In how many ways can $5$ Indians, $4$ Chinese, and $3$ Americans be assigned to $12$ stations so that no two Americans serve at consecutive stations? On a railway route from Delhi to Jaipur there are $12$ stations . A booking clerk is to be deputed for each of these stations out of $12$ candidates of whom $5$ are Indians , $4$ are Chinese and the rest are Americans . The number of ways of deputing the persons on these stations so that no two Americans serve on two  consecutive stations is 
My Approach : Since there are $12$ stations ...and the number of Americans are $3$  , so simply we can select three stations which should not be consecutive .. and the ways to select things which are not consecutive is C($n$$+r$$-$$1$, $r$).. 
After substituting value of $n$ as 12 and $r$ as 3 answer is not coming correct ...
Please tell the correct approach...

I will assume that people of each nationality are distinguishable.
There are $9!$ ways of arranging the four Chinese and five Indian candidates in a row.  To ensure no two Americans are consecutive, we must place the American candidates in the eight spaces between the other candidates or the two spaces at the ends of the row.  We choose three of these ten spaces in which to place the American candidates.  The Americans can be arranged in the three chosen spaces in $3!$ ways.  Hence, the number of ways the candidates can be assigned so that no two Americans serve at consecutive stations is 
$$9!\binom{10}{3}3!$$ 
Note: The number 
$$\binom{n + r - 1}{n - 1} = \binom{n + r - 1}{r}$$
is the number of ways $r$ objects can be selected from $n$ types of objects, which is equal to the number of solutions of the equation
$$x_1 + x_2 + \cdots + x_n = r$$
in the non-negative integers.
