station game

One just needs to use the $2^{nd}$ Kaplansky's Lemma:

The number of ways of selecting $p$ objects with no two adjacent, from $n$ objects arranged in a cycle, is: $g(n,p)=\frac{n}{n-p}{{n-p}\choose{p}}.$ Therefore: $g(10,3)=\frac{10}{7}{{7}\choose{3}}=\boxed{50}.$

Label the stations 1 through 10 going clockwise, and assume that the starting point of the train is between stations 10 and 1. So going clockwise, the train will pass stations before making its first stop, stations before making its second, stations before making its third stop and stations bef huore crossing its starting point, where

(i) are positive integers,

(ii) are non-negative integers, at least one of which is a positive integer, and

(iii) , as stations are passed over in total.

Condition (ii) is required since stations 1 and 10 are adjacent, (since the path is circular), and thus we cannot have .

This problem can be solved using the stars and bars method. We then need to look at the following cases:

: equation (iii) then becomes with each of being positive integers, which according to Theorem 1 of the link has solutions.

: by symmetry, this will yield the same result of solutions.

: equation (iii) then becomes with each of being positive integers, which according to Theorem 1 has

solutions.

So in total the number of suitable station selections is 70-20=50

Consider a circular table, where the 10 stations are the 10 seats.

One seat can be chosen by 10C1 ways=10.

Now the adjacent two seats cannot be occupied so, we select the remaining two in (10-3)+2-1C2 ways= 6C2 = 15.

And to get rid of repetition we divide the result by 3. Thus, the answer is (10x15)/3=50