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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.
| 351. |
Consider all possible permutations of the letters of the word `ENDEANOEL`.The number of permutations containing the word ENDEA isA. 5!B. `2xx5!`C. `7xx5!`D. `21xx5!` |
| Answer» Taking ENDEA as one letter we have four morem letters N, O, E, and L. These five letters can be arranged in 5! ways. So, the number of permutations containing the word ENDEA is 5!. | |
| 352. |
Consider all possible permutations of the letters of the word ENDEANOEL. The number of permutations in which none of the letters D, L and N occurs in the last positions isA. 5!B. `2xx5!`C. `7xx5!`D. `21xx5!` |
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Answer» Letters D, L, N, N occur in first four positions in `(4!)/(2!)` ways. Corresponding to each arrangement of D, L, N, N in first positions there are `(5!)/(3!)` arrangement of remaining 5 letters in last five positions. `:.` Required number of permutaions `=(4!)/(2!)xx(5!)/(3!)=2xx5!` |
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| 353. |
The value of the expression `""^(47)C_(4) + sum_(i=1)^(5) ""^(52-i)C_(3)` isA. `""^(47)C_(5)`B. `""^(52)C_(5)`C. `""^(52)C_(4)`D. None of these |
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Answer» Correct Answer - C Here, `""^(47)C_(4) + overset(5)underset(j = 1)sum ""^(52 - j)C_(3)` `= ""^(47)C_(4) + ""^(51)C_(3) + ""^(50)C_(3) + ""^(49)C_(3) + ""^(48)C_(3) + ""^(47)C_(3)` `= (""^(47)C_(4) + ""^(47)C_(3)) + ""^(48)C_(3) + ""^(49)C_(3) + ""^(50)C_(3) + ""^(51)C_(3)` [using `""^(n)C_(r) + ""^(n)C_(r -1) = ""^(n + 1)C_(r)`] `= (""^(48)C_(4) + ""^(48)C_(3)) + ""^(49)C_(3) + ""^(50)C_(3) + ""^(51)C_(3)` `= (""^(49)C_(4) + ""^(49)C_(3)) + ""^(50)C_(3) + ""^(51)C_(3)` `= (""^(50)C_(4) + ""^(50)C_(3)) + ""^(51)C_(3)` `= ""^(51)C_(4) + ""^(51)C_(3) = ""^(52)C_(4)` |
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| 354. |
Evaluate `^(47)C_(4)+sum_(j=0)^(3)""^(50-j)C_(3)+sum_(k=0)^(5) ""^(56-k)C_(53-k)`. |
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Answer» We have, `.^(47)C_(4)+underset(j=0)overset(3)(sum).^(50-j)C_(3)+underset(k=0)overset(5)(sum).^(56-k)C_(53-k)` `=.^(47)C_(4)+underset(j=0)overset(3)(sum).^(50-j)C_(3)+underset(k=0)overset(5)(.^(56)-k)C_(3)" "[because.^(n)C_(r)=.^(n)C_(n-r)]` `=.^(47)C_(4)+(.^(50)C_(3)+.^(49)C_(3)+.^(48)C_(3)+.^(47)C_(3))` `+(.^(56)C_(3)+.^(55)C_(3)+.^(54)C_(3)+.^(53)C_(3)+.^(52)C_(3)+.^(51)C_(3))` `=.^(47)C_(4)+.^(47)C_(3)+.^(48)C_(3)+.^(49)C_(3)+.^(50)C_(3)+.^(51)C_(3)+.^(52)C_(3)+.^(52)C_(3)+.^(54)C_(3)+.^(55)C_(3)+.^(56)C_(3)` `=(.^(47)C_(4)+.^(47)C_(3))+.^(48)C_(3)+.^(49)C_(3)+.^(50)C_(3)+.^(51)C_(3)+.^(52)C_(3)+.^(53)C_(3)+.^(54)C_(3)+.^(55)C_(3)+.^(56)C_(3)` `=.^(48)C_(4)+.^(48)C_(3)+.^(49)C_(3)+.^(50)C_(3)+.^(51)C_(3)+ . . .+.^(56)C_(3)` `=.^(49)C_(4)+.^(49)C_(3)+.^(50)C_(3)+ . . .+.^(56)C_(3)` `=.^(56)C_(4)+.^(56)C_(3)=.^(57)C_(4)` |
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| 355. |
If `sum_(r = 0)^(25) {""^(50)C_(r)* ""^(50-r)C_(25 - r)} = K(""^(50)C_(25)),` then, K is equal toA. `2^(24)`B. `2^(25) - 1`C. `2^(25)`D. `(25)^(2)` |
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Answer» Correct Answer - C Given, `overset(25)underset(r = 0)sum {""^(50)C_(r)*""^(50-r)C_(25 - r)} = K " "^(50)C_(25)` `rArr overset(25)underset(r = 0)sum ((50!)/(r!(50-r)!) xx ((50 - r)!)/((25 - r)!25!)) = K " "^(50)C_(25)` `rArr" "overset(25)underset(r = 0)sum ((50!)/(25!25!) xx (25!)/(r!(25 - r)!)) = K" "^(50)C_(25)` [on multiplying `25!` in numerator and denominator.] `rArr" "^(50)C_(25)overset(25)underset(r = 0)sum ""^(25)C_(r) = K" "^(50)C_(25)" " [because ""^(50)C_(25) = (50!)/(25!25!)]` `rArr" "K = overset(25)underset(r = 0)sum ""^(25)C_(r) = 2^(25)` `[because ""^(n)C_(0)+ ""^(n)C_(1) + ""^(n)C_(2) + .....+ ""^(n)C_(n) = 2^(n)]` `rArr" " K = 2^(25)` |
