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Permutation and combination level 2 questions for aptitude

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- 1. C.S.VEERARAGAVAN 98948 34264 veeraa1729@gmail.com PERMUTATION LEVEL TWO
- 2. PERMUTATION 12 June 2015C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 2 When a dice is rolled for n times, then the number of total out comes as (1) 6 (2) 6n-1(3) 6n (4)6n+1 The number of total out comes is 6n 01
- 3. PERMUTATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 3 When a coin is tossed for (n-1) times, then the number of total out comes is (1) 2 (2) 2n (3) 2n+1 (4)2n-1 the number of total out comes is 2n-1 02
- 4. PERMUTATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 4 Find the value of 35P34 (1) 35! (2) 35 (3) 34! (4) 34 05
- 5. PERMUTATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 5 If nCr = nPr then the value of r can be (1) 0 (2) 1 (3) 2 (4) More than one of the above nPr = nCr .* r! Hence r could be zero or one. 07 (4) More than one of the above
- 6. PERMUTATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 6 The number of ways of arranging 6 persons in a row is (1) 6! (2) (-1)! (3) 6 (4) 08
- 7. PERMUTATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 7 How many words can be formed using all the letters of the word GINGER? (1) 720 (2) 240 (3) 380 (4)360 Number of letters = 6 G repeated two times. Hence number of words = 6! 2! =3 x 4 x 5 x 6 =360 09
- 8. PERMUTATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 8 How many different words, which begin with N can be formed using all the letters of the word COUNTRY? (1) 720 (2) 120 (3) 24 (4)5040 The word COUNTRY has 7 letters. Since the first letter is N, the remaining 6 places can be filled with 6 letters in 6! Ways. = 720. 11
- 9. PERMUTATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 9 How many three-digit numbers can be formed using the digits {1,2,3,4,5} , so that each digit is repeated any number of times? (1) 150 (2) 200 (3) 25 (4)125 Consider 3 blanks Since there are 5 digits, each blank can be filled in 5 ways. Total number of ways is (5) (5) (5) = 125. 14
- 10. PERMUTATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 10 All possible four-digit numbers, with distinct digits are formed, using the digits {1,3,4,5,6}. How many of them are divisible by 5? (1) 8 (2) 12 (3) 24 (4)20 Consider four blanks The units place is filled with 5. The remaining three blanks can be filled with 4 digits in 4P3 ways. The number of four digit numbers required is 4 (3) (2) or 24. 16
- 11. PERMUTATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 11 In how many ways can 3 boys and 2 girls be seated in arrow, so that all girls sit together? (1) 12 (2) 24 (3) 84 (4)48 Let All the girls be one unit. There are 3 boys and 1 unit of girls arranged in 4! Ways. The two girls can be arranged among themselves in 2! Ways. Total number of arrangements 4! 2! = 24 (2) = 48. 17
- 12. PERMUTATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 12 In how many ways can 4 boys be seated in 6 chairs ? (1)180 (2) 720 (3) 360 (4)240 4 boys be seated in 6 chairs in 6P4 = 360 ways. 22
- 13. PERMUTATION WITH RESTRICTIONS 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 13 A basket contains 3 bananas and 4 apples. In how many ways can one select one or more fruits? (1) 17 (2)20 (3) 18 (4)19 Bananas can be selected in 4 ways and Apples can be selected in 5 ways Number of ways of selecting fruits is 4 x 5 = 20. Number of ways of selecting one or more = 20 – 1 = 19. 30
- 14. PERMUTATION WITH RESTRICTIONS 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 14 How many different words can be formed using the letters of the word MARKET so that they begin with K and end with R? (1) 42 (2) 24 (3) 12 (4)64 Since the first place and the last place are to be filled with K and R, the remaining four places can be arranged In 4! Or 24 ways. 26
- 15. COMBINATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 15 When a coin is tossed for n times then the number of ways of getting exactly ‘r’ heads is (1) 2r (2) nCr (3) nPr (4)2n 03
- 16. COMBINATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 16 Find the value of 120C119 (1) 119! (2) 120! (3) 120 (4)119 04
