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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. |
Thirty members attended a party. If each person shakes hands with every other person exactly once. Then find the number of handshakes made in the party.A. `.^(30)P_(2)`B. `.^(30)C_(2)`C. `.^(29).C_(2)`D. `.^(60)C_(2)` |
| Answer» Select two persons from 30 persons. | |
| 2. |
In how many ways can we select two vertices in a hexagon? |
| Answer» A hexagon has 6 vertices. Any 2 vertices can be selected in `.^(6)C_(2)` ways `=(6xx5)/(1xx2)=15` ways. | |
| 3. |
How many distinct positive integers are possible with the digits 1,3,5,7 without repetition? |
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Answer» Possible number of, Single-digit numbers =4 two-digit numbers =`4xx3=12` Three-digit numbers `=4xx3xx2=24` four-digit numbers `=4xx3xx2xx1=24`. Thus, total number of distinct positive integers without repetition `=4+12+24+24=64`. |
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| 4. |
In a plane there are 12 points, then answer the following questions. (a) Find the number of different straight lines that can be formed by joining these points, when no combination of 3 points are collinear. (b) Find the number of different straight lines that can be formed by joining these points, when 4 of these given points are collinear and no other combination of three points are collinear. (c) Find the number of different triangles that can be formed by joining these points, when no combination of 3 points are collinear. |
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Answer» (a) We know passing through two points in a plane we can draw only one line, i.e., we require to select any two points from the given 12 points which is possible in `.^(12)C_(2)` ways. `therefore` The number of different straight lines that can be formed by joining the given 12 points. `.^(12)C_(2)(12xx11)/(1xx2)=66`. (b) Given, out of the 12 points, 4 points are collinear. We know that collinear points form only one line. `therefore` These four points when they are not collinear will actually form `.^(4)C_(2)` lines, which are not forming here. `therefore` The number of the required lines `=.^(12)C_(2)-.^(4)C_(2)+1=66-6+1=61`. (c) We know, by joining three non-collinear points a triangle forms. `therefore` Three points can be selected from 12 points in `.^(12)C_(3)` ways. `therefore` The required number of triangles `=.^(12)C_(3)=(12xx11xx10)/(1xx2xx3)=220`. (d) Given 5 points are collinear. `rArr .^(5)C_(3)` triangles will not form. `therefore` The required number of triangles `=.^(12)C_(3)-.^(5)C_(3)=220-10=210`. (d) Given 5 points are collinear. `rArr .^(5)C_(3)` triangles will not form. `therefore` The required number of triangles `=.^(12)C_(3)-.^(5)C_(3)=220-10=210`. |
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| 5. |
Find `.^(n)C_(3)`, if `.^(n)C_(7)=.^(n)C_(4)`. |
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Answer» `.^(n)C_(7)=.^(n)C_(4)rArrn=7+4=11` So `.^(n)C_(3)=.^(11)C_(3)=(11xx10xx9)/(1xx2xx3)=165`. |
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| 6. |
Using the digits 0,1,2,5, and 7 how many 4-digit number that are divisible by 5 can be formed if repetition of the digits is not allowed?A. 38B. 46C. 32D. 42 |
| Answer» Unit place is 0 or 5 then the number is divisible by 5. | |
| 7. |
How many three-digit numbers that are divisible by 5, can be formed, using the digit 0,2,3,5,7, if no digit occurs more than once in each number?A. 10B. 15C. 21D. 25 |
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Answer» If a number is divisible by 5, then the units digit must be either 0 or 5. Case 1: If the units digit is 0, then the remaining two places can be done in `.^(4)P_(2)=12` ways. Case 2: If the units digits is 5, then the remaining two places can be filled by the remaining 4 digits. It can be done in `3xx3=9` ways. (`therefore` The hundreds place can not be filled with 0) `therefore` The required number of 3-digit numbers=12+9=21. |
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| 8. |
Find the number of diagonals of a polygon of 10 sides. |
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Answer» Assume that there are 10 points in a plane where no 3 of them are collinear which are the verticews of the given polygon, The number of different lines that can be formed by joining these 10 points is `.^(10)C_(2)`. We know in any polygon the lines joining non-adjacent vertices are called diagonals. Hence, the required number of diagonals=Number of lines formed -Number of sides of the polygon `=.^(10)C_(2)-10=35`. Using the formula, the number of diagonals in the above problem `=(10(10-3))/(2)=35`. |
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| 9. |
Using all the letters of the word QUESTION, how many different words can be formed?A. 8!B. 7!C. `7xx7!`D. 9! |
| Answer» `.^(n)P_(n)=n!` | |
| 10. |
In how many ways can 5 prizes be distributed to 3 students, it each students is eligible for any number of prizes?A. `3^(5)`B. `5^(3)`C. `.^(5)P_(3)`D. `.^(5)C_(3)` |