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| 356. |
The parallelogram is cut by two sets of m lines parallel to its sides. The numbers parallelogram thus formed, isA. `(""^(m)C_(2))^(2)`B. `(""^(m+1)C_(2))^(2)`C. `(""^(m+2)C_(2))^(2)`D. none of these |
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Answer» Correct Answer - C |
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| 357. |
If `sum_(i = 1)^(20) ((""^(20)C_(i - 1))/(""^(20)C_(i) + ""^(20)C_(i - 1)))^(3) = k/21`, then k equalsA. 100B. 400C. 200D. 50 |
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Answer» Correct Answer - A `overset(20)underset(i = 1)sum ((""^(20)C_(i - 1))/(""^(20)C_(i) + ""^(20)C_(i - 1)))^(3) = k/21` `rArr " "overset(20)underset(i = 1)sum ((""^(20)C_(i - 1))/(""^(21)C_(i)))^(3) = k/21 " "(because ""^(n)C_(r) + ""^(n)C_(r - 1) = ""^(n + 1)C_(r))` `rArr" "overset(20)underset(i = 1)sum ((""^(20)C_(i - 1))/(21/i""^(20)C_(i-1)))^(3) = k/21 " "(because ""^(n)C_(r) = n/r " "^(n -1)C_(r - 1))` `rArr" "overset(20)underset(i = 1)sum (i/21)^(3) = k/21` `rArr" "(1)/((21)^(3))overset(20)underset(i = 1)sum i^(3) = k/21` |
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| 358. |
There are n-points `(ngt2)` in each to two parallel lines. Every point on one line is joined to every point on the other line by a line segment drawn within the lines. The number of point (between the lines) in which these segments intersect, isA. `""^(2n)C_(2)-2.""^(n)C_(1)+2`B. `""^(2n)C_(2)-2xx""^(n)C_(2)`C. `""^(n)C_(2)xx""^(n)C_(2)`D. none of these |
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Answer» Correct Answer - C |
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| 359. |
All possible permutations of the letters of the word ENDEANOEL if A, E, O occur only in odd positions isA. 5!B. `2xx5!`C. `7xx5!`D. `21xx5!` |
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Answer» There five odd and 4 even positions. Five odd posittions can be filled with A, E, E, E, O in `(5!)/(3!)` ways and four even positions can be filled with D, N, N L in `(4!)/(2!)` ways. `:.` Required number of permutantions `=(5!)/(3!)xx(4!)/(2!)=2xx5!` |
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| 360. |
If `""^(m)C_(r)=0" for "rgtm` then the sum `sum_(r=0)^(m)""^(18)C_(r)""^(20)C_(m-r)` is maximum when m =A. 20B. 19C. 10D. 10 |
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Answer» We have, `sum_(r=0)^(m)""(18)C_(r)""^(20)xxC_(m-r)` = Number of ways of choosing m persons out of 18 men and 20 women `=""^(38)C_(m)` Clearly, it is maximum when m =19. |
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| 361. |
The maximum number of points of intersection of 8 circles isA. 16B. 24C. 28D. 56 |
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Answer» Correct Answer - D |
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| 362. |
There are `n`straight lines in a plane in which no two are parallel andno three pass through the same point. Their points of intersection arejoined. Show that the number of fresh lines thus introduced is`1/8n(n-1)(n-2)(n-3)`A. `(n(n-1)(n-1))/(8)`B. `(n(n-1)(n-2)(n-3))/(6)`C. `(n(n-1)(n-2)(n-3))/(8)`D. none of these |
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Answer» Correct Answer - C |
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| 363. |
The maximum number of points of intersection into which 4 circles and 4 straight lines intersect, isA. 26B. 50C. 56D. 72 |
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Answer» Correct Answer - B |
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| 364. |
Three letters can be posted in five letter boxes in `3^(5)` ways. |
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Answer» False Required number of ways `= 5^(3) = 125` |