- 17. COMBINATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 17 If nC5=nC7, then 2n+1C2 is (1) 250 (2) 300 (3) 240 (4)280 nCr=nCs r + s = n 5 + 7 = 12 25C2 = 25 X 24 2 = 300 06
- 18. COMBINATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 18 The number of ways of selecting 4 members from a group of 10 members so that one particular member is always included is (1) 63 (2) 72 (3) 84 (4)56 Since one particular member is always included, we have to select 3 members from 9 members. This can be done in 9C3 = 84 ways. 10
- 19. COMBINATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 19 In how many ways can two consonants be selected from the English alphabet? (1) 420 (2) 105 (3) 210 (4)300 There are 21 consonants. Two consonants can be selected from 21 consonants in 21C2 = 210 12
- 20. COMBINATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 20 A bag contains 3 white balls, 4 green balls and 5 red balls. In how many ways two balls can be selected? (1) 132 (2) 66 (3) 33 (4)76 Total balls = 12. 12C2 = 66 ways. 13
- 21. COMBINATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 21 Ten points are selected on a plane, such that no three of them are collinear. How many different straight lines can be formed by joining these points? (1) 54 (2) 45 (3) 90 (4)108 A straight line is formed by joining any two points. Two points can be selected from 10 points in 10C2 ways. i.e., 45 ways. 18
- 22. COMBINATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 22 If the number of diagonals of a polygon is 5 times the number of sides, the polygon is (1) 13 (2) 20 (3) 15 (4)17 Let the number of sides be n. Number of diagonals = n(n−3) 2 Given n(n−3) 2 = 5n, N-3 = 10 N = 13 19
- 23. COMBINATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 23 Rahul has 6 friends. In how many ways can he invite 5 or more friends for dinner? (1) 1 (2) 6 (3) 7 (4)8 5 friends can be invited in 6C5 ways. 6 friends can be invited in 6C6 ways. Total = 6C5 + 6C6 = 6 + 1 = 7 29
- 24. COMBINATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 24 In how many ways can four letters be selected from the word EDUCATION ? (1)9C8 (2)9C7 (3) 9C4 (4) 9C6 The number of letters in the word EDUCATION is 9. 4 letters can be selected from these 9 letters in 9C4 ways. 24
- 25. COMBINATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 25 In how many ways can 3 blue balls be selected from a bag which contains 4 white balls and 6 blue balls ? (1) 20 (2) 10 (3) 120 (4)210 The number of blue balls is 6. 3 balls can be selected from 6 blue balls in 6C3 ways = 20 27
- 26. CIRCULAR PERMUTATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 26 In how many ways can 5 men and 3 women be seated around a circular table? (1) 720 (2) 5040 (3) 4020 (4)2520 N persons can be positioned around a circle in (n-1)! ways. Hence 8 persons can be arranged in 7! = 5040 ways. 20
- 27. 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 27 How many odd numbers can be formed using the digits {0,2,4,6} ? (1)0 (2) 192 (3) 18 (4)20 Since all given numbers are even, we could not get any odd number. 15
- 28. CIRCULAR PERMUTATION 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 29 If n books can be arranged on an ordinary shelf in 720 ways, then in how many ways can these books be arranged on a circular shelf ? (1)120 (2) 720 (3) 360 (4)60 N books can be arranged in n! ways. Given n! = 720 = 6! N = 6 6 books can be arranged on a circular shelf in (6-1)! =5! 120 ways. 23
- 29. COMPOUND EVENT 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 30 In how many ways can 4 letters be posted into 3 letters boxes ? (1)32 (2) 64 (3) 27 (4)81 Each letter can be posted in 3 ways. Four letters can be posted in 3 x 3 x 3 x 3 or 81 ways. 25
- 30. COMPOUND EVENT 12 June 2015 C.S.VEERARAGAVAN 9894834264 VEERAA1729@GMAIL.COM 31 When two coins are tossed and a cubical dice is rolled, then the total outcomes for the compound event is? (1)42 (2) 24 (3) 28 (4)10 When two coins are tossed there are 4 possible outcomes. When a dice is rolled there are 6 possible outcomes. Hence there are 4 x 6 = 24 outcomes. 28

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