| Answer» First prize can be distributed in 3 ways similarly 2nd, 3rd, 4th and 5th prizes can be distributed each in 3ways. Now use the fundamental principle of counting. | |
| 11. |
If `.^(n)P_(r)=24.^(n)C_(r)` then r= _______A. 24B. 6C. 4D. 2 |
| Answer» `r!=(.^(n)C_(r))/(.^(nC_(r))`. | |
| 12. |
How many words can be formed from the letters of the word EDUCATION using any four letters in each word?A. 840B. 1680C. 2080D. 3050 |
| Answer» There are 8 letters in the word EQUATION . The number of words with 4 letters formed with the 8 letters is `.^(8)P_(4)=8xx7xx5=1680`. | |
| 13. |
The following are the steps involved in finding the value of `(n)/(r)` from `.^(n)P_(r)=1320`. Arrange them in sequential order.A. `.^(n)P_(4)=(12!)/(9!)=(12!)/((12-3)!)`B. `rArr(n)/(r)=(12)/(3)=4`C. `rArr.^(n)P_(r)=.^(12)P_(3)`D. `.^(n)P_(r)=1320=12xx11xx10` |
| Answer» DACB is the required sequential order. | |
| 14. |
The following are the steps invovled in solving `.^(n)C_(2)=36` for n. Arrange then in sequential order.A. `n^(2)-n-72=0`B. As `sgt0,n=9`C. `n=0,n=-8`D. `(n-9)(n+8)=0` |
| Answer» EADCB is the required sequential order. | |
| 15. |
In how many ways can a prime or an odd number be chosen from {1,2,3,4,5,6,7,8,9,10}? |
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Answer» We form two sets P and O as follows. `P={2,3,5,7}` (primes) and O={1,3,5,7,9} (odd numbers). On applying the general form of Sum Rule we get. `n(PcupO)=n(P)+n(O)-n(PcapO)=4+5-3=6`. We note that the numbers 3,5 and 7 are counted among primes and also among odd numbers. So, we discount 3 (common numbers) from the sum `n(P)+n(O)`. |
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| 16. |
A={1,2,3,4) and B={a,e,i,o,u) are two sets. In how many ways can a number from a or a letter from B be chosen? |
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Answer» As no element of A is in B, we can apply the sum rule of disjoint counting. `thereforen(AcupB)=n(A)+n(B)=4+5=9`. |
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| 17. |
How many different outcomes arise from first tossing a coin and then rolling a die? |
| Answer» There are 2 possibilities (i.e., head or tail) for the first task (tossing a coin) and after each of these outcomes there are 6 possibilities (i.e., any number from 1 to 6) for the second task (rolling a dice.) thus, by the product rule, there are `2xx6`, i.e., 12 possible outcomes, for the given compound task. | |
| 18. |
There are 10 railway stations between a station X and another station Y. Find the number of different tickets that must be printed so as to enable a passenger to travel from one station to any other. |
| Answer» Including X and Y there are 12 stations. From any one station to any other, we need 11 different types of tickets . Since there are 12 stations , the different tickets possible are (12)(11)=132. | |
| 19. |
There are 18 stations between Hyderabad and Bangalore. How many second class tickets have to be printed, so that a passanger can travel from one station to any other station?A. 380B. 190C. 95D. 100 |
| Answer» Total number of stations =20. Select 2 stations from 20 stations. | |
| 20. |
There are four different white balsl and four different black balls. The number of ways that balls can be arranged in a row so that white and black balls are placed alternately is_____A. `(4!)^(2)`B. `2(4!)^(2)`C. `4!`D. `(4!)^(3)` |
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Answer» We have to arrange 4 different white balls and 4 different balck balls as shown below (i) WB WB WB WB (ii) BW BW BW BW This can be done in `4!xx4!+4!+4!xx4!` ways, i.e., `2(4!)^(2)`. |
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| 21. |
The number of ways of selecting five members to form a committee from 7 men and 10 women is ____,A. 5266B. 6123C. 6188D. 8123 |
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Answer» Here, the total number of persons is 17. The number of ways of selecting 5 members from 17 members in `.^(17)C_(5)=6188`. |
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| 22. |
In how many ways can we select two vowels and three consonants from the letters of the word ARTICLE?A. 12B. 14C. 18D. 22 |
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Answer» There are three vowels and four consonants in the word ARTICLE. The number of ways of selecting 2 vowels and 3 consonants from 3 vowels and 4 consonants `.^(3)C_(2)xx.^(4)C_(3)=3xx4=12`. |
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| 23. |
How many different odd numbers are formed using the digits (2,4,0,6)? (Repetition digits is not allowed).A. 16B. 0C. 24D. 108 |
| Answer» All the given digits are even, so no odd number can be formed with the given digits. | |
| 24. |
A class has 20 boys and 15 girls . If one representative from each gender has to be chosen, in how many ways can this be done? |
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Answer» Task 1: Choosing a representative from boys. This can be done in 20 ways. Task 2: Choosing a representative from girls. This can be done in 15 days. By the product rule, the number of ways of performing the two tasks is `20xx15` ,i.e., 300 ways. |
