `(b) ^(6)C_(1)9!` `(c ) (6+1)!4!` `(d) ^(10)P_(4)` |
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| 89. |
Number of six-digit numbers such that any digit that appears in the number appears at least twice, where the digits of each number are from the set `{1, 2, 3, 4, 5},` is (Example 225252 is valid but 222133 is not valid)A. `1500`B. `1850`C. `1405`D. `1205` |
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Answer» Correct Answer - C `(c )` Case I : All six digits alike i.e. `111111`, `222222`……..etc. `=5` ways Case II : `2` alike `+2` other alike. Select any three in `"^(5)C_(3)` ways (i.e.`1,2,3` and take `11,22,33`) For each such selections number of ways `=(6!)/(2!2!2!)=90` `implies` Total `=10xx90=900` Case III : `2` alike `+4` other alike i.e. `11 2222` or `22 11 11` etc. Number of ways selecting `2` digits `=("^(5)C_(2))(2)=20` For each selections number of ways `=(6!)/(2!4!)=15` `implies` Total `=20xx15=300` Case IV : `3` alike `+3` other alike Select any two in `"^(5)C_(2)=10` ways For each selection number of ways `=(6!)/(3!*3!)=20` `implies` Total `=10xx20=200` Hence total`=5+900+300+200=1405` |
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| 90. |
The number of arrangments of all digits of `12345` such that at least `3` digits will not come in its position isA. `89`B. `109`C. `78`D. `57` |
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Answer» Correct Answer - B `(b)` Total number of ways such that at least `3` digits will not occur in its position. `=^(5)C_(3){3!-^(3)C_(1)2!+^(3)C_(2)1!-^(3)C_(2)0!}+^(5)C_(4)[4!-^(4)C_(1)(3!)+^(4)C_(2)(2!)-^(4)C_(3)(1!)+^(4)C_(4)(0!)}+^(5)C_(5){5!-^(5)C_(1)4!+^(5)C_(2)3!-^(5)C_(3)2!+^(5)C_(4)1!-^(5)C_(5)(0!)}` `=10(2)+5(9)+(44)` `=20+45+44=109` |
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| 91. |
The number of arrangments of all digits of `12345` such that at least `3` digits will not come in its position is |
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Answer» Correct Answer - 109 Total number of ways such that at least 3 digits will not come in its position. =Derangement of 3 digits +Derangement of 4 digits +Derangement of 5 digits `= ""^(5)C_(3)xx3!xx(1-(1)/(1!)+(1)/(2!)-(1)/(3!))+""^(5)C_(4)xx4!xx(1-(1)/(1!)+(1)/(2!)-(1)/(3!)+(1)/(4!))+""^(5)C_(5)xx5!xx(1-(1)/(1!)+(1)/(2!)-(1)/(3!)+(1)/(4!)-(1)/(5!))` `=10xx2+5xx9+44` `=20+45+44=109` |
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| 92. |
Evaluate: (i) 8! (ii) `4! - 3!` |
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Answer» (i) `lfloor8 = 8 xx 7xx 6 xx 5 xx 4 xx 3 xx 2 xx 1` `= 40320` Therefore, `8! = 40320` (ii) `lfloor4 - lfloor3 = (4 xx lfloor3) - lfloor3` `= (4-1) lfloor3` `= 3 xx xx 2 xx 1 = 18` Therefore, `lfloor4 - lfloor3 = 18` |
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| 93. |
Find the number of ways in which 16 constables can be assigned to patrol villages, 2for each. |
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Answer» Correct Answer - `(16!)/((2!)^(8))` The number of ways 16 constables can be divided into 8 bactes is `(16)!//8!(2!)^(8)`. Now, the first batch may be assigned to patrol any of the 8 batches to patrol 8 villages is `((16)!)/(8!(2!)^(8))(8!)=((16)!)/((2!)^(8))` |
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| 94. |
In how many ways can 10 different prizes be given to 5 students if oneparticular boy must get 4 prizes and rest of the students can get any numberof prizes? |
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Answer» Correct Answer - 860160 The number of selections of 4 prizes to the particular boy is `.^(10)C_(4)`. From the remaining 6 prizes, each prize can be given to any of the four students. Therefore, the number of ways is `4^(6)`. Therefore, the total number of ways is `.^(10)C_(4)xx4^(6)=210xx4096=860160`. |
