Ways to arrange $8$ people in $8$ chairs in a row if person $A$ can't sit next to $B$ or $C$. 
In how many ways can I arrange $8$ persons in a row if person $A$
  can't sit next to person $B$ and person $C$.

I have made the case that $5$  persons ($A,B,C$ are not included) are already in place and I can have such a configuration in $5!$ ways. Now, $A$ can take place between each person and also before and after the first and last person, thus $A$ can be placed in $\dbinom {6}{1}$ and $B, C$ can take place in $5\cdot4$ ways. 
So I have $5!\cdot 6 \cdot 5 \cdot 4 =1440$ ways to do that.
However I don't know if this result is correct since the solution is not available online (this problem comes from Mu Alpha Theta 1991).
Also what are other ways to look at the problem?
 A: Your idea is good, but there is a slip  of execution.
Once "the five" and A have been placed, there are $5$ allowed places for B. Now we have  $7$ objects in place, so $8$ "gaps," of which $6$ are allowed for C, for a total of $(5!)(6)(5)(6)$.  Now the answer should agree with the Inclusion/Exclusion answer.
A: Use the inclusion/exclusion principle:


*

*Include the total number of permutations, which is $8!=40320$

*Exclude the number of permutations containing $AB$ or $BA$, which is $2\cdot7!=10080$

*Exclude the number of permutations containing $AC$ or $CA$, which is $2\cdot7!=10080$

*Include the number of permutations containing $BAC$ or $CAB$, which is $2\cdot6!=1440$



Hence the number of legal permutations is:
$$40320-10080-10080+1440=21600$$
A: Total number of ways in which $8$ people can sit without any restriction = $8!$
Consider the cases where $A$ sits next to $B$ and $C$.
We consider $AB$ as a single unit.
Total number of ways in which $A$ sits next to $B$ = $7! \times 2!$
Next we consider $AC$ as a single unit.
Total number of ways in which $A$ sits next to $C$ = $7! \times 2!$
Now in the above computations we have considered the seating arrangements $BAC$ and $CAB$ twice. So we have to subtract that once.
We consider $BAC$ as a single unit.
Total number of ways in which $A$ sits next to both $B$ and $C$ = $6! \times 2!$
Total number of ways in which $A$ sits next to either $B$ or $C$ or both $= 7! \times 2!+7! \times 2!-6! \times 2!$
Required no. of ways $= 8!- (7! \times 2!+7! \times 2!-6! \times 2!)$
A: Count the ways when they are together  answer is$8!-1680$. Hope it helps you. But 'and' is confusing so i give another answer total ways are $8!$  now consider A,B,C to be 1 so total people =1+5=6 so total arrangements are $2!.8C6$ so total ways are $8!-(2!.8C6)$=$8!-56$ hope it helps you.