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| 25. |
From 8 boys and 5 girls, a delegation of 5 students is to be formed. Find the number of ways this can be done such that delegation must contain exactly 3 girls.A. 140B. 820C. 280D. 410 |
| Answer» Select 3 girls from 5 girls and 2 boys from 8 boys then apply fundamental principle. | |
| 26. |
In a class there are 20 boys and 15 girls. In how many ways can 2 boys and 2 girls be selected?A. `.^(35)C_(4)`B. `.^(35)C_(2)`C. `.^(20)C_(2)xx.^(15)C_(2)`D. `20xx15` |
| Answer» r objects can be selected from b objects in `.^(n)C_(r)` ways. Now use the fundamental principle. i.e., task `T_(1)` can be done in m ways and task `T_(2)` can be done in n ways, then the two tasks can be done simultaneously in mn ways. | |
| 27. |
In how many wasys can 3 consonents be selected from the English alphabet?A. `.^(21)C_(3)`B. `.^(26)C_(3)`C. `.^(21)C_(5)`D. `.^(26)C_(5)` |
| Answer» Select 3 consonants from 21 consonants. | |
| 28. |
There are 15 stations from New Delhi to Mumbai. How many first class tickets can be printed to travel from one station to any other station?A. 210B. 105C. 240D. 135 |
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Answer» (i) The total number of stations are 15 (say n). (ii) On a ticket, two station should be printed. (iii) The required number of ways `=.^(n)C_(2)xx2`. |
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| 29. |
In how many can 3 consonants and 3 vowels be selected from the letters of the word TRIANGLE?A. 25B. 13C. 30D. 20 |
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Answer» (i) Find the number of ways in which 3 consonants and 2 vowels can be selected from 5 consonatns and 3 vowels. (ii) Then use fundamental theorem of counting. |
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| 30. |
There are 15 points in a plane. No three points are collinear except 5 points. How many different straight lines can be formed?A. 105B. 95C. 96D. 106 |
| Answer» The number of lines that can be formed from n points in which m points are collinear is `.^(n)C_(2)-.^(m)C_(2)+1`. | |
| 31. |
In how many ways can 4 consonants be chosen from the letters of the word SOMETHING?A. `.^(9)C_(4)`B. `.^(6)C_(4)`C. `.^(4)C_(4)`D. `.^(4)C_(3)` |
| Answer» There are 6 consonants and 4 are to be chosen. | |
| 32. |
Using all the letters of the word PROBLEM, how many words can be formed such that the consonants occupy the middle place?A. 3000B. 4200C. 720D. 3600 |
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Answer» (i) There are 5 consonants and middle place can be filled in 5 ways. (ii) Remaining places can be filled with remaining letters. |
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| 33. |
How many different words can be formed using all the letters of the word SPECIAL, so that the consonants always in the odd positions?A. 112B. 72C. 24D. 144 |
| Answer» First arrange the consonants in odd places is in 1,3,5 and 7 places. Now arrange the consonants in odd places is in 1, 3,5,7 places. Now arrange the vowels in the remaining places and then use the fundamental principla. | |
| 34. |
A man has 7 trousers and 10 shirts. How many different outfits can be wear? |
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Answer» Task 1: Choosing a representative from boys. Task 2: Choosing a representative from girls. This can be done in 15 ways. By the product rule, the number of ways of performing the two tasks is `20xx15` , i.e., 300 wasys. |
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| 35. |
There are 12 points in a plane, no three points are collinear except 6 points. How many different triangles can be formed?A. 200B. 201C. 220D. 219 |
| Answer» The number of triangles that can be formed from n points in which m points are collinear is `.^(n)C_(3)-.^(m)C_(2)`. | |
| 36. |
A plane contains 20 points of which 6 are collinear. How many different triangle can be formed with these points?A. 1120B. 1140C. 1121D. 1139 |
| Answer» The number of triangles that can be formed from n points in which in points are colinear is `.^(n)C_(3)-.^(m)C_(3)`. | |
| 37. |
Using the letters of the word ENGLISH, how many five letters words can begin with G?A. 2520B. 360C. 180D. 1260 |
| Answer» As Y is filled in last blank, five more letters are to be selected from the remaining 8 letters. | |
| 38. |
Using the letters of the word PRIVATE. How many 6-letter words can be formed which begin with P and edn with E?A. 3!B. 4!C. 7!D. 5! |
| Answer» As the letters begin with P and ends with E, four more letters are to be selected from the remaining 5 letters. | |
| 39. |
How many 5-digit numbers that are divisible by 5 can be formed using the digits (0,1,3,5,7,5)? (Each digit can be repeated any number of times)A. 1080B. 2160C. 6480D. 3175 |
