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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.
| 1. |
The number of different numbers that can be formed by using all the digits 1, 2, 3, 4,3, 2,1 so that odd digits always occupy the odd places IS 1) 6 2)72 3) 60 4) 18A. 3!4!B. 34C. 18D. 12 |
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Answer» Correct Answer - C |
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| 2. |
There are 10 girls and 8 boys in a class room including Ms. Rani and Radha. A list of speakers consisting of 8 girls and 6 boys has to be prepared. Mr. Ravi refuses to speak if Ms. Rani is a speaker. Ms. Rani refuses to speak if Ms. Radha is a speaker . The number of ways the list can be prepared isA. 202B. 308C. 567D. 952 |
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Answer» Following cases arise: CASE I When Ms. Radha is a speaker In this case, Ms. Rani refuses to speak. So, we have to select 7 girls out of remaning 8 girls and 6 boys from 8 boys. The numbr of ways of doing this is `""^(8)C_(7)xx""^(8)C_(6)=224`. CASE II When Radha is not a speaker In this case, two subcases arise. (i) When Rani is a speaker (ii) When Rani is not a speaker. If Rani si a speaker, Mr Ravi refused to speak. So, we have to select 7 girls from remaining 8 girls and 6 boys from remaining 7 boys which can be done in `""^(8)C_(7)xx""^(7)C_(6)` ways = 56 ways. If Rani is not a speaker, then 8 girls are to be chosen out of remainig 8 girls and 6 boys from 8 boys which can be done in `""^(8)C_(6)xx""^(8)C_(8)=28` ways. Hence, required number of ways =224+56+28=308. |
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| 3. |
Ten persons numbered 1, ,2..,10 play a chess tournament, each player against every other player exactlyone game. It is known that no game ends in a draw. If `w_1, w_2, , w_(10)`are thenumber of games won by players 1, ,2 3, ...,10, respectively, and `l_1, l_2, , l_(10)`are thenumber of games lost by the players 1, 2, ...,10, respectively, thena. `sumw_1=suml_i=45`b. `w_1+1_i=9`c. `sumw l1 2()_=81+suml1 2`d. `sumw l i2()_=suml i2`A. `sumw_(i)^(2)+81-suml_(i)^(2)`B. `sumw_(i)^(2)+81=suml_(i)^(2)`C. `sumw_(i)^(2)=suml_(i)^(2)`D. None of these |
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Answer» Correct Answer - C Clearly, each player will play 9 games. `therefore`Total number of games=`.^(10)C_(2)=45` clearly, `w_(i)+l_(i)=9 and sumw_(i)=suml_(i)=45` `impliesw_(i)=9-l_(i)impliesw_(1)^(2)=81-18l_(i)+l_(i)^(2)` `impliesumw_(i)^(2)=sum81-18suml_(i)+suml_(1)^(2)` `=81xx10-18xx45+suml_(i)^(2)=suml_(1)^(2)`. |
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| 4. |
In how many ways 5 different balls can be distributed into 3 boxes so that no box remains empty? |
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Answer» The required number of ways `=3^(5)-.^(3)C_(1)(3-1)^(5)+.^(3)C_(2)(3-2)^(5)-.^(3)C_(3)(3-3)^(5)` `=243-96+3-0=150` Or Coefficient of `x^(5)` inn `5!(e^(x)-1)^(3)` =Coefficient of `x^(5)` in `5!(e^(3x)-3e^(2x)+3e^(x)-1)` `=5!((3^(5))/(5!)-3xx(2^(5))/(5!)+3xx(1)/(5!))=3^(5)-3*2^(5)+3=243-96+3=150` |
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| 5. |
The total number of ways in which a beggar can be given atleast one rupee from four 25 paise coins, three 50 paise coins and 2 one rupee coins are |
| Answer» No. of 25 paise coins = 4No. of 50 paise coins = 3No. of 1 rupee coins = 2Total coins = 9Total no. of ways to give some or none rupees to a beggar `= (4+1)*(3+1)*(2+1) = 5*4*3 = 60` waysCases of 0 paisa = 1Cases of 25 paise = 1Cases of 50 paise = 2Cases of 75 paise = 2Subtracting above cases as they are less than one rupee cases from the total cases:`= 60 - 1 - 1 - 2 - 2 = 54` | |
| 6. |
A lady gives a dinner party to 5 guests to be selected from nine friends. The number of ways of forming the party of 5, given that two of the friends will not attend the party together, isA. 56B. 126C. 91D. none of these |
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Answer» Correct Answer - C |
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| 7. |
There are 10 persons named `P_1, P_2, P_3 ..., P_10`. Out of 10 persons, 5 persons are to be arranged in a line such that is each arrangement `P_1` must occur whereas `P_4` and `P_5` do not occur. Find the number of such possible arrangements. |
