How many numbers greater than 24000 can be formed using the digits 7, 0, 9, 2 and 8 such that each digit is used exactly once?
Explanation:
The number must be greater than 24000 and each digit is to be used exactly once.
⇒ all the numbers will be 5 digits numbers.
So, numbers starting with 7, 8 and 9 will always be greater than 24000.
In this case the 1st digit can be selected in 3 ways. The remaining 4 digits can be selected in 4! ways.
∴ Total number of possible numbers when the 1st digit is 7, 8 or 9 = 3 × 4 × 3 × 2 × 1 = 72.
Now, when the 1st digit is 2, the second digit cannot be 0. The 2nd digit can be any of 7, 8 or 9. So, the 2nd digit can be selected in 3 ways. The remaining 3 digits can be selected in 3! ways.
So, total number of possible numbers when the 1st digit is 2 = 1 × 3 × 3 × 2 × 1 = 18
The number cannot start with 0.
∴ Total number of required numbers = 72 + 18 = 90
Hence, option (d).
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