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Answer» (i) If unit digit is 0 or 5, then the number is division by 5. (ii) If the units digit is 5, then ten thousands digit cannot be zero, now find the number of ways the other four digits can be arranged. (iii) Similarly whe the units digit is 0, the other 4 digits can be arranged in `.^(5)P_(4)` ways. use the fundamental theorem of counting. |
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| 40. |
There is a three-digit password and it is known that each digit can have four values 5,6,7 or 8 . If there is exactly one correct password, how many distinct wrong passwords are there?A. 63B. 80C. 81D. 64 |
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Answer» (i) Each digit of the password can have 4 values, hence first digit can be filled in 4 ways. (ii) Find the number of ways in which remaining digits can be filled. (iv) Use the fundamental principal of counting and find total number of passwords that rae formed. |
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| 41. |
How many four-digits even numbers can be formed using the digits (3,5,7,9,1,0) (Repetition of digits is not allowed).A. 120B. 60C. 360D. 100 |
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Answer» (i) The units digit of the required number is 0. (ii) Find the number of ways in which the remaing 5 digits can be arranged in three places by using `.^(n)P_(r)`. |
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| 42. |
Using all the letter of the work EDUCATION. How many words can be formed which begin with DU? (Repetition is not allowed).A. 8!B. 7!C. 6!D. 9! |
| Answer» Find the number of 7 letter words using the 7 letters. | |
| 43. |
A telephone number has seven digits, no number starts with 0. In a city, how many different telephone numbers the formed using the digits 0 to 6? (Each digit can occur only once)A. 6!B. 6.6!C. 7!D. 2.7! |
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Answer» (i) The first digit of the number cannot be zero, so it can filled in 6 ways. (ii) Now the second digit can be any of the 6 digits, i.e., it can be filled in 6 ways. (iii) As the digits cannot be repeated, the number of ways the third digit can be filled is 5 ways and so on. (iv) Now apply the fundamental theoram. |
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| 44. |
In how many ways can 4 letters be posted in 3 letter boxes?A. `4^(3)`B. `3^(4)`C. 6!D. 4 |
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Answer» (i) Each letter be posted in 3 ways. (ii) Now calculate the number of ways in which four letters can be posted by using fundamental theorem of counting. |
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| 45. |
Using the letters of the word PUBLIC, how many four letter words can be formed which begin with B and with P? (Repetition of letters is not allowedA. 360B. 12C. 24D. 30 |
| Answer» Take 6 blanks. First baanks can be filled with the remaining letters. | |
| 46. |
Consider a,b,c,d, List all combination taken 3 at a time. |
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Answer» The list includes abc, abd, acd, bcd. Here, abc,bca,cab are regarded the same as order is not important. The number of combination of n things taken r at a time is denoted by `.^(n)C_(r)`. |
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| 47. |
In a library there as 10 research scholars. In how many ways can we select 4 of them? |
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Answer» Out of 10 scholars, we can select 4 of them in `.^(10)C_(4)` ways. `.^(10)C_(4)=(10xx9xx8xx7)/(1xx2xx3xx4)=210` ways. |
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| 48. |
Using the letters of the word CHEMISTRY, how many six letter words can be formed, which end with Y?A. `.^(8)P_(6)`B. `.^(9)P_(6)`C. `.^(9)P_(5)`D. `.^(8)P_(5)` |
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Answer» (i) Take 6 blanks. (ii) As Y is filled in last blank, five more letters are to be selected from the remaining 8 letters. |
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| 49. |
The number of the words that can be formed using all the letters of the word BRAIN such that it starts with R and but does not end with A.A. 18B. 14C. 16D. 20 |
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Answer» Since the first place of the word always starts with R. and last place is not A. the last place can be filled by any one of the remaining three letters. The remaining 3 places can be filled with 3 letters in 3! Ways. `therefore` The total number of words=`3xx3! =18`. |
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| 50. |
How many 4-digit odd number can be formed using the digit 0,2,3,5,6,8 (each digit occurs only once)?A. 64B. 72C. 86D. 96 |
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Answer» If a number is odd, then the units digit of the number must contain any one of the numbers 1,3,5,7 or 9. Given digits are 0,2,3,5,6,8. Here units digit must contain either 3 or 5,i.e.,2 ways. First place can be filled in 4 ways (since ten thousand pace can not be filled with 0). Second place and thrid digit can be filled in 4 and 3 ways respectively. The required number of odd numbers =`4xx4xx3xx2=96`. |
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