How many 3 letter words can be formed using Abcde if repetition is not allowed?
Page 171: Letters ABCDE, forming strings of length 3 Show 34. With repetitions: 5^3 35. Without repetitions: 5*4*3 36. Beginning with A, with
repetitions: 5^2 37. Beginning with A, without repetitions: 4*3 38. Without A, with repetitions: 4^3 39. Without A, without repetitions: 4*3*2 40. With A, with repetitions: 5^3-4^3 41. With A, without repetitions: 5*4*3 � 4*3*2 = 3*4*3 Page 172: Five computer science books, three math books, and two art books 54. All orderings: 10! 55. All orderings, putting CS books on left, math books in center, and art books on right: 5!*3!*2! 56. All orderings, putting CS books on left: 5!*5! 57. 3!*5!*3!*2! a. Arrange the subjects (3!) b. Arrange the cs books (5!) c. Arrange the math books (3!) d. Arrange the art books (2!) 58. 8!*C(9,2)*2! a. Arrange the cs and math books (8!) b. Pick which two of the 9 positions between (and outside) the previously arranged books to use for art books C(9,2) c. Pick which art book goes in each position (2!) 59. 60. Functions from X to Y, with |X| = n and |Y| = m: mn Page 182: Number of strings formed
by ordering the letters ABCDE: 10. Containing ACE: 3! 11. Containing A, C, and E together in any order: a. Order A, C, E: 3! b. Order p(ACE) with B and D: 3! c. Multiplication: 3!*3! 12. Containing DB and AE: 3! 13. Either AE or EA: 4!*2 14. A appears before D: C(5,2)*3! = 5!/(3!*2!)*3! = �*5! 15. Neither AB nor CD: a. Contains AB: 4! b. Contains CD: 4! c. Contains both AB and CD: 3! d. Contains either AB or CD: 2*4! � 3! e. Contains neither: 5! � (2*4! � 3!) 16. Containing neither AB nor BE: a. Contains AB: 4! b. Contains BE: 4! c. Contains both: 3! d. Contains either AB or BE: 2*4! � 3! e. Doesn�t contain either 5! � (2*4! � 3!) 17. A before C and C before E: C(5,3)*2! 18. Either DB or BE (or both): a. DB: 4! b. BE: 4! c. Both: 3! d. Either: 2*4! � 3! =================================================== Page 183: Club of six men and six women: 31. Committee of five: C(12,5) 32. Three men, four women: C(6,3)*C(6,4) 33. Four persons with at least one woman: a. Four men: C(6,4) b. Four persons: C(12,4) c. Subtract: C(12,4) � C(6,4) 34. Four persons with at most one man: a. Four women: C(6,4) b. Three women and a man: C(6,3)*C(6,1) c. Addition: C(6,4)+C(6,3)*C(6,1) 35. Four persons with at least one of each sex: C(12,4) � 2*C(6,4) 36. Four persons, not containing both Mabel and Ralph a. Four persons, containing both: C(10,2) b. Subtract: C(12,4)-C(10,2) 37. 4/10 GOP, 3/12 Dem., 2/4 Ind: C(10,4)*C(12,3)*C(4,2) 38. C(8,3) 39. Three zeros in a row in an 8-bit string: 6 40. Poker hands 41. Four aces: 48 42. Four of a kind: 13*48 43. All spades: C(13,5) 44. Cards of exactly two suits: C(4,2)*C(26,5) 45. Cards of all suits: Not restricted to any three suits a. Cards of no suits: 0 b. Cards of 1 suit: C(4,1)*C(13,5) c. Cards of two suits: C(4,2)*C(26,5) d. Cards of three suits: C(4,3)*C(39,5) e. None of the above: C(52,5) � C(4,3)*C(39,5) � C(4,2)*C(26,5) � C(4,1)*C(13,5) 46. A2345 in one suit: 4 47. Consecutive and of the same suit: 9*4 (A low) 48. 9*4^4 49. 13*C(4,2)*12*C(4,2)*11 ?*C(3,2) Bridge hands 50. Page 209: Number of strings by re-ordering letters
Comics are Action, Superman, Captain Marvel, Archie, X-man, Nancy
Categories
Bicycles
Books
Red, blue, and green balls
x1+x2+x3=15
How many threeSo, the required number of 3-letter words =(5×4×3)=60.
How many 3 letter combinations are possible from ABCD?Total possible arrangement of letters a b c d is 24. together. as one entity so we have total 3 letters.
How many words of 3 letters no repeat are there?There are 15,600 different 3-letter passwords, with no letters repeating, that can be made using the letters a through z.
How many ways are there to form 3 letter sequence using the letters A B C D E F without repetition and not containing a *?Without any restrictions on the number of repetitions, we found 216 three-letter words.
|