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Answer» Given that `P_(1), P_(2),…,P_(10)`, are 10 persons, out of which 5 persons are to be arranged but `P_(1)` must occur whereas `P_(4) " and " P_(5)` never occur. `therefore` Selection depends on only 10 - 3 = 7 persons As, we have already occur `P_(1)`, Therefore, we have to select only 4 persons out of 7. Number of selection `= ""^(7)C_(4) = (7"!")/(4"!"(7-4)"!") = (7"!")/(4"!" 3"!") = (5040)/(24 xx 6) = 35` `therefore` Required number of arrangement of 5 persons `= 35 xx 5"!" = 35 xx 120 = 4200` |
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| 8. |
There are 10 lamps in a hall.Each one of them can be switched on independently.Find the number of ways in which the hall can be illuminated.A. 102B. 1023C. 210D. 10! |
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Answer» Correct Answer - B |
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| 9. |
There are 10 lamps in a hall.Each one of them can be switched on independently.Find the number of ways in which the hall can be illuminated.A. `10^(2)`B. 1023C. `2^(10)`D. 10! |
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Answer» Correct Answer - B |
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| 10. |
A hall has 12 gates. In how many ways can a man enter the hall through one gate and come out through a different gate? |
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Answer» Since, there are 12 ways of entering into the hall. After entering into the hall, the man come out through a different gate in 11 ways. Hence, by the fundamental principal of multiplication, total number of ways is `12xx11=132` ways. |
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| 11. |
The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given byA. `7!xx5!`B. `6!xx5!`C. 30D. 50 |
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Answer» Correct Answer - B |
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| 12. |
There are 5 gentlemen and 4 ladies to dine at a round table. In how many ways can they as themselves so that no that ladies are together ?A. 2880B. 576C. 1440D. 480 |
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Answer» Five gentlemen can be seated at a round table in (5-1)!=4! Ways. Now 5 places are created in which 4 ladies can be arranged in `""^(5)P_(4)` ways. Hence, the total number of ways in which no two ladies sit together `=4!xx""^(5)P_(4)xx4!xx5!""=2880`. |
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| 13. |
From 6 different novels and 3 differentdictionaries, 4 novels and 1 dictionary are to be selected and arranged in arow on a shelf so that the dictionary is always in the middle. Then thenumber of such arrangements is(1) less than 500(2) at least 500 but less than 750(3) at least 750 but less than 1000(4) at least 1000A. less than 500B. at least 500 but less than 750C. at least 750 but less than 1000D. at least 1000 |
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Answer» Out of 6 novels, 4 novels can be selected in `""^(6)C_(4)` ways and 1 dictionary can be chosen out of 3 dictionaries in `""^(3)C_(1)` ways. Since 1 dictionary is to be fixed in the middle and a pair of novels on its either side. This can be done in `""^(4)C_(2)xx2!xx2!` ways. `:.` Required number of arrangements `""^(6)C_(4)xx""^(3)C_(1)xx""^(4)C_(2)xx2!xx2!=1080` |
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| 14. |
A person goes in for an examination in which there are four papers with a maximum of m marks from each paper. The number of ways in which one can get 2m marks isA. `.^(2m+3)C_(3)`B. `((1)/(3))(m+1)(2m^(2)+4m+1)`C. `((1)/(3))(m+1)(2m^(2)+4m+3)`D. None of these |
| Answer» Correct Answer - C | |
| 15. |
How many different numbers which are smallerr than `2xx10^(8)` and are divisible by 3, can be written by means of the digits 0,1 and 2? |
