I am currently looking at the following question:
Find the number of possible distributions of 6 distinguishable balls in 3 distinct boxes, in such a way that each box contains at least one ball?
The following method was my attempt to solve this problem:
- First, choose 1 ball to go in each box so that no box is empty. This can be done in $P(6,3) = 120$ ways.
- With a ball in every box, we know need to allocate the 3 remaining balls between the 3 boxes. By the multiplication principle, there are $3 \times 3 \times 3 = 27$ ways to do this.
- Thus, by the multiplication principle, the number of ways of allocating the balls is $120 \times 27 = 3240$
However, I have the solutions to this problem and know that the correct answer is 540.
Can anyone tell me why my method gave the wrong answer?