You are given 20 identical balls and 5 bins that are coloured differently ( so that any two of the bins can be distinguished from one another). In how many ways can the balls be distributed into the bins in such a way that each bin has at least two balls?
My attempt: First of all , 2 balls are distributed in each bin. . Then I think that the remaining 10 balls can be distributed into either 1 bin or 2 bins or 3 bins and so on. Now if all 10 balls are distributed into 1 bins then there are 5 distint ways of doing so . If two bins are selected (10 ways) , then for each of this selection , the 10 balls can be distributed in the following way (9+1) , (8+2), (7+3) upto (5+5) and then permuting those two bins. Overall , my strategy is to decompose 10 as the sum of 1 , 2 , 3,.. 5 natural numbers in unique ways . Obviously the process is tedious , but doing this way my answer is 981 (the correct ans is 1001) . Is that calculation mistake ? or my method is wrong ? Please help