# Probability that $j$ persons to get off on the same floor, and $k-j$ persons to get off on separate floors

There are $k$ persons and $n$ floors. Assuming that the probability of any person to get off on any floor is $\frac{1}{n}$, and the decisions taken by the persons are independent, what is the probability that $j$ persons get off on the same floor, and the rest $k-j$ persons to get off on different floors?

I know that the answer is $\dfrac{C_k^j \cdot A_n^{k-j+1}}{n^k}$ but I don't know how to get there.

Thanks!

• What have you tried? Are there any ideas you think might come into play in a solution? The more of your thoughts you can share with us, the more helpful we are able to make our hints/solutions. – Brett Frankel Nov 5 '13 at 16:27

The thing to keep in mind is that it doesn't matter which $j$ people get off on a floor, nor does it matter which floor this happens to be. So think about the number of ways a `good' outcome can happen.
First, by definition of the choose functions, there are $C_k^j$ ways to choose $j$ people which get off together, and $n$ different floors for them to get off on. Then, once these people have gotten off, the next person has $(n-1)$ floors to choose from, the next has $(n-2)$, and so on.
So there are $n C_k^j A_{n-1}^{k-j+1}$ good outcomes, in which $j$ people get off at the same floor, and everyone else gets off on a separate one. There are $n^k$ total ways of assigning floors, so you get $$\frac{n C_k^j A_{n-1}^{k-j+1}}{n^k} = \frac{C_k^j A_n^{k-j+1}}{n^k}$$