Assuming that 9 players are needed as follows: one pitcher, one catcher, 7 other field positions, there is a choice of 3 for the pitcher, 2 for the catcher, making 3*2=6 combinations of these two positions alone. There is a choice of 4 out of 7 for the remaining positions: 7*6*5*4/(4*3*2*1)=35, but there are insufficient players to fill all 7 positions.
In all, then, combining the two combinations there are 6*35=210 ways.
To explain why there are 35 ways of choosing the remaining 4 players, call the 4 other players A, B, C, D. A is picked first and can be assigned to any one of the remaining 7 positions; then B is picked to fill one of the remaining 6; then C for one of the remaining 5 and finally D for one of the remaining 4. That makes 7*6*5*4=840. However, we picked A, B, C and D in order, but any order would suffice, and there are 4*3*2*1=24 ways of ordering the four players. So we need to reduce 840 by this amount: 840/24=35.