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Answer» 12,21 . . . 122222222 are form the required numbers we can assume all of them to be nine digit in the form. `a_(1),a_(2),a_(3),a_(4),a_(5),a_(6),a_(7),a_(8),a_(9)` and can use 0 for `a_(1),a_(2)` and `a_(0)` and `a_(0),a_(1),a_(2)` and `a_(3)` . . and so on to get 8-digit, 7-digit, 6digit numbers etc. `a_(1)` can assume one of the 2 values of 0 or 1. `a_(2),a_(3),a_(4),a_(5),a_(6),a_(7),a_(8)` can assume any of 3 values 0,1,2. the number for which `a_(1)=a_(2)=a_(3)=a_(4)=a_(5)=a_(6)=a_(7)=a_(8)=a_(9)=0` must be eleminated. the sum of first 8-digits i.e., `a_(1)+a_(2)+ . . .+a_(8)` can be in the form of `3n-2` or 3n-1 or 3n. in each case `a_(9)` can be chosen from 0,1,2 in only 1 way so that the sum of all 9-digits in equal to 3n. `therefore`Total numbers=`2xx3^(7)xx1-1=4374-1=4373`. |
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| 16. |
Let `n and k` be positive such that `n leq (k(k+1))/2`.The number of solutions `(x_1, x_2,.....x_k), x_1 leq 1, x_2 leq 2, ........,x_k leq k`, all integers, satisfying `x_1 +x_2+.....+x_k = n`, is ....... |
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Answer» We have, `x_(1)+x_(2)+ . . +x_(k)=n` . . . (i) Now, let `y_(1)=x_(1)-1,y_(2)=x_(2)-2, . . ,y_(k)=x_(k)-k` `thereforey_(1)ge0,y_(2)ge0, . . ,y_(k)ge0` On substituting the values `x_(1),x_(2), . . ,x_(k)` in terms of `y_(1),y_(2), . . ,y_(k)` In Eq. (i), we get `y_(1)+1+y_(2)+2+ . . +y_(k)+k=n` `impliesy_(1)+y_(2)+ . . +y_(k)=n-(1+2+3+ . .+k)` `thereforey_(1)+y_(2)+ . . +y_(k)=n-(k(k+1))/(2)=A` (say) . . . (ii) The number of non-negative integral solutions of the Eq. (ii) is `=.^(k+A-1)C_(A)=((k+A-1)!)/(A!(k-1)!)` where, `A=n-(k+1))/(2)` |
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| 17. |
If the letters of the word SACHIN are arranged inall possible ways and these words are written out as in dictionary, then theword SACHIN appears at serial number602 (2) 603(3) 600(4) 601A. 603B. 602C. 601D. 600 |
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Answer» Correct Answer - C Words starting with A,C,H,I,N are each equals to 5! `therefore`Total words `=5xx5!=600` the first word starting with S is SACHIN. `therefore`SACHIN appears in dictionary at serial number 601. |
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| 18. |
Prove that the no. of all permutations of n different objects taken r at a time when a particular object is to be always included in each arrangement is `r.(n-1)P_(r-1)`A. `""^(n)C_(r)xxr!`B. `""^(n-1)C_(r-1)xx(r-1)!`C. `""^(n-1)C_(r-1)xxr!`D. `""^(n-1)C_(r)xxr!` |
| Answer» In order to arrange n distinct items by taking r at a time when a particular object is to be always included in each arrangement, let us first put aside the sepecified item and select (r-1) items from the remaining (n-1) items. This can be done in `""^(n-1)C_(r-1)` ways. Now, we have r items, namely, one specified item and (r-1) selected items. These r items can be arranged in r! ways. Hence, the required number of arrangements is `""^(n-1)C_(r-1)xxr!` | |
| 19. |
There are unlimited number of identical balls of four different colours. How many arrangements of at most 8 balls in row can be made by using them?A. 21845B. 87380C. 262140D. none of these |
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Answer» Since there are balls of four different colours. Therefore, Number of arrangements of one ball =4 Number of arrangements of two balls `=4xx4=4^(2)` Number of arrangements of three balls `=4xx4xx4=4^(3)` etc. `:.` Required number of arrangements `=4+4^(2)+4^(3)+......+4^(8)` `=4((4^(8)-1)/(4-1))=(4)/(3)(4^(8-1))=87380` |
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| 20. |
Let p=2520, x=number of divisors of p which are multiple of 6, y=number of divisors of p which are multiple of 9, thenA. x=12B. x=24C. y=12D. y=16 |
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Answer» Correct Answer - B::D `becausep=2520=2^(3)*3^(2)*5^(1)*7^(1)` `=6*2^(2)*3^(1)*5^(1)*7^(1)=9*2^(3)*5^(1)*7^(1)` `thereforex=(2+1)(1+1)(1+1)(1+1)=24` and `y=(3+1)(1+1)(1+1)=16`. |
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| 21. |
The number of ways of selecting 10 balls from unlimited number of red, black, white and green balls, isA. 286B. 84C. 715D. none of these |
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Answer» Correct Answer - A |
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| 22. |
