In a train of n wagons, at random m passengers enter choosing a wagon.. Find the probability that: $(a)$ In every wagon there is at most one passenger, if $m\leq n;$ $(b)$ In the $n$th wagon, there is exactly one passenger, if $m\leq n;$ $(c)$ In every wagon there is at least one passenger, if $m\geq n;$ My thoughts: $(a)$ ${\binom{n}{m}m!}\over n^m$ (choose m wagons of the n, permute the m wagons..) $(b)$ ${m (n-1)^{m-1}}\over n^m$ $(c)$ $\binom{n}{1}\binom{n-1}{1}..\binom{n-m}{1}?$ thats to arrange one person in ever wagon, not sure how to make it arrange the remaining passengers... Need help with this last one.

For the last one $$A={(x_1,x_2,...x_n); x_1+x_2+..+x_n=m; x_1,..x_n > 1 } $$ Therefore: $$A={(x_1,x_2,...x_n); x_1+x_2+..+x_n=m-n; x_1,..x_n > 0 } $$ $$|A|= \binom {m-1}{m-n}=\binom {m-1}{n-1}$$