# Choosing a committee from two people who are not sitting beside each other.

Assume that $10$ people are sitting around a table. Determine the number of ways to choose a committee, where the committee is made up of two people who are NOT sitting next to each other.

Take ${10 \choose 2}$ and take away the pairs with partners, which is still the $10$ which gives me the same result. ${10 \choose 2}-10=35$.

Assume that $10$ people are sitting around a table. Determine the number of ways to choose a committee, where the committee is made up of three people of which NONE are sitting next to each other.

${10\choose 3}=130$ take away the pairs that are sitting beside each other: (still 10??)

$120$

Your answer to the first question is fine. Your answer to the second question has problems. First, $\binom{10}3=120$. Second, subtracting $10$ pairs from $\binom{10}3$ triples makes no sense. What you want to subtract is the number of triples, at least two of whom are sitting beside each other. There are two different kinds of bad triples, which you have to count separately.
I. Three people in consecutive seats. There are $10$ of those.
II. Two people sitting beside each other, the third not sitting beside either of then. There are $10\times6=60$ triples like that: $10$ ways to seat two beside each other, then $6$ places to seat the third.
So the answer is $120-10-60=50$.