How many distinct patterns can be formed using 3 chesspieces on a 8x8 standard board?

Question

Scenario 1:

You have 3 same colored pawns, and there are no restrictions as to their placement. How many unique patterns can be formed? By unique, I mean that the patterns can't coincide by mere rotation of the board.

Scenario 2:

Same as above, but all 3 pieces are different from one another.

Thanks for the help.

Edit: please provide a formula and explain your thinking. If the problem had involved only one piece, the answer would be 20, but that can be achieved through simple logic.

Answer 1

How do you came up with 20 possibilities for one piece? The board has 64 squares and regardless where you place it, the obtained position will be equivalent to exactly three other positions modulo rotations, so you have $64/4 = 16$ possibilities if you count equivalent positions only once.

Now I think every position with three pieces is equivalent to exactly three different positions as well, so the answers should be ${64 \choose 3} / 4$ (place 3 pieces on 64 squares = ${64 \choose 3}$ possibilities, but each position is counted 4 times, so divide by 4) and $\bigl({64 \choose 3} / 4\bigr)*3*2$ for the second part.