1. $[\mathbf{1 0}]$ How many ways are there to place pawns on an $8 \times 8$ chessboard, so that there is at most 1 pawn in each horizontal row? Express your answer in the form $p_{1}^{e_{1}} \cdot p_{2}^{e_{2}} \cdots$, where the $p_{i}$ are distinct primes and the $e_{i}$ are positive integers.
Answer: $3^{16}$ If there is at most 1 pawn in each row, then each row of the chessboard may have either 0 or 1 pawn somewhere in the row. There is 1 case if there are no pawns in the row. There are 8 possible cases if there is 1 pawn in the row, one case for each square in the row. Hence for each row, there are 9 possible pawn arrangements. There are 8 rows, thus we have $9^{8}=3^{16}$.