The Numbers $1,2,\dots,12$ are arranged in a circle in that order. In how many ways can five numbers be chosen from this arrangement so that no two adjacent numbers are selected?
I can't think of a good way to think about this problem. I tried picking a random number and looking at the numbers that are possible, but there are cases where the $2$ numbers are just seperated by $1$ number.
On
Start with the most basic configuration: $$1,3,5,7,9$$ If we rotate this configuration by $30^0$, we get:
$$2,4,6,8,10$$ then $$3,5,7,9,11$$ and so forth until we reach $$12,2,4,6,8$$ totalling $12$ seperate layouts with this particular configuration from this.
In short, the total is the number of different configurations, including the one I've demonstrated on, multiplied by $12$ for the $12$ different numbers it can start on. See if you can go from there.
In each configuration 5/12 of the numbers are chosen, so 5/12 of the possibilities will include the number $1$. Let's concentrate on these.
There are five gaps between numbers, adding to a total of $7$, and none of the gaps can be zero - so $1+1+1+1+3$ or $1+1+1+2+2$. The first gives $5$ ways of choosing the remaining four numbers (the gap of $3$ can be in one of five places). The second gives [?] ways.
There are therefore [?] possibilities which include the number $1$ and hence [?] altogether.