There are $10$ halting stations on a circular road in a city. On the route, a bus will stop at any three stations so that no two stations are adjacent. The number of such possible bus routes are
I made $7$ places left to be filled in a circle which were not adjacent that gave $35$ ways of filling those up. But, the answer is $50$. I can't get myself to the correct answer.
On
To each stop ($S$), attach a non-stop ($N$) clockwise, e.g.$\quad\boxed{SN}$
So we have $3$ boxes and $4$ others, total $7$ entities.
The boxes can now be placed in $\binom73$ ways,
but since you are allowing each entity only $7$ positions instead of $10$,
multiply by $\frac{10}{7}$ to get ans $= \frac{10}{7}\times\binom{7}{3} = 50$
Allow a 1 to represent a stop and a 0 to represent a pass. If the ten are on a line you get a code like 1001000100. We can get all codes that don't end in 1 by taking the strings 10 10 10 0 0 0 0 and putting them together. This is like taking seven spaces and choosing 3 to be 10. Thus there are $C(7,3)=35$ such strings. However, we left out the ones that end in 1. Then we have to rearrange 10 10 0 0 0 0 0 and end in an automatic 1. There are $C(7,2)=21$ such strings. So there are a total of 35+21=56 strings.
Now put this on a circle. The only ones which would have two 1's in a row are the ones that both begin and end in a 1. These as strings start with 10, end with 01 and have a 1 in any of the 6 remaining spots. Thus there are 6. We get 56-6=50 for our answer.