How Many Numbers Less Than 1000 Can Be Formed Using Specific Digits?

The problem at hand is to determine how many unique numbers less than 1000 can be formed using the digits 5, 6, 7, 8, and 9 without repeating any digit. This question can be approached by considering the number of 1-digit, 2-digit, and 3-digit numbers that can be formed. Let's break it down step-by-step.

1-Digit Numbers

Since we are only using one digit, we can straightforwardly count the possibilities. We have 5 choices (5, 6, 7, 8, and 9).

Count

5 5 6 7 8 9

2-Digit Numbers

For 2-digit numbers, we can choose any of the 5 digits for the first position, and then any of the remaining 4 digits for the second position. The total number of combinations can be calculated as:

5 times; 4 20

3-Digit Numbers

For 3-digit numbers, we can choose any of the 5 digits for the first position, then any of the remaining 4 digits for the second position, and any of the remaining 3 digits for the third position. The total number of combinations is:

5 times; 4 times; 3 60

Total Count

Adding up the counts of 1-digit, 2-digit, and 3-digit numbers:

5 (1-digit) 20 (2-digit) 60 (3-digit) 85

Brute Force Approach Using J Programming Language

To verify this result, we can use the J programming language to perform a brute force calculation:

 5{5 6 7 8 9n./:~~.10.3{.85

The result is 85, confirming our earlier calculation.

Listing the Formed Numbers

The list of all numbers less than 1000 formed using the digits 5, 6, 7, 8, and 9 without repetition includes:

n5 6 7 8 9 56 57 58 59 65 67 68 69 75 76 78 79 85 86 87 89 95 96 97 98 567 568 569 576 578 579 586 587 589 596 597 598 657 658 659 675 678 679 685 687 689 695 697 698 756 758 759 765 768 769 785 786 789 795 796 798 856 857 859 865 867 869 875 876 879 895 896 897 956 957 958 965 967 968 975 976 978 985 986 987

Understanding the Mathematical Approach

We can also approach this problem from a mathematical perspective, considering combinations and permutations. The total natural numbers less than 1000 that can be formed from 5 digits (5, 6, 7, 8, and 9) without repetition can be calculated as the sum of combinations of 5 numbers taken 1, 2, and 3 at a time.

The formula for combinations is given by:

(binom{n}{k} frac{n!}{k!(n-k)!})

Where (n! 1) for (n 0) and (n! n times (n-1)!) for (n in Z^ ).

Summing these up:

[text{Total} binom{5}{1} binom{5}{2} binom{5}{3}]

Calculating individually:

(binom{5}{1} 5) (binom{5}{2} 10) (binom{5}{3} 10)

The total is:

(5 10 10 25)

However, this approach counts only the combinations, not considering all permutations. The actual count for numbers less than 1000 is 85, as derived through the step-by-step combination of digits.

Note: The mathematical approach using combinations only counts distinct number sets, while the combinatorial approach considers all permutations of the digits, leading to the correct count of 85.

Conclusion

In conclusion, the total number of unique numbers less than 1000 that can be formed using the digits 5, 6, 7, 8, and 9 without repeating any digit is 85. This can be verified through both a step-by-step counting method and a programming approach.

Further Reading

To explore similar problems and techniques in combinatorics, you may want to look into:

Combinatorial mathematics Permutations and combinations Number theory

Understanding these concepts will help in solving more complex problems and optimizing search engine optimization (SEO) strategies for web content.