How many ways 3 persons A, B , C can be put in a row of six empty chair such that A nd B don't sit together

How many ways 3 persons A,B and C can seat in a row of six empty chairs such that A and B don't sit together?
My Attempt
Let us introduce three more persons D, E and F. Then these six persons can be placed in six chair in $$6!$$ ways. Now in these placements if we tie together A and B then number of permutation where A and B seat together is $$2 \cdot 5!$$ ways. Hence there are $$(6! -2\cdot 5!)$$ permutations with six persons which do not have A nd B together. Now we remove D , E , F to create six chairs and three persons A , B and C seating with condition A and B are not seating together. But empty chairs are indistinguishable and D , E , F can be permuted in $$3!$$ ways. Hence final answer is $$\frac{(6! -2\cdot 5!)}{3!}=80$$.

• Does $A$ and $B$ seating together mean they are adjacent? For example, does $- A -BC-$ count as $A$ and $B$ seating together? – Thomas Bladt Nov 6 '18 at 5:10
• $-A-BC--$ is acceptable but $ABC---$ or $C-BA--$ are not acceptable. – rugi Nov 6 '18 at 5:12

• When you bind $$A,B$$ together you can seat them in $$5$$ ways beside each other on $$6$$ chairs and permute the two: $$\color{blue}{2\cdot 5}$$
• From the remaining $$4$$ chairs you choose one for $$C$$: $$\color{blue}{4}$$
• All possible ways of placing $$3$$ onto the $$6$$ chairs (choose $$3$$ seats from $$6$$ and arrange $$A,B,C$$ in them): $$\color{blue}{\binom 63 \cdot 3!}$$
$$\color{blue}{\binom 63 \cdot 3! - 2\cdot 5 \cdot 4 = 80}$$