There are three teams x,x+1 and y childrens and total number of childrens in the teams is 24. if two childrens of the same team do not fight, thenA. maximum number of fights is 190B. maximum number of fights is 191C. maximum number of fights occur when x=7D. maximum number of fights occur when x=8 |
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Answer» Correct Answer - B::C::D `becausex+x+1+y=24` `impliesy=23-2x` Let N=Total number of fights subject to the condition that any two children of one tem do not fight. `thereforeN=2.^(24)C_(2)-(.^(x)C_(2)+.^(x+1)C_(2)+.^(y)C_(2))` `=.^(24)C_(2)-(.^(x)C_(2)+.^(x+1)C_(2)+.^(23-2x)C_(2))` `=23-3x^(2)+45x` `therefore(dN)/(dx)=0-6x+45` for maximum or minimum, put `(dN)/(dx)=0impliex=7.5` `impliesx=7" "[because x in I]` Now, `(d^(2)N)/(dx^(2)) lt0` `therefore` N will be maximum when x=7 and N=23-`3(7)^(2)+45xx7=191` |
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| 23. |
All possible products are formed from the numbers `1, 2, 3, 4, ..., 200` by selecting any two without repetition. The number of products out of the total obtained which are multiples of 5 is :A. 5040B. 7180C. 8150D. none of these |
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Answer» Correct Answer - B |
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| 24. |
If `n`denotes thenumber of ways of selecting `r`objects ofout of `n`distinctobjects `(rgeqn)`withunlimited repetition but with each objet included at least once in selection,then `n`m is equal isa. `^r-1C_(r-n)`b. `^r-1C_n`c.`^r-1C_(n-1)`d. none of theseA. `.^(r-1)C_(r-n)`B. `.^(r-1)C_(n)`C. `.^(r-1)C_(n-1)`D. `.^(r-1)C_(r-n-1)` |
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Answer» Correct Answer - A::C `becausex_(1)+x_(2)+x_(3)+ . . .+x_(n)=r, Aa x_(i)ge1,(1 le I le )` Total number of such solutions `=.^(r-1)C_(n-1)=.^(r-1)C_(r-n)` |
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| 25. |
If the number of ways of selecting n cards out of unlimited number of cards bearing the number 0,9,3, so that they cannot be used to write the number 903 is 96, then n is equal toA. 3B. 4C. 5D. 6 |
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Answer» Correct Answer - C We cannnot write 903. If in the selection of n cards, we get either (9 or 3), (9 or 0), (0 or 3), (only 0), (only 3) or (only 9). For (9 or 3) can be selected `=2xx2xx2xx . . .xxn` factors `=2^(n)` Similarly, (9 or 0) or (0 or 3) can be selected `=2^(n)` In the above selection (only 0) or (only 3) or (only 9) is repeated twice. `therefore`Total ways `=2^(n)+2^(n)+2^(n)-3=93` `implies3*2^(n)=96implies2^(n)=32=2^(5)` `thereforen=5`. |
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| 26. |
The total number of ways in which 4 boys and 4 girls can form a line, with boys and girls alternating, isA. (4!)2B. 8!C. 2(4!)2D. `4!.""^(5)P_(4)` |
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Answer» Correct Answer - C |
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| 27. |
The number of ways in which seven persons can be arranged at a round table if two particular persons may not sit together, isA. 480B. 120C. 80D. none of these |
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Answer» Correct Answer - A |
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| 28. |
Five boys and five girls from a line. Find the number of ways of making the seating arrangement under the following condition: |
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Answer» (i) Boys and girls alternate total arrangement `= (5"!")^(2) + (5"!")^(2)` (ii) No two girls sit together `= 5"!"6"!"` (iii) All the girls sit together `= 2"!" 5"!" 5"!"` (iv) All the girls are never together `= 10"!" - 5"!" 6"!"` |
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| 29. |
Five boys and five girls form a line with the boys and girlsalternating. Find the number of ways of making the line.A. `(5!)^(2)`B. `10!`C. `5!+5!`D. `2(5!)^(2)` |
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Answer» 5 boys can be arranged in a linen in 5! Ways. Since the boys and girls are alternating. So, corresponding each of the 5! Ways of arrangements of 5 boys we obtain 5 places marked by cross as shown below: (i) `B_(1)xxB_(2)xxB_(3)xxB_(4)xxB_(5)` (ii) `xxB_(1)xxB_(2)xxB_(3)xxB_(4)xxB_(5)` Clearly, 5 girls can be arranged in 5 places marked by cross in `(5!+5!)` ways. Hence, the total number of ways of making the line `=5!xx(5!+5!)=2(5!)^(2)` |
