Final answer:
To find the number of integers greater than 2000 with the digits 0, 1, 2, 3, 4, 5 without repetition, we calculate permutations for both four-digit and five-digit numbers. There are 240 four-digit numbers and 120 five-digit numbers, totaling 360 possible integers.
Step-by-step explanation:
The question asks us to calculate the number of integers greater than 2000 that can be formed using the digits 0, 1, 2, 3, 4, 5 without repetition. To ensure the integers are greater than 2000, the first digit must be either 2, 3, 4, or 5. Once the first digit is chosen, there are 5 remaining digits for the second place, 4 for the third place, and 3 for the fourth place, considering that zero cannot be the first digit of the integer.
Therefore, we calculate the number of combinations as follows:
4 options for the first digit (since the number has to be greater than 2000)
5 options for the second digit
4 options for the third digit
3 options for the fourth digit
The total number of combinations is the product of these options: 4 × 5 × 4 × 3 = 240.
However, this accounts only for four-digit numbers. Since we can also form five-digit numbers with the provided digits, we need to consider that scenario as well. For five-digit numbers, all five non-zero digits can be used as the first digit, followed by permutations of the remaining four digits.
Therefore, we can form additional combinations as follows:
5 options for the first digit
4 options for the second digit (after choosing the first digit)
3 options for the third digit
2 options for the fourth digit
1 option for the fifth digit
The total number of five-digit combinations is the product of these options: 5 × 4 × 3 × 2 × 1 = 120.
Adding both four-digit and five-digit combinations gives us the total number of integers: 240 + 120 = 360.