In how many ways can we place 10 identical red balls (R) and 10 identical blue balls (B) into 4 distinct urns if each urn has at least 1 ball?
My attempt: Put the minimum number of balls into each urn:
case 1: R R R R (4C0, or 4 choose 0) ways to place 4 red balls, remaining 6R balls can be placed in 4 urns in (6+4-1)C(4-1) or (9 choose 3) ways, and 10B can be placed in (10+4-1)C(4-1) or (13 choose 3) ways, then total ways for this case 4C0*9C3*13C3 = 1*84*286 = 24024
case 2: B R R R, R B R R , etc, total of 4C1 ways to have 1 blue ball and 3 red ones, then for remaining 7R and 9B - 4C1*10C3*12C3 = 4*120*220 = 105600
case 3: B B R R , B R B R etc - following same logic 4C2*11C3*11C3 = 163350
case 4: 4C3*12C3*10C3 = 105600
case 5: 4C4*13C3*9C3 = 24024
Total ways = 24024+105600+163350+105600+24024 = 422598 However, the answer doesn't seem to be correct. What am I doing wrong?