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| 30. |
The number of ways in which four persons be seated at a round table, so that all shall not have the same neighbours in any two arrangements,isA. 24B. 6C. 3D. 4 |
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Answer» Correct Answer - C |
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| 31. |
In how many ways can 24 persons be seated round a table, if there are 13 sets? |
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Answer» In case of circular table, the clockwise and anti-clockwise order are different, the required number of circular permutations=`(.^(24)P_(13))/(13)=(24!)/(13xx11!)` `impliesn!=nxx`number of circular arrangements of n different things `implies`Number of circular arrangements of n different things `=(n!)/(n)=(n-1)!` Hence, the number of circular permutations of n different things taken all at a time is (n-1)!, if clockwise and anti-clockwise orders are taken as different. |
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| 32. |
Consider 21 different pearls on a necklace. How many ways can the pearls be placed in on this necklace such that 3 specific pearls always remain together? |
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Answer» After fixing the places of three pearls, treating 3 specific pearsl=1 units. So , we have now 18 pearls+1 unit=19 and the number of arrangement will be (19-1)!=18! also, number of ways of 3 pearls can be arranged between themselves is 3!=6. since, there is no distinction between the clockwise and anti-clockwise arrangement. So, the required number of arrangements`=(1)/(2)18!*6(18!)` |
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| 33. |
If 11 members of a committee sit at a round table so that the president and secretary always sit together, then the number of arrangements, isA. `9!xx2`B. `10!`C. `10!xx2`D. `11!` |
| Answer» Correct Answer - A | |
| 34. |
In how many ways 7 men and 7 women can be seated around a round table such that no two women can sit togetherA. 7!B. `7!xx6!`C. `(6!)^(2)`D. `(7!)^(2)` |
| Answer» Correct Answer - B | |
| 35. |
How many necklace of 12 beads, each can be made from 18 beads of various colours? |
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Answer» In the case of necklace, there is no distinction between the clockwise and anti-clockwise arrangement, the required number of circular permutations. `=(.^(18)P_(12))/(2xx12)=(18!)/(6!xx24)=(18xx17xx16xx15xx14xx13!)/(6xx5xx4xx3xx2xx1xx24)=(119xx13!)/(2)` |
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| 36. |
Find number of ways that 8 beads o different colors bestrung as a necklace.A. 2520B. 2880C. 4320D. 5040 |
| Answer» Correct Answer - A | |
| 37. |
Find the number of ways in which 12 different beads can be arranged to form a necklace. |
| Answer» 12 different beads can be arranged among themselves in a circular order in (12-1)!=11! Ways. Now, in the case of necklace, there is not distinction between clockwise and anti-clockwise arrangements. So the required number of arrangements=`(1)/(2)(11!)`. | |
| 38. |
Suppose that six students, including Madhu and Puja, are having six beds arranged in a row. Further, suppose that Mudhu does not want a bed adjacent to Puja. Then the number of ways, the beds can be allotted to students isA. 384B. 264C. 480D. 600 |
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Answer» Six students, including Madhu and Puja, can be adjusted on six beds in 6! ways. Puja and Madhu can be adjusted on two adjacent beds in 2! ways. Considering these adjacent beds as one and the remaining 4 beds can be arranged beds in `5!xx2!` ways. Hence, Number of ways in which Puja and Madhu are not on adjacent beds `=6!-5!xx2!""=480`. |
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| 39. |
Six identical coins are arranged in a row. The total number of ways in which the number of heads is equal to the number of tails isA. 9B. 20C. 40D. 120 |
| Answer» Correct Answer - B | |
| 40. |
The number of words of that can be made by writing down the letters of the word CALCULATE such that each word starts and ends with a consonant, isA. `(3)/(2)(7)!`B. `2(7)!`C. `(5)/(2)(7)!`D. `3(7)!` |
