How many four-digit odd numbers can be formed using the digits $0,2,3,5,6,8$ (each digit occurs only once)?

The correct option is D $96$
The digits from which we can choose digits is $0,2,3,5,6,8$, that is a total of $6$ digits
Since each desired number has to be odd , we must have any of $3,5$ at the units place. This can be done in $2$ ways
Also $0$ has to occupy only the tens or the hundreds place, else the number would not be a $4$ digit number. This can be done in $2$ ways.
Now, the thousands, hundred's , tens digit can be formed by any of the digits $2,6,8,$ and one of $3$ or $5$, that is by $4$ digits in ${}^4{P}_3=\frac{4!}{(4-3)!}=\frac{4!}{1!}=4\times 3\times 2=24$ ways
So, total number of $4$ digit odd numbers that can be formed $=24\times 2\times 2=96$ ways