Question

Let S = 1; 2; 3; 4; 5(a) List all the 3-permutations of S.(b) List all the 3-combinations of S.

Answer 1

To list all the 3-permutations of S, select 3 numbers from the set S and arrange them in different orders. To list all the 3-combinations of S, select 3 numbers from the set S without considering their order.

Step-by-step explanation:

To list all the 3-permutations of S, we need to select 3 numbers from the set S = {1, 2, 3, 4, 5} and arrange them in different orders. The 3-permutations are:

1. 123
2. 124
3. 125
4. 132
5. 134
6. 135
7. 142
8. 143
9. 145
10. 152
11. 153
12. 154
13. 213
14. 214
15. 215
16. 231
17. 234
18. 235
19. 241
20. 243
21. 245
22. 251
23. 253
24. 254
25. 312
26. 314
27. 315
28. 321
29. 324
30. 325
31. 341
32. 342
33. 345
34. 351
35. 352
36. 354
37. 412
38. 413
39. 415
40. 421
41. 423
42. 425
43. 431
44. 432
45. 435
46. 451
47. 452
48. 453
49. 512
50. 513
51. 514
52. 521
53. 523
54. 524
55. 531
56. 532
57. 534
58. 541
59. 542
60. 543

To list all the 3-combinations of S, we need to select 3 numbers from the set S = {1, 2, 3, 4, 5} without considering their order. The 3-combinations are:

1. 123
2. 124
3. 125
4. 134
5. 135
6. 145
7. 234
8. 235
9. 245
10. 345

Answer 2

To list all the 3-permutations of S, we select 3 elements from the set S without repetition and order matters. The number of possible outcomes is 60. To list all the 3-combinations of S, we select 3 elements from the set S without repetition and order does not matter. The number of possible outcomes is 10.

Step-by-step explanation:

(a) To list all the 3-permutations of S, we need to select 3 elements from the set S without repetition and order matters. We can use the formula for permutations to calculate the number of possible outcomes: P(S,3) = 5! / (5-3)! = 5! / 2! = 5 × 4 × 3 = 60.

The 3-permutations of S are: (1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 3, 2), (1, 3, 4), (1, 3, 5), (1, 4, 2), (1, 4, 3), (1, 4, 5), (1, 5, 2), (1, 5, 3), (1, 5, 4), (2, 1, 3), (2, 1, 4), (2, 1, 5), (2, 3, 1), (2, 3, 4), (2, 3, 5), (2, 4, 1), (2, 4, 3), (2, 4, 5), (2, 5, 1), (2, 5, 3), (2, 5, 4), (3, 1, 2), (3, 1, 4), (3, 1, 5), (3, 2, 1), (3, 2, 4), (3, 2, 5), (3, 4, 1), (3, 4, 2), (3, 4, 5), (3, 5, 1), (3, 5, 2), (3, 5, 4), (4, 1, 2), (4, 1, 3), (4, 1, 5), (4, 2, 1), (4, 2, 3), (4, 2, 5), (4, 3, 1), (4, 3, 2), (4, 3, 5), (4, 5, 1), (4, 5, 2), (4, 5, 3), (5, 1, 2), (5, 1, 3), (5, 1, 4), (5, 2, 1), (5, 2, 3), (5, 2, 4), (5, 3, 1), (5, 3, 2), (5, 3, 4), (5, 4, 1), (5, 4, 2), (5, 4, 3).

(b) To list all the 3-combinations of S, we need to select 3 elements from the set S without repetition and order does not matter. We can use the formula for combinations to calculate the number of possible outcomes:

C(S,3) = 5! / (3! × (5-3)!) = 5! / (3! × 2!) = 5 × 4 / 2 = 10.

The 3-combinations of S are: (1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 3, 4), (1, 3, 5), (1, 4, 5), (2, 3, 4), (2, 3, 5), (2, 4, 5), (3, 4, 5).