Question -
Answer -
Given:
The digits 0, 1, 1, 5,9
Total number of digits= 5
So now, number greaterthan 50000 will have either 5 or 9 in the first place and will consists of 5digits.
Number of 5 digit numbersat first place = 4! / 2! [Since, 1 is repeated]
= [4×3×2!] / 2!
= 4 × 3
= 12
Similarly, number of 9digit numbers at first place = 4! / 2! = 12
The required number ofNumbers = 12 + 12 = 24
Hence, 12 differentnumbers can be formed.