Math Problem Statement

Using the digits 1,2,3,4,5 to create a 4 digit number sequence by combining the last two digits at the end of the sequence in order to get a 4 number result....examples.. 1, 2, 3, 45 or 2, 3, 4, 51 or 4, 5, 1, 23...or 4,5, 1, 32 list all possible combinations

Solution

We are tasked with finding all possible 4-digit combinations using the digits 1, 2, 3, 4, 5 where the last two digits can be combined to form a two-digit number. Here, the first two digits are kept separate, and the last two digits are combined into a two-digit number.

Let's list all the possible combinations systematically:

First digit possibilities:

  • 1, 2, 3, 4, 5

Second digit possibilities:

  • 1, 2, 3, 4, 5 (but different from the first)

Combined last two digits possibilities:

  • 12, 13, 14, 15, 21, 23, 24, 25, 31, 32, 34, 35, 41, 42, 43, 45, 51, 52, 53, 54

Total possible combinations:

  • Start with two digits, then combine the last two digits. We calculate all combinations.

Here is a systematic list of all valid combinations:

  1. 1, 2, 3, 45

  2. 1, 2, 4, 35

  3. 1, 2, 5, 34

  4. 1, 3, 2, 45

  5. 1, 3, 4, 25

  6. 1, 3, 5, 24

  7. 1, 4, 2, 35

  8. 1, 4, 3, 25

  9. 1, 4, 5, 23

  10. 1, 5, 2, 34

  11. 1, 5, 3, 24

  12. 1, 5, 4, 23

  13. 2, 1, 3, 45

  14. 2, 1, 4, 35

  15. 2, 1, 5, 34

  16. 2, 3, 1, 45

  17. 2, 3, 4, 15

  18. 2, 3, 5, 14

  19. 2, 4, 1, 35

  20. 2, 4, 3, 15

  21. 2, 4, 5, 13

  22. 2, 5, 1, 34

  23. 2, 5, 3, 14

  24. 2, 5, 4, 13

  25. 3, 1, 2, 45

  26. 3, 1, 4, 25

  27. 3, 1, 5, 24

  28. 3, 2, 1, 45

  29. 3, 2, 4, 15

  30. 3, 2, 5, 14

  31. 3, 4, 1, 25

  32. 3, 4, 2, 15

  33. 3, 4, 5, 12

  34. 3, 5, 1, 24

  35. 3, 5, 2, 14

  36. 3, 5, 4, 12

  37. 4, 1, 2, 35

  38. 4, 1, 3, 25

  39. 4, 1, 5, 23

  40. 4, 2, 1, 35

  41. 4, 2, 3, 15

  42. 4, 2, 5, 13

  43. 4, 3, 1, 25

  44. 4, 3, 2, 15

  45. 4, 3, 5, 12

  46. 4, 5, 1, 23

  47. 4, 5, 2, 13

  48. 4, 5, 3, 12

  49. 5, 1, 2, 34

  50. 5, 1, 3, 24

  51. 5, 1, 4, 23

  52. 5, 2, 1, 34

  53. 5, 2, 3, 14

  54. 5, 2, 4, 13

  55. 5, 3, 1, 24

  56. 5, 3, 2, 14

  57. 5, 3, 4, 12

  58. 5, 4, 1, 23

  59. 5, 4, 2, 13

  60. 5, 4, 3, 12

This gives 60 possible combinations in total.

Would you like details on any specific cases or further explanations?

Relative Questions:

  1. How would this process change if we could repeat digits?
  2. Can we generate 5-digit sequences by combining more than two digits?
  3. How many combinations would we get if we could combine any two digits (not necessarily the last two)?
  4. How do these combinations relate to permutations and combinations in combinatorics?
  5. What would happen if we allowed zeros in our sequences?

Tip:

In problems involving sequences or arrangements, organizing the choices systematically can help avoid overlooking possibilities and ensure all combinations are covered.

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Math Problem Analysis

Mathematical Concepts

Permutations
Combinatorics

Formulas

nPr = n! / (n - r)! (Permutation formula)

Theorems

Permutation theorem

Suitable Grade Level

Grades 9-12