Find the number of ways of selecting three people out of $10$ people sitting at a round table, such that no two people selected are consecutive.
I recently learned about the stars and bars technique and thought I'd do a slightly changed version of it. I went to convert the question into its binary equivalent (using $0$s and $1$s to describe the same situation to make it mathematically more tangible) and thought I'd give stickers with $0$ and $1$ printed on them to those people sitting at the table assuming those who get $0$ are selected such that no two consecutive people get a $0$. One of the combinations could be this
$0, 1, 1, 0, 1, 0, 1, 1, 1, 1$
Note that they are sitting in circle, so first and last terms together cannot be $0$ because this would make them consecutive.
Another combination could be
$1, 1, 1, 0, 1, 0, 1, 0, 1, 1$
Since the order of selection obviously doesn't matter, I can choose to fix the $1$s and then count the number of ways I can fill the spots in-between them which would make sure a distance between two selections is taken care of.
Hence, let $_$ represent the possible positions $0$ can take:
$_ 1 _ 1 _ 1 _ 1 _ 1 _ 1 _ 1$
Note that first and last spots are actually the same if you join the string of numbers and dashes to form a loop.
I have $7$ spots to place $3$ $0$s, order irrelevant, which makes it $\binom73$ choices, $35$ ways
However, I saw the solution being done with Inclusion-Exclusion Principle, to which my answer does not match. What combinations did I skip in this, because I'm getting a smaller answer.
Best Answer
To see what goes wrong, consider the version of the problem where you are choosing three of six people seated around a round table so that no two chosen people are consecutive. Evidently there are two way to do this, $010101$ and $101010$. But applying your method, we have to choose three of the blanks in the configuration
which can only be done in one way (the way that yields $010101$).
For a table of $n$ people the problem with your method is that it omits all configurations in which the $n^\text{th}$ person is among the chosen. In the six-person case you can add these in by also considering the configurations
where two blanks must be chosen. This gives the missing one of the two choices. In the ten-person case, add in configurations like
from which two blanks again must be chosen. This yields the missing $15$ choices.