| Answer» Correct Answer - C | |
| 41. |
The number of ways in which thirty five apples can be distributed among 3 boys so that each can have any number of apples, is :A. 1332B. 666C. 333D. none of these |
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Answer» Correct Answer - B |
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| 42. |
The total number of ways in which 11 identical apples can be distributed among 6 children, isA. 252B. 462C. 42D. none of these |
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Answer» Correct Answer - D |
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| 43. |
Four boys picked 30 apples. The number of ways in which they can divide then if all the apples are identical is |
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Answer» Clearly, 30 mangoes can be distributed among 4 boys such that each boy can receive any number of mangoes. Hence, total number of ways=`.^(30+4-1)C_(4-1)` `=.^(33)C_(3)=(33*32*31)/(1*2*3)=5456` |
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| 44. |
Find the number of ways in which 8 non-identical apples can bedistributed among 3 boys such that every boy should get at least 1 apple andat most 4 apples. |
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Answer» no of ways the apples can be distributed into 3 (1,3,4) ; (2,3,3) ; (2,2,4) now, applying COMBINATION, `(3!.^8C_1 * .^7C_3*.^4C_4)+ (3!*.^8C_2*.^6C_3*.^3C_3) + (3!*.^8C_2 * .^6C_2 * .^4C_4)` `= 3!*((8!7!)/(7!*3!*4!) + (8!*6!)/(2!*6!*3!*3!) + (8!*6!)/(2!*6!*2!*4!)` `= 3!*(8!)/(3!*4!) + (8!)/(2!*3!*3!) + (8!)/(2!*2!*4!)` |
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| 45. |
How many 10-digit numbers can be formed by using digits 1 and 2 |
| Answer» there are total `2^10` numbers. | |
| 46. |
A student is allowed to select at most n books from a collection of (2n+1) books. If the total number of ways in which a student selects at least one book is 63. then n equals to -A. 6B. 3C. 4D. none of these |
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Answer» Correct Answer - B |
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| 47. |
The total number of ways in which 5 balls of differert colours can be distributed among 3 persons so thai each person gets at least one ball is |
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Answer» arrangement ways: (1,1,3), (1,2,2) total no of ways : `5!/((1!)^2*(3!)*(2!))*3! + 5!/(1!*2!^2*2!)*3!` =60+90 = 150 |
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| 48. |
There are 10 points in a plane, out of these 6 are collinear. The number of triangles formed by joining these points, isA. `N gt 190`B. `N le 100`C. `100 lt N le 140`D. `140 lt N le 190` |
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Answer» Correct Answer - B Number of triangles`=.^(10)C_(3)-.^(6)C_(3)` `impliesN=(10*9*8)/(1*2*3)-(6*5*4)/(1*2*3)impliesN=120-20impliesN=100` `thereforeN le 100` |
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| 49. |
Five persons entered the lift cabin on the ground floor of a 8 floor house. Suppose each of them can leave the cabin independently at any floor beginning with the first. The total number of ways in which each of the five person can leave the cabin at any one of the 7 floor, isA. `5^(7)`B. `7^(5)`C. `35`D. 2520 |
| Answer» Correct Answer - B | |
| 50. |
The total number of ways in which 5 balls of differert colours can be distributed among 3 persons so thai each person gets at least one ball isA. 75B. 150C. 210D. 243 |
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Answer» Correct Answer - B `{:("Objects", "Groups", "Objects", "Groups"),("Distinct", "Distinct", "Identical", "Identical"),("Distinct", "Identical", "Identical", "Distinct"):}` Description of Situation Here, 5 distinct balls are distributed amongst 3 persons so that each gets at least one ball. i.e. `"Distinct " to " Distinct"` So, we should make cases `"Case I " {:(A, B, C),(1, 1, 2):}} " ""Case II " {:(A, B, C),(1, 2, 2):}}` Number of ways to distribute 5 balls `= (""^(5)C_(1) * ""^(4)C_(1) * ""^(3)C_(3) xx (3!)/(2!)) + (""^(5)C_(1) * ""^(4)C_(2) * ""^(2)C_(2) xx (3!)/(2!))` `= 60 + 90 = 150` |
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