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) .


Which method is correct and why the other one is wrong


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:

$$\frac{8!}{8} - 6!\times2 = 5040 - 1440 = 3600$$


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 $= 5\cdot6!$


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