Partitioning Pawns

Relevant wiki: Combinations - Problem Solving

Because each of the 3 bags must hold at least 1 white pawn and 1 black pawn, we distribute it now, so there remain 5 white pawns and 5 black pawns (10 in total).

• Firstly, we calculate the number of possibilities, ignoring the maximum capacities. For each colour, there are 5 undistributed pawns which will be distributed into 3 bags, so there are $^{(5+3-1)}C_5=^{7}C_5=21$ possibilities. Therefore, for both colours, there are $21^{2}=441$ possibilities in total.

• Secondly, we calculate the number of possibilities which has a bag overfilled. Each bag has been filled with 2 pawns, so the red, blue, and yellow bag can hold 10, 9, and 8 more pawns, respectively.

  1. The red bag can contain all 10 remaining pawns, so it cannot be overfilled.
  2. Both the blue bag and the yellow bag will be overfilled if filled with all 10 remaining pawns. It is clear that there is only 1 possibility for each bag, so in total there are $2$ possibilities in this case.
  3. The yellow bag will also be overfilled if filled with 9 of the remaining pawns. The remaining pawn has 2 colour possibilities (black and white) and 2 place possibilities (the red bag and the blue bag), so in total there are $4$ possibilities in this case.

  Therefore, in total, there are $6$ possibilities which has a bag overfilled.

• Finally, from the $441$ possibilities obtained in the first part, we subtract the $6$ possibilities obtained in the second part, obtaining $435$ .

Therefore, there are $\boxed{435}$ ways to put the pawns in the bags.