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The total number of ways in which 5 balls of different colour can be distributed among three persons so that each person gets at least one ball?
A book on combinatorics. Was asked in an entrance exam.
To simplify the problem, we can distribute $3$ of the $5$ balls beforehand so we can get rid of the constraint. Let us select 3 balls and permute them within these three persons, so the number of ways of doing so $$ N_1 = C(5,3) + P(3,3) = 60$$
Let both the remaining balls be with one person, so we have $$(B_1 B_2, 0, 0)$$ and there are $3$ ways of doing so.
Let the remaining balls be distributed one by one, so we have $$(B_1, B_2, 0)$$ and there are $3! = 6$ ways of doing so.
We can take the total number ways ($N$) by: $$N = N_1 \times 3 + N_1 \times 6$$ $$N = 60 \times 3 + 60 \times 6$$ $$N = 540$$
This approach seems correct to me but still it gets the wrong answer...
The right answer is $150$ ways
Please guide me and correct me if I'm wrong. All help appreciated!