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| 365. |
The greatest possible number of points of intersection of 8 straight lines and 4 circles is:A. 32B. 64C. 76D. 104 |
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Answer» Correct Answer - D |
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| 366. |
In a steamer there are stalls for 12 animals and there are cows, horses and calves (not less than 12 of each) ready to be shipped, the total number of ways in which the shipload can be made, is |
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Answer» True There are three types of animals and stalls available for 12 animals. Number of ways of loading `= 3^(12)` |
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| 367. |
In the permutations of n things r, taken together, the number of permutations in which m particular things occur together is `""^(n-m)P_(r-m) xx ""^("r")P_(m)*` |
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Answer» False Arrangement of n things, taken r at a time in which m things occur together, we considered these m things as 1 group. Number of object excluding those m objects = (r - m) Now, first we have to arrange (r-m+1) objects. Number of arrangements `=(r-m+1)"!"` and m objects which we consider as 1 group, can be arranged in `m"!"` ways. `therefore` Required number of arrangements `=(r-m+1)"!" xx m"!"` |
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| 368. |
Find the number of permutations of `n`distinct things taken `r`together, in which 3 particular things must occur together. |
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Answer» Total number of things = n We have to arrange r things out of n in which three things must occur together. Therefore, combination of n things taken r at a time in which3 things always occurs `= ""^(n-3)C_(r-3)` If three things taken together, then it is considered as 1 group. Arrangement of these three things `=3"!"` Now, we have to arrange `= r - 3 + 1 = (r - 2)` objects `therefore" "` Arranged of (r-2) objects `= r - 2"!"` `therefore" "` Total number of arrangements `= ""^(n - 3)C_(r - 3) xx r - 2"!" xx 3"!"` |
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| 369. |
Find the number of arrangements of the letters of the INDEPENDENCE. In how many of these arrangements,(i) do the words start with P(ii) do all the vowels always occur together(iii) do the vowels never occur together(iv) do the words begin with I and end in P? |
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Answer» total arrangements = `(12!)/(3!2!4!)` (i) `P -----------` `= (11!)/(3!2!4!)` (ii) `= (8!)/(3!2!)*(5!)/(4!)` (iii) `(12!)/(3!2!4!) - (8!5!)/(3!2!4!)` (iv) `I ---------- P` `= (10!)/(3!2!4!)` answer |
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| 370. |
In a city all telephone numbers have six digits. The first two digits are always 41 or 42 or 46 or 62. How many telephone numbers have all six distinct digits? |
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Answer» If first two digit is 41, the remaining 4 digits can be arranged in `= ""^(8)P_(4) = (8"!")/(8 - 4"!") = (8"!")/(4"!")` `= (8 xx 7 xx 6 xx 5 xx 4"!")/(4"!")` ` = 8 xx 7 xx 6 xx 5 = 1680` Similarly, if first two digit is 42, 46, 62 or 64, the remaining 4 digits can be arranged in `""^(8)P_(4)` ways i.e., 1680 ways. `therefore` Total number of telephone numbers have all six digits distinct `= 5 xx 1680 = 8400` |
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| 371. |
A number of 18 guests have to be seated, half oneach side of a long table. Four particular guests desire to sit on oneparticular side and three others on the other side. Determine the number ofways in which the sitting arrangements can be made. |
