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by GMATGuruNY » Thu Jun 28, 2012 7:52 pm
abcd1111 wrote:How many Ways 8 people can sit together around a round circular table If A and B cannot seat together .

Pls explain how to solve this !!
Good arrangements = total arrangements - bad arrangements.

Total arrangements:
The number of ways to arrange n elements in a circle = (n-1)!.
Thus, the total number of ways to arrange the 8 people = (8-1)! = 7! = 5040.

Bad arrangements:
In a bad arrangement, AB sit next to each other.
Seat AB at the table first.
Now count the number of ways to arrange the 6 other people around AB:
6! = 720.
Since AB can be reversed to BA, multiply by 2:
2*720 = 1440.

Good arrangements = 5040-1440 = 3600.
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by tutorphd » Thu Jun 28, 2012 8:05 pm
Circular permutations are the same as linear permutation but fix the position of one of the people and arrange the other people after that person. The actual physical position of the fixed first person doesn't matter because the table is round and every circular shift of the linear permutaion is considered equivalent.

Let's take the A as the fixed-position person so we have the 8 slots:
A _ _ _ _ _ _ _

B cannot be in the second slot or the last slot because he will be next to A so we have 6 possibilities for the second slot (8 people - A - B) and 5 possibilities for the last slot (8 people - A - B - person in second slot). For the middle 5 slots you have 5 people remaining (one of them is B) in 5 places so the possibilities for rearranging the middle 5 slots are 5!. Multiplying all the possiblities together:

1 (for A in first slot) * 6 (for second slot) * 5 (for last slot) * 5! (for 5 middle slots) = 30*5! = 30* 120 = 3600
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by eagleeye » Thu Jun 28, 2012 8:12 pm
Mitch wrote a good way to do those.

Here's another using fundamental principal of counting.

8 people, A and B can't sit together.

1. Fix A. Then we have something like this. (I can't draw a circle here so here goes.)
_ _ _ _ A _ _ _

2. Then the position immediately to the left and right of A can't be B. So let's start filling up using this condition.
From the seven people remaining, we have 6 options for the left of A. Once that is filled, from the remaining six, we have 5 options for the right of A (Because we can't have B in those spots).

Once we have filled these 2 spots, the remaining 5 people can be arranged in 5! ways.

So the total number of ways of sitting 8 people on a circular table such that A and B don't sit together = 6*5*5! = 30*120 = 3600.

Let me know if this helps :)
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