Correct Answer - Option 2 : 6
For numbers greater than 30000, 2 will not be placed as the first digit. So there are 4 places to fill with 2, 2, 3, 3.
Distinct numbers greater than 30000 formed = \(\frac{4!}{2!\times 2!}\)
= \(\frac{24}{2\times 2}\) = 6
Hence, option 2 is correct.
The number of ways of arranging n objects, of which p of one type are alike, q of a second type are alike, r of a third type are alike, is \(\frac{n!}{p!q!r!}\)