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Answer» Correct Answer - `""^(9)P_(4) xx ""^(9)P_(3)(11)!` Let the two sides be A and B. Assume that four particular guests wish to sit on side A. Four guests who wish to sit on side A can be accommodated on nine chairs in `""^(9)P_(4)` ways and three guests who wish to sit on side B can be accommodated in `""^(9)P_(3)` ways. Now, the remaining guests are left who can sit on 11 chairs on both sides of the table in `(11!)` ways. Hence, the total number of ways in which 18 persons can be seated `= ""^(9)P_(4) xx ""^(9)P_(3) xx (11)!.` |
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| 372. |
A number of 18 guests have to be seated, half oneach side of a long table. Four particular guests desire to sit on oneparticular side and three others on the other side. Determine the number ofways in which the sitting arrangements can be made.A. `9!xx9!`B. `""^(11)C_(5)xx9!xx9!`C. `(11!)/(5!)xx9!xx9!`D. `""^(11)C_(5)` |
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Answer» Since four particular guests want to sit on a particular side A (say) and three others on the other side B (say). So, we are left with 11 guests out of which we choose 5 for side A in `""^(11)C_(5)` ways and the remaining 6 for side B in `""^(6)C_(6)` ways. Hence, the number of selections for the two sides is `""^(11)C_(5)xx""^(6)C_(6)`. Now, 9 persons on each side of the table can be arranged among themselves in 9! ways. `:.` Total number of arrangements `""^(11)C_(5)xx""^(6)C_(6)xx9!xx9` |
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| 373. |
If n lines are drawn in a plane such that no two of them are parallel and no three of them are concurrent, such that these lines divide the planein 67 parts, then find number of different points at which these lines will cut.A. `underset(k=1)overset(n-1)(sum)k`B. `n(n-1)`C. `n^(2)`D. None of these |
| Answer» Correct Answer - A | |
| 374. |
The total number of arrangements of the letters in the expression `a^(3)b^(2)c^(4)` when written at full length, isA. 1260B. 2520C. 610D. none of these |
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Answer» Correct Answer - A |
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| 375. |
If n lines are drawn in a plane such that no two of them are parallel and no three of them are concurrent, such that these lines divide the planein 67 parts, then find number of different points at which these lines will cut. |
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Answer» Given number of straight lines=n, then `1+underset(k=1)overset(n)(sum)k=67implies(n^(2)+n+2)/(2)=67` `impliesn^(2)+n-132=0 implies(n+12)(n-11)=0` `thereforen=11,n ne-12` Hence, required number of points`=.^(n)C_(2)=.^(11)C_(2)=(11*10)/(2)` |
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| 376. |
Find the number of straight lines that can be drawn through any two points out of 10 points, of which 7 are collinear.A. 26B. 21C. 25D. none of these |
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Answer» Correct Answer - C |
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| 377. |
If `20` lines are drawn in a plane such that no two of them are parallel and so three are concurrent, in how many points will they intersect each other? |
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Answer» It is given that no two lines are parallel means all line are intersecting and no three lines are concurrent means three lines intersect at a point. `therefore` Number of point of intersection `=""^(20)C_(2) = (20"!")/(2"!" 18"!") = (20 xx 19 xx 18"!")/(2 xx 1 xx 18"!")` `= (20 xx 19)/(2) = 19 xx 10 = 190` |
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| 378. |
There are m copies each ofn different books in a university library. The number of ways in which one or more than one book can be selected isA. `m^(n)+1`B. `(m+1)^(n)-1`C. `(n+1)^(n)-m`D. m |
