This question was asked a couple weeks ago but closed
Suppose 6 passengers board a train Train in Kitchener. On its way to Montreal the train will make stops at 7 different stations (numbered Toronto = 1, 2, .. .,7 = Montreal) where passengers may get off. Assuming passengers are equally likely to get off at any station, how many different ways can the passengers disembark?
Answer
I know the total possibilites, but I was trying to enumerate the different outcomes, but my numbers do not quite add up. Here is what I tried.
There should be $7^6$ total possibilities. Since the stations and the people are distinct, and if we have n stations and m people, we have m decisions at each station for how the people will be placed. We have n choices(stations) for where each person goes, and then there are m people who each then decide, and their decision is linked to the ones before, so we multiply $\underbrace{n*n*\ldots*n}_{\text{m-times}}=\boxed{n^m}$.
This is might be a complicated way of doing this but here goes with every possible combination. Note 6,0,0,0,0,0,0 means that 6 people hop off at one station. Since there are 7 stations, there is $7*{6\choose 6}$ ways of that happening. (${6\choose 6}$ because we have to choose 6 people for the one stop out of our total, which happens to be 6). Next, we can have 5 at one stop, and 1 at one of the others. There are 7 stops which could have the 5 people, we choose the 5 people from our 6 total. That leaves us to split up our remaining person over 6 possible stops. The next complicated scenario occurs when we have three people at two stops. This means from the 7 stations, we choose 2 that will house the 3 people, and then we choose the people from our people list as before.
The red indicates how many different stations can hold the configuration, and the black indicates we are choosing which combination of people to put in each station from the people list.
$$ \begin{align} 6,0,0,0,0,0,0&\longrightarrow \color{red}{7\choose 1}*{6\choose 6}=7\\ 5,1,0,0,0,0,0 &\longrightarrow \color{red}{7\choose 1}*{6\choose 5}*\color{red}{6\choose 1}*{1\choose 1}=252\\ 4,1,1,0,0,0,0&\longrightarrow \color{red}{7\choose 1}*{6\choose 4}*\color{red}{6\choose 2}*{2\choose 1}*{1\choose 1}=3150\\ 4,2,0,0,0,0,0&\longrightarrow \color{red}{7\choose 1}*{6\choose 4}*\color{red}{6\choose 1}*{2\choose 2}=630\\ 3,3,0,0,0,0,0&\longrightarrow \color{red}{7\choose 2}*{6\choose 3}*{3\choose 3}=420\\ 3,1,1,1,0,0,0&\longrightarrow \color{red}{7\choose 1}*{6\choose 3}*\color{red}{6\choose 3}*{3\choose 1}*{2\choose 1}*{1\choose 1} = 16800\\ 3,2,1,0,0,0,0&\longrightarrow \color{red}{7\choose 1}*{6\choose 3}*\color{red}{6\choose 1}*{3\choose 2}*\color{red}{5\choose 1}*{1\choose 1}=12600\\ 2,2,2,0,0,0,0&\longrightarrow \color{red}{7\choose 3}*{6\choose 2}*{4\choose 2}*{2\choose 2}=3150\\ 2,2,1,1,0,0,0&\longrightarrow \color{red}{7 \choose 2}*{6\choose 2}*{4\choose 2}*\color{red}{5\choose 2}*{2\choose 1}*{1\choose 1}=37800\\ 2,1,1,1,1,0,0 &\longrightarrow \color{red}{7\choose 1}*{6\choose 2}*\color{red}{6\choose 4}*{4\choose 1}*{3\choose 1}*{2\choose 1}*{1\choose 1}=37800\\ 1,1,1,1,1,1,0&\longrightarrow \color{red}{7\choose 6}* {6\choose 1}*{5\choose 1}*{4\choose 1}*{3\choose 1}*{2\choose 1}*{1\choose 1}=5040\\ &\boxed{\sum=117649=7^6} \end{align} $$
Thank you to the comments, the answer is correct now.
Your logic is correct. The only error was a computational mistake in the case $3, 2, 1, 0, 0, 0, 0$. You should have obtained $$\binom{7}{1}\binom{6}{3}\binom{6}{1}\binom{3}{2}\binom{5}{1}\binom{1}{1} = 7 \cdot 20 \cdot 6 \cdot 3 \cdot 5 \cdot 1 = 12600$$ You lost the factor of $3$.