Six white and six black balls of the same size are distributed among ten urns so that there is at least one ball in each urn. What is the number of different distributions of the balls?
Case 1: (6 w , 4 bl) -->Assume 6 white balls are taking the first six urns (urn 1-6) , 4 black balls are taking the next four Urns (urn 7-10), out of 2 same black balls remaining each has 10 urns Options to go in .
One eg : w w w w w w b b b b (Ten urns filled ) Now two white balls from urn 5 and urn 6 if they go to urn 7 & 8 (4c2 ) , then there is another new arrangement can be w w w w b b w w b b .....SO
4c1 + 4c2 + 4c3 + 4c4 - > these are all counting when white balls are shifting from their original position to some other Urns .
No of ways : (4c1 + 4c2 + 4c3 + 4c4) * (10 * 10 ) /2 = .. ...
Case 2 : (4w , 6 bl) --> No of ways : (4c1 + 4c2 + 4c3 + 4c4) * (10 * 10 ) /2 = .. ...
Case 3 : 5w 5bl --> (5c1 + 5c2 + 5c3 + 5c4) * (10 * 10 ) = ...
Answer is not coming right.