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Answer» There are (m+1) choices for each of n different boods. So, the total number of choices is `(m+1)^(n)` including one choice in which we do not select any book. Hence, the required number of ways `=(m+1)^(n)-1`. |
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| 379. |
There are `p`copies each of `n`different books. Find the number of ways in which a nonempty selection can bemade from them.A. npB. pnC. `(p+1)^(n)-1`D. `(n+1)p-1` |
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Answer» Correct Answer - C |
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| 380. |
A library has two books each having three copies and three other books each having tow copies. In how many ways can all these books be arranged in a shelf so that copies of the same book are not separated.A. `((a+b+c+d)!)/(a!b!c!)`B. `((a+2b+3c+d)!)/(a!(b!)^(2)(c!)^(3))`C. `((a+2b+3c+d)!)/(a!b!c!)`D. none of these |
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Answer» Correct Answer - B |
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| 381. |
A committee of 10 members is formed with members chosen from the faculties of arts, Economics, education, engineering, medicine and science. Number of possible ways in which the faculties representation be distributed on this committee, is : |
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Answer» Here, we have to select `10` positions from the `6` departments. So, first position can be filled in 6 ways. Similarly, second position can be filled in 6 ways. Similarly, all positions can be filled in `6` ways. `:.` Number of possible ways `10` positions can be filled ` = 6^10`. |
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| 382. |
Find the number of zeros at the end of 100!. |
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Answer» In terms of prime factors, 100! Can be written as `2^(a)*3^(b)*5^(c)*7^(d)` . . . Now, `E_(1)(100!)=[(100)/(2)]+[100/2^(2)]+[(100)/(2^(3))]+[(100)/(2^(4))]+[(100)/(2^(5))]+[(100)/(2^(6))]` `=50+25+12+6+3+1=97` and `E_(5)(100!)=[(100)/(5)]+[(100)/(5^(2))]` `=20+4=24` `therefore100!=2^(97)*3^(b)*5^(24)*7^(d)=2^(73)*3^(b)*(2xx5)^(24)*7^(d)` . . . `=2^(73)*3^(b)*(10)^(24)*7^(d)` Hence, number of zeros at the end of 100! is 24. or exponent of 10 in 100!=min(97,24)=24. |
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| 383. |
The exponent of 7 in `100C_50` is |
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Answer» We have, `""^(100)C_(50)=(1000!)/(50!50!)` Now, `E_(7)(100!)=[(100)/(7)]+[(100)/(7^(2))]=14+2=16` and, `E_(7)(50!)=[(50)/(7)]+[(50)/(7^(2))]=7+1=8` Thus, exponent of 7 in `""^(100)C_(50)=16-2xx8=0` |
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| 384. |
If a,b,c and d are odd natural numbers such that a+b+c+d=20, the number of values of the ordered quadruplet (a,b,c,d) isA. 165B. 455C. 310D. None of these |
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Answer» Correct Answer - A Let `a=2x-1,b=2y-1,c=2z-1,d=2w-1` where, x,y,z,w`inN` then, `a+b+c+d=20` `implies x+y+z+w=12` `therefore`Number of ordered quadruplet`=.^(12-1)C_(4-1)` `=.^(11)C_(3)=(11*10*9)/(1*2*3)=165` |
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| 385. |
The exponent of 7 in 100!, isA. 14B. 15C. 16D. none of these |
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Answer» We have, `E_(7)(100!)=[(100)/(7)]+[(100)/(7^(2))]=14+2=16` |
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| 386. |
If `a, b, c` are three natural numbers in AP such that `a + b + c=21` and if possible number of ordered triplet `(a, b, c)` is `lambda`, then the value of `(lambda -5)` is |
