How many odd numbers can be formed using the digits $0, 4, 5, 7$? How many odd numbers can be formed using digits $0,4,5,7$.
I am getting answer $12$ but the actual answer is $14$. 

Suppose the number has to be of $n$ digits. Then the last digit has to be $5$ or $7$. Hence the solution is 
$$
4^{n-1} \cdot 2=2^{2n-1}.
$$
If the number has at most $n$ digits then the answer is
$$
\sum_{i=1}^n2^{2i-1}=2\sum_{i=0}^{n-1}4^i=\frac{2}{3}(4^n-1).
$$

Probably the correct interpretation:
We have to use exactly one each digit $0,4,5,7$. Then the last digit can be $5$ or $7$. The other can be chosen in $3!$ ways. Therefore the answer $2\cdot 3!=12$.
[Moreover, if we add the number has to have exactly $4$ digits (i.e. it doesn't start with $0$), then the answer would be $8$]
Ps. I still don't get why it could be $14$ :P
