How many ways are there to distribute 6 passengers into three different hotels? 6 individuals want to go to 3 different hotels such that each hotel can select zero through 6 people all states are possible. From the passenger's angle, we know that there are $3^6=729$ different ways to do this task. But from the perspective of hotels, how can solve be this problem?
We know that hotel 1 can get 6's and hotel 2,3 zero's i.e. $(6,0,0)$ OR hotel 1 5's and hotel 2 1's and hotel 3 nothing $(5,1,0)$ OR hotel 1 3's hotel 2 2's and hotel 3 1's $(3,2,1)$ OR so on. But this counting method does not yield the correct answer.
Thanks in advance for your help.
 A: Let the three hotels be A, B, C. Suppose hotel A gets $m$ passengers with $0 \le m \le 6$. There are $\binom{6}{m}$ ways for this to happen. Then hotel B has to get $n$ of the remaining $6 - m$ passengers. There are $\binom{6 - m}{n}$ ways for them to do this. By default, hotel C gets the remaining $6 - m - n$ passengers.
Thus, the total number of ways for the hotels to do this is given by
$$ \begin{align*} \sum_{m=0}^6\sum_{n=0}^{6-m}\binom{6}{m}\binom{6 - m}{n} &= \sum_{m=0}^6\binom{6}{m}\sum_{n=0}^{6-m}\binom{6 - m}{n} \\
&= \sum_{m=0}^6 \binom{6}{m}2^{6 - m} = (1 + 2)^6 = 729 \end{align*} $$
as earlier given.
A: The number of ways to choose $a$ people for the first hotel, $b$ for the second hotel, and $c$ for the third hotel, with $a+b+c=6$, is the multinomial coefficient
$$\binom{6}{a,b,c}= \frac{6!}{a! b! c!}$$
so the total number of possible arrangements is
$$\sum_{a+b+c = 6} \binom{6}{a,b,c}$$
where the summation is over all integer triples $(a,b,c)$ with $a+b+c = 6$ and $a,b,c \ge 0$.  We could work this out, but there is a shortcut.
By the multinomial theorem,
$$(x+y+z)^6 = \sum_{a+b+c = 6} \binom{6}{a,b,c} x^a y^b z^c$$
where, as before, the summation is over all integer triples $(a,b,c)$ with $a+b+c = 6$ and $a,b,c \ge 0$. Now let $x=y=z=1$, and we have
$$3^6 = \sum_{a+b+c = 6} \binom{6}{a,b,c}$$
which reproduces the previous answer of $3^6 = 729$.