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| 95. |
Find the number of ways in which the birthday of six different personswill fall in exactly two calendar months. |
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Answer» Correct Answer - `.^(12)C_(2)xx(2^(6)-2)` Any two months can be chosen in `.^(12)C_(2)` ways. The six birthdays can fall in these two months in `2^(6)` ways. Out of these `2^(6)` ways there are two ways when all the six birthdays fall in one month. So, the required number of ways is `.^(12)C_(2)xx(2^(6)-2)`. |
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| 96. |
In how many ways can 30 marks be allotted to 8 questionif each question carries at least 2 marks? |
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Answer» Correct Answer - `.^(21)C_(14)` Required number `="coeff. Of" p^(30) " in" (p^(2)+p^(3)+..+ p^(16))^(8)` `= "coeff.of" p^(30) "in" p(16)(1+p+..+ p^(14))^(8)` `="coeff.of" p^(14) " in" ((1-p^(15))/(1-p))^(8)` `="coeff.of" p^(14) " in" (1-p)^(-8)` `=.^(21)C_(14)` |
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| 97. |
In how many ways 3 boys and 15 girls can sits together in a row suchthat between any 2 boys at least 2 girls sit. |
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Answer» Correct Answer - `3!xx15!xx.^(14)C_(3)` B B B x y z w First 3 boys can be arranged in 3! Ways. After arranging the boys, 4 gaps are created. Let in these gaps x,y,z, and w girls sit as shown in the figure. Let us first find out the distribution ways of girls in the four gaps. As given in question, `y,z ge 2 " and" x,w ge 0`, we have to find the integral solutions of the equation x+y+z+w=15 with the above condition. Let `y=y_(1)+2 " and" z=z_(1)+2 {"where" y_(1),z_(1) ge0)` `implies x+y_(1)+z_(1)+w=11` Number of solutions of above equation is `.^(11+4-1)C_(4-1)= .^(14)C_(3)`. After it is decided as in which gap how many girls will sit, they can be arranged in 15! ways. Hence, the total number of ways is `3!15! .^(14)C_(3)` |
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| 98. |
There are six teachers. Out of them tow are primary teacher, two aremiddle teachers, and two are secondary teachers. They are to stand in a row,so as the primary teachers, middle teacher, and secondary teachers areaalways in a set. Find the number of ways in which they can do so. |
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Answer» Correct Answer - 48 There are 2 primary teachers. They can stand in a row together in 2!=2 ways. There are 2 middle teachers. They can stand in a row together in 2!=2 ways. There are 2 secondary teachers. They can stand in a row together in 2!=2 ways. These three sets can be arranged in 3!=6 ways Hence, the required number of ways `=2xx2xx2xx6=48` |
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| 99. |
Find the number of ways in which the number 94864 can be resolved as aproduct of two factors. |
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Answer» `94864=2^(4)xx7^(2)=11^(2)` (perfect square) Hence, the number of ways is `(1)/(2)[(4+1)(2+1)+1]=23` |
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| 100. |
If the 11 letters `A,B,..... K` denote an arbitrary permutation of the integers `(1,2.....11),` then `(A-1)(B-2)(C-3).....(K-11)` will be |
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Answer» Correct Answer - c Given set of numbers is {1,2,..11} in which 5 are even and six are odd, which signifies that in the given product, it is not possible to arrange but only to subtract the even numbers from odd numbers. There must be at least one factor involving subtraction of an odd number from another odd number. So at least one of the factors is even. hence, product is always even. |
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