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Answer» Correct Answer - 8 `a+b+c=21implies3b=21impliesb=7" "[because a+c=2b]` `impliesa+b+c=21impliesa+c=14` `implieslamda=.^(14-1)C_(2-1)=.^(13)C_(1)=13` hence, `lamda-5=13-5=8` |
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| 387. |
The number of numbers divisible by 3 that can be formed by four different even digits, isA. 18B. 36C. 24D. 48 |
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Answer» There are 6 even digits viz. 0, 2, 4, 6, 8. We know that a numbers is divisible by 3 if the sum of its digits is divisible by 3. Therefore, numbers must be formed by the digits 0, 2, 4, 6 or by 0, 4, 6, 8. `:.` Required number of numbres `=(4!-3!)+(4!-3!)=36`. |
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| 388. |
The number of five digit even numbers that can be made with the digits, 0, 1, 2 and 3, isA. 384B. 192C. 768D. 576 |
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Answer» An even number formed with the digits 0, 1, 2 and 3 will have either 0 or 2 at units places. So uints place can be filled in `""^(2)C_(1)` ways. Now, remaining four places can be filled in `""^(3)C_(1)xx""^(4)C_(1)xx""^(4)C_(1)xx""^(4)C_(1)` ways. Hence, required number of numbers `=""^(2)C_(1)xx(""^(3)C_(1)xx""^(4)C_(1)xx""^(4)C_(1)xx""^(4)C_(1))=396` |
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| 389. |
If a,b, and c are positive integers such that a+b+c`le8`, the number of possible values of the ordered triplet (a,b,c) isA. 84B. 56C. 83D. None of these |
| Answer» Correct Answer - B | |
| 390. |
Statement-1: The expression `""^(40)C_(r)xx""^(60)C_(0)+""^(40)C_(r-1)xx""^(60)C_(1)+""^(40)C_(r-2)xx""^(60)C_(2)+....` attains its maximum volue when r = 50 Statement-2: `""^(2n)C_(r)` is maximum when r=n.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. |
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Answer» We know that when n is even `""^(n)C_(r)` is maximum for r=n. Therefore, `""^(2n)C_(r)` is maximum when r=n. So, statement-2 is true. Now, `""^(40)C_(r)xx""^(60)C_(0)+""^(40)C_(r-1)xx""^(60)C_(1)+""^(40)C_(r-2)xx""^(60)C_(2)+....` = Number of ways selecting r objects from a group of 40 objects of one kind adn 60 objects of other kind `=""^(100)C_(r)` This is maximum when r=50. So, statement-1 is also true. Clearly, statement-2 is a coorect explanation for statement-1. |
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| 391. |
If x, y and r are positive integers, then `""^(x)C_(r)+""^(x)C_(r-1)+""^(y)C_(1)+""^(x)C_(r-2)""^(y)C_(2)+......+""^(y)C_(r)=`A. `(x!y!)/(r!)`B. `((x!y!))/(r!)`C. `""^(x+y)C_(r)`D. `""^(xy)C_(r)` |
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Answer» Correct Answer - C |
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| 392. |
A box contains 2 white balls, 3 black balls and 4 red balls. The number of ways three balls be drawn from the box, if atleast one black ball is to be included in the draw is …….. .A. 129B. 84C. 64D. none of these |
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Answer» Correct Answer - C |
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| 393. |
A box contains 2 white balls, 3 black balls and 4 red balls. The number of ways three balls be drawn from the box, if atleast one black ball is to be included in the draw is …….. .A. 74B. 64C. 84D. 20 |
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Answer» Correct Answer - B |
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| 394. |
A box contains 2 white balls, 3 black balls and 4 red balls. The number of ways three balls be drawn from the box, if atleast one black ball is to be included in the draw is …….. . |
