It would have helped if you had given some reason why you "think" those things!
When you don't understand something, look at simple examples. If A is a 2 by 2 matrix, it can be written as
[tex]\left[\begin{array}{cc}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right][/tex]
If no other conditions on A are given, you would be free to choose the four numbers to be anything you want- you have "four independent choices".
Now suppose A is symmetric- that means [itex]a_{12}= a_{22}[/itex]. You can still choose [itex]a_{11}[/itex] and [itex]a_{22}[/itex] to be anything you want and you can choose either [itex]a_{12}[/itex] or [itex]a_{21}[/itex] to be anything you want but as soon as you choose one of them the other must be the same- you have three independent choices.
Suppose A is skew-symmetric. Then [itex]a_{ij}= -a_{ji}[/itex]. No, you are NOT free to choose all entries independently! For one thing, on the diagonal you must have [itex]a_{ii}= -a_{ii}[/itex] which means each diagonal element must be 0! In our 2 by 2 case, [itex]a_{11}= a_{22}= 0[/itex]. We can choose either [itex]a_{12}[/itex] or [itex]a_{21}[/itex] but once you have chosen one of them, the other is the negative of that. You have only one independent choice.
Some of the things you will have to figure out to do this: in an n by n matrix, how many entries are there on the main diagonal? How many entries are there above the main diagonal? How many entries are the below the main diagonal? Think about the definitions of "symmetric" and "skew-symmetric" to see why you need to know those numbers and the answers to these questions should be easy.