Counting possible outcomes in k trials with n possible outcomes in each trial

Question

Suppose we need to carry out k trials with n possible outcomes in each trail. The order of the trials matters. It is easy to calculate the quantity of possible outcomes as N=n^k (^ is exponent sign). Now, we review this process in a differet method. Suppose we need to put k balls which are numbered as "1,2,...,k" into n bins. Each bin could accomodate as many as all the balls together. The experiment is essentially same as the above k trials with n possible outcomes because when you place a ball into a particular bin, it is like choosing an outcome for that trial. The n different bins could be viewed as n-1 walls lining up together (we disregard the end walls). The process of placing the balls into these bins can be viewed as arranging the orders of the k balls and n-1 walls (k+n-1 objects altogether). The order of the balls matter while the order of the walls doesn't matter. Thus, the possible outcomes could be calculated as (k+n-1)!/(n-1)!=(k+n-1)(k+n-2)...(n)>n^k. The second thread is wrong. Where is wrong with my thread? Could anyone please help? Thank you very much.

Answer 1

Consider the case of $2$ trials and $2$ outcomes. $n^k$ gives $4$ but your expression gives $6$. This is because you are over-counting. If $|$ represents a wall, and $1,2$ represent the trials, then you are counting $1,2,|$ and $2,1,|$ as different but they are actually the same. In both cases, trials $1$ and $2$ give the first outcome. Similar for $|,1,2$ and $|,2,1$.