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Answer» Since, there are 2 white, 3 black and 4 red balls. It is given that atleast one black ball is to be included in the draw. `therefore` Required number of ways `= ""^(3)C_(1) xx ""^(6)C_(2) + ""^(3)C_(2) xx ""^(6)C_(1) + ""^(3)C_(3)` `= 3 xx 15 + 3 xx 6 + 1` `= 45 + 18 + 1 = 64` |
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| 395. |
Assuming the balls to be identical except for difference in colours, thenumber of ways in which one or more balls can be selected from 10 white, 9green and 7 black balls is(1) 880 (2) 629 (3) 630 (4) 879 |
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Answer» total ways= `11 xx 10 xx 8 = 880` ans`-> 880 - 1`(no one is selected) = 879 option 3 is correct |
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| 396. |
Howmany different words can be formed by jumbling the letters in the wordMISSISSIPPI in which no two S are adjacent?(1)`8""dot^6C_4dot^7C_4`(2) `6""dot""7""dot^8C_4`(3) `6""dot""8""dot^7C_4`(4) `7""dot^6C_4dot^8C_4` |
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Answer» 11 L`->`4I,4S,2P,1M `=> .^8C_4*(7!)/(4!*2!)` `= .^8C_4 *(7 xx 6!)/(4!*2!)` `= .^8C_4 .^6C_4 xx 7` option 4 is correct |
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| 397. |
Howmany different words can be formed by jumbling the letters in the wordMISSISSIPPI in which no two S are adjacent?(1)`8""dot^6C_4dot^7C_4`(2) `6""dot""7""dot^8C_4`(3) `6""dot""8""dot^7C_4`(4) `7""dot^6C_4dot^8C_4`A. `8xx""^(6)C_(4)xx""^(7)C_(4)`B. `6xx7xx""^(8)C_(4)`C. `6xx8xx""^(7)C_(4)`D. `7xx""^(6)C_(4)xx""^(8)C_(4)` |
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Answer» Correct Answer - D |
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| 398. |
A question paper consists of two sections having respectively, 3 and 5 questions. The followinng note is given on the paper "it is not necessory to attempt all the questions one questions from each section is compulsory". In how many ways can candidate select the questions? |
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Answer» Here, we have two sections A and B (say), the section A has 3 questions and section B has 5 questions and one question from each section is compulsory, according to the given direction. `therefore`Number of ways selecting one or more than one question from section A is `2^(3)-1=7` and number of ways selecting one or more than one questions from section B is `2^(5)-1=31` Hence, by the principle multiplication, the required number of ways in which a candidate can select the questions. `=7xx31=217`. |
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| 399. |
In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions? |
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Answer» selection for I & II when 3,5 selected `.^5C_3.^7C_5` when 4, 4 selected `.^5C_4.^7C_4` when 5,3 slected `.^5C_5.^7C_3` required = `(5!)/(3!2!) * (7!)/(5!2!) + (5!)/(4!1!)*(7!)/(4!3!) + (5!)/(5!0!)*(7!)/(3!4!)` = `7*6*5 + 5*7*5 + 7*5` `= 210 + 175 + 35` `= 210+210 = 420` answer |
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| 400. |
There are three sections in a question paper, each containing 5 questions. A candidate has to solve any 5 questions, choosing at least one from each section. Find the number of ways in which the candidate can choose the questions.A. `""^(15)C_(5)`B. `""^(3)C_(1)xx""^(12)C_(4)`C. 2250D. 2253 |
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Answer» We observe that Total number of ways of choosing at least one question from each section = Total number of ways of choosing 5 questions out of 15 questions - Number of ways of selecting 5 questions from 2 sections only - Number of ways of choosing 5 questions from one section only. `=""^(15)C_(5)-""^(3)C_(2)(""^10C_(5)-""^(2)C_(1)xx""^(5)C_(5))-""^(3)C_(1)xx""^(5)C_(5)` `=""^(15)C_(5)-3xx""^(10)C_(5)+6-3=2250` |
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