Selection with condition The problem statement is :
16 people are occupying seats round a table. If ever person refuses to work with any of his neighbours, in how many ways can a committee of 6 can be made with those persons? 
How do I approach this type of problem? 
 A: Mainly to facilitate the explaning in the answer let's give the persons numbers $1,2,\dots,16$ clockwise.
Pick one person out - let's say number $1$ - to be the somehow virtual chairman of the committee.
Now we start looking for tuples $\left(n_{1},\dots,n_{6}\right)$
where the $n_{i}$ are positive integers and $n_{1}+\cdots+n_{6}=10$.
These $n_{i}$ stand for the cardinality of the gaps, i.e. consecutive
persons that are not in the comittee.
If e.g. $n_{1}=2$ and $n_{2}=1$ then persons with number $4$ and
number $6$ are the next in the committee.
Applying stars and bars we find $\binom{4+5}{5}=126$ possibilities and each tuple $\left(n_{1},\dots,n_{6}\right)$
represents a committee having person $1$ as chairman.
There are $16$ choices for the chairman so we come to $16\times126=2016$
possibilities.
However, every possible committee has been counted $6$ times because there are
$6$ choices for the chairman, so we end up with: $$\frac{2016}{6}=336$$
possibilities.
A: Since you asked for another way
Glue an "unchosen" to the right of every "chosen", $\fbox {cu}$
thus number of entities becomes $6+4=10$
Place the glued entities in $\binom{10}{6} = 210$ ways,
but since you are allowing entities only $10$ starting points instead of 16, multiply by $\frac{16}{10}$,
thus number of committees = $\frac{16}{10}\cdot 210 = 336$ 
