In how many way 8 can be seated in a round table if two of them (say A and B) must not sit together.
1) total minus when AB sit together .
7! - 6! * 2
2) selecting one from A and B and arranging it with other 6 person . Then arranging the one person left in 4 seats available (not adjacent) .
2*6!$^4$$C_1$
Which method is correct and why the other one is wrong
On
I take it (as is usual) that seats are unnumbered. To answer your specific questions:
Method 1 is correct.
Method 2 needs some amendment to get the correct answer, as explained below.
Only one of the two specified persons needs to be fixed to take care of unnumbered seats.
Fix $A$, $B$ has choice of $5$ positions non-adjacent to $A$ and the others can be permuted in $6!$ ways
Thus # of ways $= 5\cdot6!$
It'll be much easier to count the "bad" combinations and subtract them from the total number of combinations. We could think of them two as one person, so there will be $7$ people to sit down.
This means that there are $7!$ combination but note there are 7 people and we can start counting from anyone so every combination will be included seven times, so we divide by $7$. But note that when we "split" the two women the can sit in two different ways. So we'll have to multiply the number of combinations by $2$. So the number of "bad" combinations is:
$$\frac{7!}{7} \cdot 2 = 6!\times2$$
And the total number of combinations that satisfy the condition is:
$$\frac{8!}{8} - 6!\times2 = 5040 - 1440 = 3600$$