How many ways can five different people M,N,O,Q,P sit together such that NQ are not together?
You can do 5!-unfavorable cases.
Pretend [NQ] is one person [NQ]_ _ _ and there are four positions. Three people have the choices $3!$
N and Q may re-arrange positions which is a double.
Is this multiplication principle or binomial coefficient $4$ choose $1$ for [NQ] either goes in the first, between the others, or last?
Is the answer $5!-2\cdot 3! \cdot {4 \choose 1}$ or $5!-2\cdot 3! \cdot 4$?