How many $6$ digits odd numbers greater than $60,0000$ can be formed from the digits $5,6,7,8,9,0$ if Repetitions are not allowed:

For greater than $600000$, the starting digit should be greater than or equal to $6$, and for odd numbers, the ending digit will be $5,7,8,9$.
Consider 
6_ _ _ _ 5
The middle digits can be filled by remaining $4$ numbers $(0,7,8,9)$ in $4!$ ways.
Similarly the ending digits can be $7,9.$
Hence total number of ways will be $3\times 4!$.
Similarly for 8_ _ _ _5
Total number of ways will be $3\times 4!$ ways.
Now if the starting digit is odd and greater than $6$.
7_ _ _ _ 5.
The ending digit can be only $9$, other than $5$. The middle digits can be filled by $4$ numbers is $4$ ways. Hence total number of permutations will be 
$=2\times (4!)$.
Similarly for 9_ _ _ _ 5 and so on.
Permutations will be $2\times 4!$.
Hence the required number of odd numbers greater than $600000$ will be 
$=(3\times 4!)+(2\times 4!)+(3\times 4!)+(2\times 4!)$
$=(2\times 3\times 4!)+(2^2\times 4!)$
$=4!(6+4)$
$=10\times 4!$
$=240$.