In how many ways can 6 players be assianed \( t_{6} 6 \) positions on a base ball tram assuming that any plager can play any position?

To determine the number of ways to assign 6 players to 6 positions on a baseball team, we can think of this as a permutation problem. Since each player can occupy any position and there are no restrictions, we can assign the players to the positions in the following way: 1. The first position can be filled by any of the 6 players. 2. The second position can then be filled by any of the remaining 5 players. 3. The third position can be filled by any of the remaining 4 players. 4. The fourth position can be filled by any of the remaining 3 players. 5. The fifth position can be filled by any of the remaining 2 players. 6. The sixth position can be filled by the last remaining player. Thus, the total number of ways to assign the players to the positions is given by the product: \[ 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 6! \] Calculating \( 6! \): \[ 6! = 720 \] Therefore, the total number of ways to assign 6 players to 6 positions is \( \boxed{720} \).